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Ha, tells you if the test is left, right, or two-tailed. It is the key to conducting the appropriate test. • Ha never has a symbol that contains an equal sign. • Thinking about the meaning of the p-value: A data analyst should have more confidence that he made the correct decision to reject the null hypothesis with a smaller p-value (for example, 0.001 as opposed to 0.04) even if using the 0.05 level for alpha. Similarly, for a large p-value such as 0.4, as opposed to a p-value of 0.056 (alpha = 0.05 is less than either number), a data analyst should have more confidence that she made the correct decision in not rejecting the null hypothesis. This makes the data analyst use judgment rather than mindlessly applying rules. The following examples illustrate a left-, right-, and two-tailed test. Example 9.11 H0: μ = 5 Test of a single population mean. Ha tells you the test is left-tailed. The picture of the p-value is as follows: Ha: μ < 5 Figure 9.3 9.11 H0: μ = 10 Assume the p-value is 0.0935. What type of test is this? Draw the picture of the p-value. Ha: μ < 10 Example 9.12 H0: p ≤ 0.2 Ha: p > 0.2 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 533 This is a test of a single population proportion. Ha tells you the test is right-tailed. The picture of the p-value is as follows: Figure 9.4 9.12 H0: μ ≤ 1 Assume the p-value is 0.1243. What type of test is this? Draw the picture of the p-value. Ha: μ > 1 Example 9.13 Ha: p ≠ 50 H0: p = 50 This is a test of a single population mean. Ha tells you the test is two-tailed. The picture of the p-value is as follows. Figure 9.5 9.13 H0: p = 0.5 Assume the p-value is 0.2564. What type of test is this? Draw the picture of the p-value. Ha: p ≠ 0.5 534 Chapter 9 |
| Hypothesis Testing with One Sample Full Hypothesis Test Examples Example 9.14 Jeffrey, as an eight-year-old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. His dad, Frank, thought that Jeffrey could swim the 25-yard freestyle faster using goggles. Frank bought Jeffrey a new pair of expensive goggles and timed Jeffrey for 15 25-yard freestyle swims. For the 15 swims, Jeffrey's mean time was 16 seconds. Frank thought that the goggles helped Jeffrey to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset α = 0.05. Assume that the swim times for the 25-yard freestyle are normal. Solution 9.14 Set up the hypothesis test: Since the problem is about a mean, this is a test of a single population mean. H0: μ = 16.43 For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed. Ha: μ < 16.43 Determine the distribution needed: ¯ Random variable: X = the mean time to swim the 25-yard freestyle. ¯ Distribution for the test: X is normal (population standard deviation is known: σ = 0.8) with mean μ = 16.43 and standard error of 0.8 15 ; μ = 16.43 comes from H0 and not the data. σ = 0.8, and n = 15. Using a table or a calculator, we can calculate the p-value as the area to the left of 16 under the normal curve: p-value = P( x¯ < 16) = 0.0187 where the sample mean in the problem is given as 16. p-value = 0.0187. The p-value is the area to the left of the sample mean given as 16. Graph: Figure 9.6 μ = 16.43 comes from H0. Our assumption is μ = 16.43. Interpretation of the p-value: If H0 is true, there is a 0.0187 probability (1.87 percent), that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87 percent chance is small, the mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event. Compare α and the p- |
value: α = 0.05 p-value = 0.0187 α > p-value This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 535 Make a decision: Since α > p-value, reject H0. An alternative approach is to find the z-test corresponding to the sample mean x¯ = 16. This is z-test = x¯ − μ X σ X n = 16 − 16.43 0.8 15 = - 2.081729. The critical z-value = –1.645 for this test has probability 0.05 to its left tail, according to the Normal Table (see Appendices). Because the z-test is to the left of the critical z-value, we reject the null hypothesis. This means that you reject μ = 16.43. In other words, you do not think Jeffrey swims the 25-yard freestyle in 16.43 seconds but instead that he swims faster with the new goggles. Conclusion: At the 5 percent significance level, we conclude that Jeffrey swims faster using the new goggles. The sample data show there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. The p-value can easily be calculated. Press STAT and arrow over to TESTS. Press 1:z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 16.43 for μ0 (null hypothesis),.8 for σ, 16 for the sample mean, and 15 for n. Arrow down to μ : (alternate hypothesis) and arrow over to < μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0187) but it also calculates the test statistic (z-score) for the sample mean. μ < 16.43 is the alternative hypothesis. Do this set of instructions again except arrow to Draw(instead of Calculate). Press ENTER. A shaded graph appears with z = -2.08 (test statistic) and p = 0.0187 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. When the calculator does a z-Test, the z-Test function finds the p-value by doing a normal probability calculation: |
P( x¯ < 16) = 2nd DISTR normcdf ⎛ ⎝ − 10 ^ 99, 16, 16.43, 0.8 / 15⎞ ⎠. The Type I and Type II errors for this problem are as follows: The Type I error is to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually swims the 25-yard freestyle, on average, in 16.43 seconds. (Reject the null hypothesis when the null hypothesis is true.) The Type II error is that there is not evidence to conclude that Jeffrey swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, he actually does swim the 25-yard freestyle, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.) HISTORICAL NOTE (EXAMPLE 9.11) The traditional way to compare the two probabilities, α and the p-value, is to compare the critical value (z-score from α) to the test statistic (z-score from data). The calculated test statistic for the p-value is –2.08. (From the central limit theorem, the test statistic formula is z = x¯ − μ X σ X n ) (. For this problem, x¯ = 16, μX = 16.43 from the null hypothesis, σX = 0.8, and n = 15.) You can find the critical value for α = 0.05 in the normal table (see Appendix H: Tables). The z-score for an area to the left equal to 0.05 is midway between –1.65 and –1.64 (0.05 is midway between 0.0505 and 0.0495). The z-score is –1.645. Since –1.645 > –2.08 (which demonstrates that α > p-value), reject H0. Traditionally, the decision to reject or not reject was done in this way. Today, comparing the two probabilities α and the p-value is 536 Chapter 9 | Hypothesis Testing with One Sample very common. For this problem, the p-value, 0.0187, is considerably smaller than α, 0.05. You can be confident about your decision to reject. The graph shows α, the p- |
value, and the test statistic and the critical value. Figure 9.7 9.14 The mean throwing distance of a football by Marco, a high school freshman quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marco to adjust his grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, Marco’s mean distance was 45 yards. The coach thought the different grip helped Marco throw farther than 40 yards. Conduct a hypothesis test using a preset α = 0.05. Assume the throw distances for footballs are normal. First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion. Press STAT and arrow over to TESTS. Press 1: z-Test. Arrow over to Stats and press ENTER. Arrow down and enter 40 for μ0 (null hypothesis), 2 for σ, 45 for the sample mean, and 20 for n. Arrow down to μ: (alternative hypothesis) and set it either as <, ≠, or >. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value but it also calculates the test statistic (z-score) for the sample mean. Select <, ≠, or > for the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with test statistic and p-value. Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. Example 9.15 A college football coach records the mean weight that his players can bench press as 275 pounds, with a standard deviation of 55 pounds. Three of his players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1). Conduct a hypothesis test using a 2.5 percent level of significance to determine if the bench press mean is more than 275 pounds. Solution 9.15 Set up the hypothesis test: |
This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 537 Since the problem is about a mean weight, this is a test of a single population mean. H0: μ = 275 Calculating the distribution needed: Ha: μ > 275 This is a right-tailed test. Random variable: X ¯ = the mean weight, in pounds, lifted by the football players. Distribution for the test: It is normal because σ is known. ¯ X ~ N ⎛ ⎝275, 55 30 ⎞ ⎠ x¯ = 286.2 pounds (from the data). σ = 55 pounds. Always use σ if you know it. We assume μ = 275 pounds unless our data shows us otherwise. First, we compute the sample mean: x− = 205 + 205 + 205 + 215 + ⋯ + 385 30 = 286.2. Next, we compute the z-test: z-test = 286.2 − 275 55 30 = 1.115362 Finally, the p-value is the probability to the right tail of the z-test, which we can compute from the table of z-scores as 0.5 –- 0.36650 = 0.1335. () p-value = P( x¯ > 286.2) = 0.1323 Interpretation of the p-value: If H0 is true, then there is a 0.1331 probability, 13.23 percent, that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23 percent chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event. Figure 9.8 Compare α and the p-value: α = 0.025 p-value = 0.1323 Make a decision: Since α < p-value, do not reject H0. Conclusion: At the 2.5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds. The p-value can easily be calculated. 538 Chapter 9 | Hypothesis Testing with One Sample Put the data and frequencies into lists. Press STAT and arrow over to TESTS. Press 1:z-Test. Arrow over to Data and press ENTER. Arrow down and enter 275 for μ0, |
55 for σ, the name of the list where you put the data, and the name of the list where you put the frequencies. Arrow down to μ: and arrow over to > μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.1331, a little different from the previous calculation—in it we used the sample mean rounded to one decimal place instead of the data), but also the test statistic (z-score) for the sample mean, the sample mean, and the sample standard deviation. μ > 275 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with z = 1.112 (test statistic) and p = 0.1331 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. Example 9.16 Statistics students believe that the mean score on the first statistics test is 65. A statistics instructor thinks the mean score is higher than 65. He samples 10 statistics students and obtains the scores 65; 65; 70; 67; 66; 63; 63; 68; 72; 71. He performs a hypothesis test using a 5 percent level of significance. The data are assumed to be from a normal distribution. Solution 9.16 Set up the hypothesis test: A 5 percent level of significance means that α = 0.05. This is a test of a single population mean. H0: μ = 65 Since the instructor thinks the average score is higher, use a ">". The ">" means the test is right-tailed. Ha: μ > 65 Determine the distribution needed: Random variable: X ¯ = average score on the first statistics test. Distribution for the test: If you read the problem carefully, you will notice that there is no population standard deviation given. You are only given n = 10 sample data values. Notice also that the data come from a normal distribution. This means that the distribution for the test is a Student's t-distribution. Use t-distribution. Therefore, the distribution for the test is t with nine degrees of freedom. Calculate the p-value using the Student's t-distribution: First, we compute the sample mean as Next, we compute the t-test as x¯ = 65 + 65 + ⋯ + 71 10 = 67. t-test = x¯ − |
μ X s X n = 67 − 65 3.12 10 ≈ 1.98. The p-value is the probability to the right tail of 1.98 in a t-distribution with nine degrees of freedom. p-value = P( x¯ > 67) = 0.0396 where the sample mean and sample standard deviation are calculated as 67 and 3.1972 from the data. Interpretation of the p-value: If the null hypothesis is true, then there is a 0.0396 probability— (3.96 percent—) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 539 that the sample mean is 65 or more. Figure 9.9 Compare α and the p-value: Since α = 0.05 and p-value = 0.0396, α > p-value. Make a decision: Since α > p-value, reject H0. Alternatively, according to a Student's t-distribution table (see Appendices), the critical t-value is 1.833. Since the t-test (1.98) is to the right of the critical t-value 1.833, we reject the null hypothesis. This decision means we reject μ = 65. In other words, we believe the average test score is more than 65. Conclusion: At a 5 percent level of significance, the sample data show sufficient evidence that the mean (average) test score is more than 65, just as the math instructor thinks. The p-value can easily be calculated. Put the data into a list. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 65 for μ0, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to > μ0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator not only calculates the p-value (p = 0.0396) but it also calculates the test statistic (t-score) for the sample mean, the sample mean, and the sample standard deviation. μ > 65 is the alternative hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph appears with t = 1.9781 (test statistic) and p = |
0.0396 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. 9.16 It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock won’t grow as quickly. The changes in stock price are recorded for 10 weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5 percent level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors. 540 Chapter 9 | Hypothesis Testing with One Sample Example 9.17 Joon believes that 50 percent of first-time brides in the United States are younger than their grooms. She performs a hypothesis test to determine if the percentage is the same or different from 50 percent. Joon samples 100 firsttime brides and 53 reply that they are younger than their grooms. For the hypothesis test, she uses a 1 percent level of significance. Solution 9.17 Set up the hypothesis test: The 1 percent level of significance means that α = 0.01. This is a test of a single population proportion. H0: p = 0.50 The words is the same or different from tell you this is a two-tailed test. Ha: p ≠ 0.50 Calculate the distribution needed: Random variable: P′ = the percentage of first-time brides who are younger than their grooms. Distribution for the test: The problem contains no mention of a mean. The information is given in terms of percentages. Use the distribution for P′, the estimated proportion. P′ follows a normal distribution with mean value μ = p, and standard error σ = p ⋅ q n. In our example, p = q = 0.5, and n = 100, where p = 0.50, q = 1 – p = 0.50, and n = 100. Calculate the p-value using the normal distribution for proportions: First, we compute the sample proportion as p^ = 53 100 = 0.53. Next, the z-test is given by z-test = p^ − p p ⋅ q n = 0.53 − 0.50 0.50 |
×0.50 100 = 0.6. is positive, we compute the area to the right Since the z-test tail of 0.6 in a normal distribution, P(Z > 0.6) = 0.2742531. Finally, because this is a two-sided test of significance, we multiply this probability times two to account for the left tail, and obtain () p-value = 2×0.2742531 = 0.5485062 where x = 53, p′ = x n = 53 100 = 0.53. Interpretation of the p-value: If the null hypothesis is true, there is 0.5485 probability, (54.85 percent) that the sample (estimated) proportion p′ is 0.53 or more OR 0.47 or less (see the graph in Figure 9.9). Figure 9.10 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 541 μ = p = 0.50 comes from H0, the null hypothesis. p′ = 0.53. Since the curve is symmetrical and the test is two-tailed, the p′ for the left tail is equal to 0.50 – 0.03 = 0.47 where μ = p = 0.50. (0.03 is the difference between 0.53 and 0.50.) Compare α and the p-value: Since α = 0.01 and p-value = 0.5485, α < p-value. Make a decision: Since α < p-value, you cannot reject H0. Conclusion: At the 1 percent level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50 percent. The p-value can easily be calculated. Press STAT and arrow over to TESTS. Press 5:1-PropZTest. Enter.5 for p0, 53 for x and 100 for n. Arrow down to Prop and arrow to not equals p0. Press ENTER. Arrow down to Calculate and press ENTER. The calculator calculates the p-value (p = 0.5485) and the test statistic (z-score). Prop not equals.5 is the alternate hypothesis. Do this set of instructions again except arrow to Draw (instead of Calculate). Press ENTER. A shaded graph |
appears with z = 0.6 (test statistic) and p = 0.5485 (p-value). Make sure when you use Draw that no other equations are highlighted in Y = and the plots are turned off. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50 percent when, in fact, the proportion is actually 50 percent. Reject the null hypothesis when the null hypothesis is true. The Type II error is there is not enough evidence to conclude that the proportion of first-time brides who are younger than their grooms differs from 50 percent when, in fact, the proportion does differ from 50 percent. Do not reject the null hypothesis when the null hypothesis is false. 9.17 A teacher believes that 85 percent of students in the class will want to go on a field trip to the local zoo. She performs a hypothesis test to determine if the percentage is the same or different from 85 percent. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1 percent level of significance. First, determine what type of test this is, set up the hypothesis test, find the p-value, sketch the graph, and state your conclusion. Example 9.18 Suppose a consumer group suspects that the proportion of households that have three cell phones is 30 percent. A cell phone company has reason to believe that the proportion is not 30 percent. Before the cell phone company starts a big advertising campaign, it conducts a hypothesis test. The company's marketing people survey 150 households with the result that 43 of the households have three cell phones. Solution 9.18 Set up the hypothesis test: 542 Chapter 9 | Hypothesis Testing with One Sample H0: p = 0.30 Determine the distribution needed: Ha: p ≠ 0.30 The random variable is P′ = proportion of households that have three cell phones. The distribution for the hypothesis test is P'~ N ⎛ ⎝0.30, (0.30) ⋅ (0.70) 150 ⎞ ⎠. a. The value that helps determine the p-value is p′. Calculate p′. Solution 9.18 a. p′ = x n where x is the number of successes and n is the total number in the sample. x = 43, n = 150 p′ = |
43 150. b. What is a success for this problem? Solution 9.18 b. A success is having three cell phones in a household. c. What is the level of significance? Solution 9.18 c. The level of significance is the preset α. Since α is not given, assume that α = 0.05. d. Draw the graph for this problem. Draw the horizontal axis. Label and shade appropriately. Calculate the p-value. Solution 9.18 d. First we compute the sample proportion p^ = 43 150 = 0.287. Next, the z-test is given by z-test = p^ − p p ⋅ q n = 0.287 – 0.30 0.30×0.70 150 ≈ –0.36. is negative, we compute the area to the left tail of –0.36 in a normal distribution, Since the z-test P(Z < − 0.36) ≈ 0.3607902. Finally, because this is a two-sided test of significance, we multiply this probability times two to account for the right tail, and obtain p-value = 2×0.3607902 = 0.7215804. e. Make a decision. _____________(Reject/Do not reject) H0 because____________. Solution 9.18 e. Assuming that α = 0.05, α < p-value. The decision is do not reject H0 because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30 percent. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 543 9.18 Marketers believe that 92 percent of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. Two hundred American adults are surveyed, of which 174 report having cell phones. Use a 5 percent level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors. The next example is a poem written by a statistics student named Nicole Hart. The solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion. This means that the null and alternate hypotheses use the parameter p. The distribution for the test is normal. The estimated proportion p′ is the proportion of fle |
as killed to the total fleas found on Fido. This is sample information. The problem gives a preconceived α = 0.01, for comparison, and a 95 percent confidence interval computation. The poem is clever and humorous, so please enjoy it! Example 9.19 My dog has so many fleas, They do not come off with ease. As for shampoo, I have tried many types Even one called Bubble Hype, Which only killed 25 percent of the fleas, Unfortunately I was not pleased. I've used all kinds of soap, Until I had given up hope Until one day I saw An ad that put me in awe. A shampoo used for dogs Called GOOD ENOUGH to Clean a Hog Guaranteed to kill more fleas. I gave Fido a bath And after doing the math His number of fleas Started dropping by 3's! Before his shampoo I counted 42. At the end of his bath, I redid the math And the new shampoo had killed 17 fleas. So now I was pleased. Now it is time for you to have some fun With the level of significance being.01, You must help me figure out Use the new shampoo or go without? Solution 9.19 Set up the hypothesis test: H0: p ≤ 0.25 Determine the distribution needed: Ha: p > 0.25 544 Chapter 9 | Hypothesis Testing with One Sample In words, clearly state what your random variable X ¯ or P′ represents. P′ = The proportion of fleas that are killed by the new shampoo State the distribution to use for the test. Normal: ⎛ ⎝0.25, N (0.25)(1 − 0.25) 42 ⎞ ⎠ The z-test is given by z-test = p¯ − p p ⋅ q n = 0.4048 − 0.25 42 ≈ 2.316834. Because this is a hypothesis test one-sided to the right, we compute the p-value as the area to the right tail of the z-test in a standard normal distribution, P(Z > 3.32) ≈ 0.0103. In one to two complete sentences, explain what the p-value means for this problem. If the null hypothesis is true (the proportion is 0.25), then there is a 0.0103 probability that the sample (estimated) proportion is 0.4048 ⎛ ⎝ or |
more. ⎞ ⎠ 17 42 Use the previous information to sketch a picture of this situation. Clearly label and scale the horizontal axis and shade the region(s) corresponding to the p-value. Figure 9.11 Compare α and the p-value: Indicate the correct decision (reject or do not reject the null hypothesis) and the reason for it, and write an appropriate conclusion, using complete sentences. Alpha Decision Reason for Decision 0.01 Do not reject H0 α < p-value Table 9.3 Conclusion: At the 1 percent level of significance, the sample data do not show sufficient evidence that the percentage of fleas that are killed by the new shampoo is more than 25 percent. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 545 Construct a 95 percent confidence interval for the true mean or proportion. Include a sketch of the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. Figure 9.12 Confidence Interval: (0.26, 0.55). We are 95 percent confident that the true population proportion p of fleas that are killed by the new shampoo is between 26 percent and 55 percent. NOTE This test result is not very definitive since the p-value is very close to alpha. In reality, one would probably do more tests by giving the dog another bath after the fleas have had a chance to return. Example 9.20 The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass: 1.11, 1.07, 1.11, 1.07, 1.12, 1.08, 0.98, 0.98, 1.02, 0.95, 0.95 Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal. Solution 9.20 Let’s follow a four-step process to answer this statistical question. 1. State the question: We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be as follows: a. H0: μ ≤ 1 b. Ha: μ > 1 2. Plan: We are |
testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's t-distribution. Assume the underlying population is normal. 3. Do the calculations: We will input the sample data into the TI-83 as follows. 546 Chapter 9 | Hypothesis Testing with One Sample Figure 9.13 Figure 9.14 Figure 9.15 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 547 Figure 9.16 4. State the conclusions: Since the p-value (p = 0.036) is less than our alpha value, we will reject the null hypothesis. It is reasonable to state that the data support the claim that the average conductivity level is greater than one. Example 9.21 In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users. The rate of brain cancer for noncell phone users is 0.0340 percent. Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error. Solution 9.21 We will follow the four-step process. 1. We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be as follows: a. H0: p ≤ 0.00034 b. Ha: p > 0.00034 If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancercausing environments, we want to minimize the chances of incorrectly identifying causes of cancer. 2. We will be testing a sample proportion with x = 172 and n = 420,019. The sample is sufficiently large because we have np = 420,019(0.00034) = 142.8, nq = 420,019(0.99966) = 419,876.2, two independent outcomes, and a fixed probability of success p = 0.00034. Thus we will be able to generalize our results to the population. 3. The associated TI results are shown in the following figures. 548 Chapter 9 | Hypothesis Testing with One Sample Figure 9.17 Figure 9.18 4. Since the p-value = 0.0073 is greater than our alpha value = 0. |
005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher brain cancer rates for the cell phone users. 9.6 | Hypothesis Testing of a Single Mean and Single Proportion This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 549 9.1 Hypothesis Testing of a Single Mean and Single Proportion Student Learning Outcomes • The student will select the appropriate distributions to use in each case. • The student will conduct hypothesis tests and interpret the results. Television Survey In a recent survey, it was stated that Americans watch television on average four hours per day. Assume that σ = 2. Using your class as the sample, conduct a hypothesis test to determine if the average for students at your school is lower. 1. H0: _____________ 2. Ha: _____________ 3. In words, define the random variable. __________ = ______________________ 4. The distribution to use for the test is _______________________. 5. Determine the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 9.19 b. Determine the p-value. 7. Do you or do you not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Language Survey About 42.3 percent of Californians and 19.6 percent of all Americans over age five speak a language other than English at home. Using your class as the sample, conduct a hypothesis test to determine if the percentage of the students at your school who speak a language other than English at home is different from 42.3 percent. 1. H0: ___________ 2. Ha: ___________ 550 Chapter 9 | Hypothesis Testing with One Sample 3. In words, define the random variable. __________ = _______________ 4. The distribution to use for the test is ________________. 5. Determine the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: Figure 9.20 b. Determine the p-value. 7. Do you or do you not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Jeans Survey You've read in an article that young adults own |
an average of three pairs of jeans. Survey eight people from your class to determine if the average is higher than three. Assume the population is normal. 1. H0: _____________ 2. Ha: _____________ 3. In words, define the random variable. __________ = ______________________ 4. The distribution to use for the test is _______________________. 5. Determine the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 551 Figure 9.21 b. Determine the p-value. 7. Do you or do you not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. 552 Chapter 9 | Hypothesis Testing with One Sample KEY TERMS binomial distribution independent trials Independent means that the result of any trial (for example, trial 1) does not affect the results of the following trials, and all trials are conducted under the same conditions. Under these circumstances the binomial RV Χ is defined as the number of successes in n trials. The notation is: X ~ B(n, p) μ = np and the standard deviation is σ = npq. a discrete random variable (RV) that arises from Bernoulli trials; there are a fixed number, n, of The probability of exactly x successes in n trials is P(X = x) = ⎞ ⎠p x qn − x. n ⎛ ⎝ x confidence interval (CI) an interval estimate for an unknown population parameter This depends on the following: • The desired confidence level. • Information that is known about the distribution (for example, known standard deviation). • The sample and its size. hypothesis a statement about the value of a population parameter; in the case of two hypotheses, the statement assumed to be true is called the null hypothesis (notation H0) and the contradictory statement is called the alternative hypothesis (notation Ha) hypothesis testing based on sample evidence, a procedure for determining whether the hypothesis stated is a reasonable statement and should not be rejected, or is unreasonable and should be rejected level of significance of the test probability of a Type I error (reject the null hypothesis when it is true) Notation: α. In hypothesis testing, the level |
of significance is called the preconceived α or the preset α. normal distribution a bell-shaped continuous random variable X, with center at the mean value (μ) and distance from the center to the inflection points of the bell curve given by the standard deviation (σ) We write X ~ N(μ, σ). If the mean value is 0 and the standard deviation is 1, the random variable is called the standard normal distribution, and it is denoted with the letter Z. p-value the probability that an event will happen purely by chance assuming the null hypothesis is true; the smaller the p-value, the stronger the evidence is against the null hypothesis standard deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation Student's t-distribution investigated and reported by William S. Gosset in 1908 and published under the pseudonym Student The major characteristics of the random variable (RV) are as follows • It is continuous and assumes any real values. • The pdf is symmetrical about its mean of zero. However, it is more spread out and flatter at the apex than the normal distribution. • It approaches the standard normal distribution as n gets larger. • There is a family of t-distributions: every representative of the family is completely defined by the number of degrees of freedom, which is one less than the number of data items. Type 1 error the decision is to reject the null hypothesis when, in fact, the null hypothesis is true Type 2 error the decision is not to reject the null hypothesis when, in fact, the null hypothesis is false CHAPTER REVIEW 9.1 Null and Alternative Hypotheses In a hypothesis test, sample data are evaluated in order to arrive at a decision about some type of claim. If certain conditions This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 553 about the sample are satisfied, then the claim can be evaluated for a population. In a hypothesis test, we do the following: 1. Evaluate the null hypothesis, typically denoted with H0. The null is not rejected unless the hypothesis test shows otherwise. The null statement must always contain some form of equality (=, ≤, or ≥). 2. Always write the alternative hypothesis, typically denoted with Ha or H1, using |
less than, greater than, or not equals symbols, i.e., (≠, >, or <). 3. If we reject the null hypothesis, then we can assume there is enough evidence to support the alternative hypothesis. 4. Never state that a claim is proven true or false. Keep in mind the underlying fact that hypothesis testing is based on probability laws; therefore, we can talk only in terms of non-absolute certainties. 9.2 Outcomes and the Type I and Type II Errors In every hypothesis test, the outcomes are dependent on a correct interpretation of the data. Incorrect calculations or misunderstood summary statistics can yield errors that affect the results. A Type I error occurs when a true null hypothesis is rejected. A Type II error occurs when a false null hypothesis is not rejected. The probabilities of these errors are denoted by the Greek letters α and β, for a Type I and a Type II error respectively. The power of the test, 1 – β, quantifies the likelihood that a test will yield the correct result of a true alternative hypothesis being accepted. A high power is desirable. 9.3 Distribution Needed for Hypothesis Testing In order for a hypothesis test’s results to be generalized to a population, certain requirements must be satisfied. When testing for a single population mean: 1. A Student's t-test should be used if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with an unknown standard deviation. 2. The normal test will work if the data come from a simple, random sample and the population is approximately normally distributed, or the sample size is large, with a known standard deviation. When testing a single population proportion use a normal test for a single population proportion if the data come from a simple, random sample, fill the requirements for a binomial distribution, and the mean number of success and the mean number of failures satisfy the conditions: np > 5 and nq > n where n is the sample size, p is the probability of a success, and q is the probability of a failure. 9.4 Rare Events, the Sample, and the Decision and Conclusion When the probability of an event occurring is low, and it happens, it is called a rare event. Rare events are important to consider in hypothesis testing because they can inform your willingness not to reject or to reject a null hypothesis. To test a null hypothesis, find the p-value for the sample data and graph the results. When |
deciding whether or not to reject the null the hypothesis, keep these two parameters in mind: 1. α > p-value, reject the null hypothesis. 2. α ≤ p-value, do not reject the null hypothesis. 9.5 Additional Information and Full Hypothesis Test Examples The hypothesis test itself has an established process. This can be summarized as follows: 1. Determine H0 and Ha. Remember, they are contradictory. 2. Determine the random variable. 3. Determine the distribution for the test. 4. Draw a graph, calculate the test statistic, and use the test statistic to calculate the p-value. (A z-score and a t-score are examples of test statistics.) 5. Compare the preconceived α with the p-value, make a decision (reject or do not reject H0), and write a clear conclusion using English sentences. Notice that in performing the hypothesis test, you use α and not β. β is needed to help determine the sample size of the data that are used in calculating the p-value. Remember that the quantity 1 – β is called the Power of the Test. A high power is 554 Chapter 9 | Hypothesis Testing with One Sample desirable. If the power is too low, statisticians typically increase the sample size while keeping α the same. If the power is low, the null hypothesis might not be rejected when it should be. FORMULA REVIEW 9.1 Null and Alternative Hypotheses H0 and Ha are contradictory. If H0 has: equal (=) then Ha has: not equal (≠) or greater than (>) or less than (<) Table 9.4 greater than or equal to (≥) less than or equal to (≤) less than (<) greater than (>) If α ≤ p-value, then do not reject H0. If α > p-value, then reject H0. α is preconceived. Its value is set before the hypothesis test starts. The p-value is calculated from the data. 9.2 Outcomes and the Type I and Type II Errors α = probability of a Type I error = P(Type I error) = probability of rejecting the null hypothesis when the null hypothesis is true. β = probability of a Type II error = P(Type II error) = PRACTICE 9.1 Null and Alternative Hypotheses probability of not rejecting the null hypothesis when the null hypothesis is false. 9.3 Distribution Needed for Hyp |
othesis Testing If there is no given preconceived α, then use α = 0.05. Types of Hypothesis Tests • Single population mean, known population variance (or standard deviation): Normal test. • Single population mean, unknown population variance (or standard deviation): Student's t-test. • Single population proportion: Normal test. • For a single population mean, we may use a normal distribution with the following mean and standard deviation. Means: μ = μ x¯ and σ x¯ =. σ x n • For a single population proportion, we may use a normal distribution with the following mean and standard deviation. Proportions: µ = p and σ = pq n. 1. You are testing that the mean speed of your cable internet connection is more than three megabits per second. What is the random variable? Describe it in words. 2. You are testing that the mean speed of your cable internet connection is more than three megabits per second. State the null and alternative hypotheses. 3. The American family has an average of two children. What is the random variable? Describe in words. 4. The mean entry level salary of an employee at a company is $58,000. You believe it is higher for IT professionals in the company. State the null and alternative hypotheses. 5. A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the proportion is actually less. What is the random variable? Describe in words. 6. A sociologist claims the probability that a person picked at random in Times Square in New York City is visiting the area is 0.83. You want to test to see if the claim is correct. State the null and alternative hypotheses. 7. In a population of fish, approximately 42 percent are female. A test is conducted to see if, in fact, the proportion is less. State the null and alternative hypotheses. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 555 8. Suppose that a recent article stated that the mean time students spend doing homework each week is 2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students. The mean length of time the students spent on homework was 3 hours |
with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. If you were conducting a hypothesis test to determine if the mean length of homework has increased, what would the null and alternative hypotheses be? The distribution of the population is normal. a. H0: ________ b. Ha: ________ 9. A random survey of 75 long-term marathon runners revealed that the mean length of time they've been running is 17.4 years with a standard deviation of 6.3 years. If you were conducting a hypothesis test to determine if the population mean time for these runners could likely be 15 years, what would the null and alternative hypotheses be? a. H0: __________ b. Ha: __________ 10. Researchers published an article stating that in any one-year period, approximately 9.5 percent of American adults suffer from a particular type of disease. Suppose that in a survey of 100 people in a certain town, seven of them suffered from this disease. If you were conducting a hypothesis test to determine if the true proportion of people in that town suffering from this disease is lower than the percentage in the general adult American population, what would the null and alternative hypotheses be? a. H0: ________ b. Ha: ________ 9.2 Outcomes and the Type I and Type II Errors 11. The mean price of mid-sized cars in a region is $32,000. A test is conducted to see if the claim is true. State the Type I and Type II errors in complete sentences. 12. A sleeping bag is tested to withstand temperatures of –15 °F. You think the bag cannot stand temperatures that low. State the Type I and Type II errors in complete sentences. 13. For Exercise 9.12, what are α and β in words? 14. In words, describe 1 – β for Exercise 9.12. 15. A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H0, is: the surgical procedure will go well. State the Type I and Type II errors in complete sentences. 16. A group of doctors is deciding whether or not to perform an operation. Suppose the null hypothesis, H0, is: the surgical procedure will go well. Which is the error with the greater consequence? 17. The power of a test is 0.981. What is the probability of a Type II error? 18. A group of divers is exploring |
an old sunken ship. Suppose the null hypothesis, H0, is the sunken ship does not contain buried treasure. State the Type I and Type II errors in complete sentences. 19. A microbiologist is testing a water sample for E. coli. Suppose the null hypothesis, H0, is the sample does not contain E. coli. The probability that the sample does not contain E. coli, but the microbiologist thinks it does is 0.012. The probability that the sample does contain E. coli, but the microbiologist thinks it does not is 0.002. What is the power of this test? 20. A microbiologist is testing a water sample for E. coli. Suppose the null hypothesis, H0, is the sample contains E-coli. Which is the error with the greater consequence? 9.3 Distribution Needed for Hypothesis Testing 21. Which two distributions can you use for hypothesis testing for this chapter? 22. Which distribution do you use when the standard deviation is not known? Assume sample size is large. 23. Which distribution do you use when the standard deviation is not known and you are testing one population mean? Assume sample size is large. 24. A population mean is 13. The sample mean is 12.8, and the sample standard deviation is two. The sample size is 20. What distribution should you use to perform a hypothesis test? Assume the underlying population is normal. 25. A population has a mean of 25 and a standard deviation of five. The sample mean is 24, and the sample size is 108. What distribution should you use to perform a hypothesis test? 556 Chapter 9 | Hypothesis Testing with One Sample 26. It is thought that 42 percent of respondents in a taste test would prefer Brand A. In a particular test of 100 people, 39 percent preferred Brand A. What distribution should you use to perform a hypothesis test? 27. You are performing a hypothesis test of a single population mean using a Student’s t-distribution. What must you assume about the distribution of the data? 28. You are performing a hypothesis test of a single population mean using a Student’s t-distribution. The data are not from a simple random sample. Can you accurately perform the hypothesis test? 29. You are performing a hypothesis test of a single population proportion. What must be true about the quantities of np and nq? 30. You are performing a hypothesis test of a single population proportion. You find out that |
np is less than five. What must you do to be able to perform a valid hypothesis test? 31. You are performing a hypothesis test of a single population proportion. The data come from which distribution? 9.4 Rare Events, the Sample, and the Decision and Conclusion 32. When do you reject the null hypothesis? 33. The probability of winning the grand prize at a particular carnival game is 0.005. Is the outcome of winning very likely or very unlikely? 34. The probability of winning the grand prize at a particular carnival game is 0.005. Michele wins the grand prize. Is this considered a rare or common event? Why? 35. It is believed that the mean height of high school students who play basketball on the school team is 73 inches with a standard deviation of 1.8 inches. A random sample of 40 players is chosen. The sample mean was 71 inches, and the sample standard deviation was 1.5 inches. Do the data support the claim that the mean height is less than 73 inches? The p-value is almost zero. State the null and alternative hypotheses and interpret the p-value. 36. The mean age of graduate students at a university is at most 31 years with a standard deviation of two years. A random sample of 15 graduate students is taken. The sample mean is 32 years and the sample standard deviation is three years. Are the data significant at the 1 percent level? The p-value is 0.0264. State the null and alternative hypotheses and interpret the p-value. 37. Does the shaded region represent a low or a high p-value compared to a level of significance of 1 percent? Figure 9.22 38. What should you do when α > p-value? 39. What should you do if α = p-value? 40. If you do not reject the null hypothesis, then it must be true. Is that statement correct? State why or why not in complete sentences. Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time students spend doing homework each week is 2.5 hours. A study was then done to see if the mean time has increased in the new century. A random sample of 26 students was taken. The mean length of time they did homework each week was three hours with a standard deviation of 1.8 hours. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of time doing |
homework each week has increased. Assume This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 557 the distribution of homework times is approximately normal. 41. Is this a test of means or proportions? 42. What symbol represents the random variable for this test? 43. In words, define the random variable for this test. 44. Is σ known and, if so, what is it? 45. Calculate the following: x¯ _______ a. b. σ _______ sx _______ c. d. n _______ 46. Since both σ and s x are given, which should be used? In one to two complete sentences, explain why. 47. State the distribution to use for the hypothesis test. 48. A random survey of 75 long-term marathon runners revealed that the mean length of time they have been running is 17.4 years with a standard deviation of 6.3 years. Conduct a hypothesis test to determine if the population mean time is likely to be 15 years. Is this a test of one mean or proportion? a. b. State the null and alternative hypotheses. H0: ____________________ Ha : ____________________ Is this a right-tailed, left-tailed, or two-tailed test? c. d. What symbol represents the random variable for this test? e. f. g. Calculate the following: In words, define the random variable for this test. Is the population standard deviation known and, if so, what is it? x¯ = _____________ i. ii. s = ____________ iii. n = ____________ h. Which test should be used? i. State the distribution to use for the hypothesis test. j. Find the p-value. k. At a pre-conceived α = 0.05, give your answer for each of the following: i. Decision: ii. Reason for the decision: iii. Conclusion (write out in a complete sentence): 9.5 Additional Information and Full Hypothesis Test Examples 49. Assume H0: μ = 9 and Ha: μ < 9. Is this a left-tailed, right-tailed, or two-tailed test? 50. Assume H0: μ ≤ 6 and Ha: μ > 6. Is this a left-tailed, right-tailed, or two-tailed test? 51. Assume H |
0: p = 0.25 and Ha: p ≠ 0.25. Is this a left-tailed, right-tailed, or two-tailed test? 52. Draw the general graph of a left-tailed test. 53. Draw the graph of a two-tailed test. 54. A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use? 55. Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use? 56. A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use? 57. You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50 percent, but you think it is less for this particular coin. What type of test would you use? 558 Chapter 9 | Hypothesis Testing with One Sample 58. If the alternative hypothesis has a not equals ( ≠ ) symbol, you know to use which type of test? 59. Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test? 60. Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test? 61. Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test? HOMEWORK 9.1 Null and Alternative Hypotheses 62. Some of the following statements refer to the null hypothesis, some to the alternate hypothesis. State the null hypothesis, H0, and the alternative hypothesis. Ha, in terms of the appropriate parameter (μ or p). a. The mean number of years Americans work before retiring is 34. b. At most 60 percent of Americans vote in presidential elections. c. The mean starting salary for San Jose State University graduates is at least $100,000 per year. d. Twenty-nine percent of high school students take physical education daily. e. Less than 5 percent of adults ride the bus to work in Los Angeles. f. The mean number of cars a person owns in her lifetime is not more than 10. g |
. About half of Americans prefer to live away from cities, given the choice. h. Europeans have a mean paid vacation each year of six weeks. i. The chance of developing breast cancer is under 11 percent for women. j. Private universities' mean tuition cost is more than $20,000 per year. 63. A recent survey of 273 randomly selected teens living in Massachusetts asked about social media. Sixty-three said that they routinely use a certain app to share pictures. The researchers want to determine if there is good evidence that more than 30 percent of teens use this app. The alternative hypothesis is as follows: a. p < 0.30 b. p ≤ 0.30 c. p ≥ 0.30 d. p > 0.30 64. A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 attended the midnight showing. An appropriate alternative hypothesis is as follows: a. p = 0.20 b. p > 0.20 c. p < 0.20 d. p ≤ 0.20 65. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The null and alternative hypotheses are as follows: a. Ho: x¯ = 4.5, Ha : x¯ > 4.5 b. Ho: μ ≥ 4.5, Ha: μ < 4.5 c. Ho: μ = 4.75, Ha: μ > 4.75 d. Ho: μ = 4.5, Ha: μ > 4.5 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 559 9.2 Outcomes and the Type I and Type II Errors 66. State the Type I and Type II errors in complete sentences given the following statements. a. The mean number of years Americans work before retiring is 34. b. At most 60 percent of Americans vote in presidential elections. c. The mean starting salary for San Jose State University graduates is at least $100,000 per year |
. d. 29 percent of high school students take physical education every day. e. Less than 5 percent of adults ride the bus to work in Los Angeles. f. The mean number of cars a person owns in his or her lifetime is not more than 10. g. About half of Americans prefer to live away from cities, given the choice. h. Europeans have a mean paid vacation each year of six weeks. i. The chance of developing breast cancer is under 11 percent for women. j. Private universitie' mean tuition cost is more than $20,000 per year. 67. For Statements A–J in Exercise 9.66, answer the following in complete sentences. a. State a consequence of committing a Type I error. b. State a consequence of committing a Type II error. 68. When a new drug is created, the pharmaceutical company must subject it to testing before receiving the necessary permission from the U.S. Food and Drug Administration (FDA) to market the drug. Suppose the null hypothesis is the drug is unsafe. What is the Type II error? a. To conclude the drug is safe when, in fact, it is unsafe. b. Not to conclude the drug is safe when, in fact, it is safe. c. To conclude the drug is safe when, in fact, it is safe. d. Not to conclude the drug is unsafe when, in fact, it is unsafe. 69. A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. The Type I error is to conclude that the percent of EVC students who attended is ________. a. at least 20 percent, when, in fact, it is less than 20 percent. b. 20 percent, when, in fact, it is 20 percent. c. d. less than 20 percent, when, in fact, it is at least 20 percent. less than 20 percent, when, in fact, it is less than 20 percent. 70. It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5 percent, do LTCC Intermediate Algebra students get less |
than seven hours of sleep per night, on average? The Type II error is not to reject that the mean number of hours of sleep LTCC students get per night is at least seven when, in fact, the mean number of hours a. b. c. d. is more than seven hours. is at most seven hours. is at least seven hours. is less than seven hours. 71. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. The Type I error is a. b. c. d. to conclude that the current mean hours per week is higher than 4.5, when, in fact, it is higher. to conclude that the current mean hours per week is higher than 4.5, when, in fact, it is the same. to conclude that the mean hours per week currently is 4.5, when, in fact, it is higher. to conclude that the mean hours per week currently is no higher than 4.5, when, in fact, it is not higher. 560 Chapter 9 | Hypothesis Testing with One Sample 9.3 Distribution Needed for Hypothesis Testing 72. It is believed that Lake Tahoe Community College (LTCC) Intermediate Algebra students get less than seven hours of sleep per night, on average. A survey of 22 LTCC Intermediate Algebra students generated a mean of 7.24 hours with a standard deviation of 1.93 hours. At a level of significance of 5 percent, do LTCC Intermediate Algebra students get less than seven hours of sleep per night, on average? The distribution to be used for this test is X ¯ ~ ________________ ) a. N(7.24, 1.93 22 b. N(7.24, 1.93) c. d. t22 t21 9.4 Rare Events, the Sample, and the Decision and Conclusion 73. The National Institute of Mental Health published an article stating that in any one-year period approximately 9.5 percent of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion |
of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population. a. Is this a test of one mean or proportion? b. State the null and alternative hypotheses. H0: ____________________ Ha: ____________________ Is this a right-tailed, left-tailed, or two-tailed test? c. d. What symbol represents the random variable for this test? e. f. Calculate the following: In words, define the random variable for this test. i. x = ________________ ii. n = ________________ iii. p′ = _____________ g. Calculate σx = __________. Show the formula setup. h. State the distribution to use for the hypothesis test. i. Find the p-value. j. At a pre-conceived α = 0.05, give your answer for each of the following: i. Decision: ii. Reason for the decision: iii. Conclusion (write out in a complete sentence): 9.5 Additional Information and Full Hypothesis Test Examples For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E, Solution Sheets. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the.doc or the.pdf files. NOTE If you are using a Student's-t-distribution for one of the following homework problems, you may assume that the underlying population is normally distributed. In general, you must first prove that assumption, however. 74. A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8,000. A survey of owners of that tire design is conducted. From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9,800 miles. Using alpha = 0.05, are the data highly inconsistent with the claim? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 561 75. In 2009, President Barack Obama announced a new national fuel economy and emissions policy for cars and light trucks. It stated that the combined fleet fuel economy for an auto manufacturer of cars and light trucks will have to average 35.5 m |
pg or better by 2016. From past studies on fuel economy, it is known that the standard deviation of a typical fleet is 7.6 mpg. An auto manufacturer selects a random sample of 55 cars and light trucks and finds the sample mean fuel economy to be 34.6 mpg with a standard deviation of 10.3 mpg. Can the manufacturer claim that their fleet meets the fuel economy standard in the 2016 policy at the 5 percent level? 76. The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 20¢. A study was done to test the claim that the mean cost of a daily newspaper is $1.00. Twelve costs yield a mean cost of 95¢ with a standard deviation of 18¢. Do the data support the claim at the 1 percent level? 77. An article in the San Jose Mercury News stated that students in the California state university system take 4.5 years, on average, to finish their undergraduate degrees. Suppose you believe that the mean time is longer. You conduct a survey of 49 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1 percent level? 78. The mean number of sick days an employee takes per year is believed to be about 10. Members of a personnel department do not believe this figure. They randomly survey eight employees. The number of sick days they took for the past year are as follows: 12; 4; 15; 3; 11; 8; 6; 8. Let x = the number of sick days they took for the past year. Should the personnel team believe that the mean number is 10? 79. In 1955, Life Magazine reported that the 25-year-old mother of three worked, on average, an 80-hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the mean work week has increased. Eighty-one women were surveyed with the following results. The sample mean was 83; the sample standard deviation was 10. Does it appear that the mean work week has increased for women at the 5 percent level? 80. Your statistics instructor claims that 60 percent of the students who take her Elementary Statistics class go through life feeling more enriched. For some reason that she can't quite figure out |
, most people don't believe her. You decide to check this out on your own. You randomly survey 64 of her past Elementary Statistics students and find that 34 feel more enriched as a result of her class. Now, what do you think? 81. A Nissan Motor Corporation advertisement read, “The average man’s I.Q. is 107. The average brown trout’s I.Q. is 4. So why can’t man catch brown trout?” Suppose you believe that the brown trout’s mean I.Q. is greater than four. You catch 12 brown trout. A fish psychologist determines the I.Q.s as follows: 5, 4, 7, 3, 6, 4, 5, 3, 6, 3, 8, 5. Conduct a hypothesis test of your belief. 82. Refer to Exercise 9.81. Conduct a hypothesis test to see if your decision and conclusion would change if your belief were that the brown trout’s mean I.Q. is not four. 83. According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100: 114 (46.7 percent girls). Suppose you don’t believe the reported figures of the percentage of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percentage of girls born in China is 46.7? 84. A group of researchers research a common contagious disease. A newspaper found that 13 percent of Americans have been diagnosed with the disease in the last year. The researchers doubt that the percentage is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had been diagnosed with the disease. Would you agree with the newspaper's poll? In complete sentences, give three reasons why polls might give different results. 85. The mean work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it’s shorter. She asks 10 engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should she count on the mean work week to be shorter than 60 hours? Data (length of mean work week): 70, 45, 55, 60, 65, 55, 55, 60 |
macaroni and cheese. It's high in taste and low in cost and nutritional value. One day, as I sat down to determine the meaning of life, I got a serious craving for this, oh, so important, food of my life. So I went down the street to Greatway to get a box of macaroni and cheese, but it was SO expensive! $2.02!!! Can you believe it? It made me stop and think. The world is changing fast. I had thought that the mean cost of a box (the normal size, not some super-gigantic-family-value-pack) was at most $1, but now I wasn't so sure. However, I was determined to find out. I went to 53 of the closest grocery stores and surveyed the prices of macaroni and cheese. Here are the data I wrote in my notebook: Price per box of Mac and Cheese • 5 stores @ $2.02 • 15 stores @ $0.25 • 3 stores @ $1.29 • 6 stores @ $0.35 • 4 stores @ $2.27 • 7 stores @ $1.50 • 5 stores @ $1.89 • 8 stores @ $0.75 I could see that the cost varied but I had to sit down to figure out whether or not I was right. If it does turn out that this mouth-watering dish is at most $1, then I'll throw a big cheesy party in our next statistics lab, with enough macaroni and cheese for just me. After all, as a poor starving student I can't be expected to feed our class of animals! 566 Chapter 9 | Hypothesis Testing with One Sample 92. "William Shakespeare: The Tragedy of Hamlet, Prince of Denmark," by Jacqueline Ghodsi THE CHARACTERS (in order of appearance): • HAMLET, Prince of Denmark and student of statistics • POLONIUS, Hamlet’s tutor • HORATIO, friend to Hamlet and fellow student Scene: The great library of the castle, in which Hamlet does his lessons Act I The day is fair, but the face of Hamlet is clouded. He paces the large room. His tutor, Polonius, is reprimanding Hamlet regarding the latter’s recent experience. Horatio is seated at the large table at right stage. POLONIUS: My Lord, how cans’t thou admit that thou |
hast seen a ghost! It is but a figment of your imagination! HAMLET: I beg to differ; I know of a certainty that five-and-seventy in one hundred of us, condemned to the whips and scorns of time as we are, have gazed upon a spirit of health, or goblin damn’d, be their intents wicked or charitable. POLONIUS: If thou dost insist upon thy wretched vision then let me invest your time; be true to thy work and speak to me through the reason of the null and alternate hypotheses. (He turns to Horatio.) Did not Hamlet himself say, “What a piece of work is man, how noble in reason, how infinite in faculties”? Then let not this foolishness persist. Go, Horatio, make a survey of three-and-sixty and discover what the true proportion be. For my part, I will never succumb to this fantasy, but deem man to be devoid of all reason should thy proposal of at least five-and-seventy in one hundred hold true. HORATIO (to Hamlet): What should we do, my Lord? HAMLET: Go to thy purpose, Horatio. HORATIO: To what end, my Lord? HAMLET: That you must teach me. But let me conjure you by the rights of our fellowship, by the consonance of our youth, but the obligation of our ever-preserved love, be even and direct with me, whether I am right or no. Horatio exits, followed by Polonius, leaving Hamlet to ponder alone. Act II The next day, Hamlet awaits anxiously the presence of his friend, Horatio. Polonius enters and places some books upon the table just a moment before Horatio enters. POLONIUS: So, Horatio, what is it thou didst reveal through thy deliberations? HORATIO: In a random survey, for which purpose thou thyself sent me forth, I did discover that one-and-forty believe fervently that the spirits of the dead walk with us. Before my God, I might not this believe, without the sensible and true avouch of mine own eyes. POLONIUS: Give thine own thoughts no tongue, Horatio. (Polonius turns to Hamlet.) But look to’t I charge you, my Lord. Come Horatio, let us go together, for this is |
not our test. (Horatio and Polonius leave together.) HAMLET: To reject, or not reject, that is the question: whether ‘tis nobler in the mind to suffer the slings and arrows of outrageous statistics, or to take arms against a sea of data, and, by opposing, end them. (Hamlet resignedly attends to his task.) (Curtain falls) 93. "Untitled," by Stephen Chen I've often wondered how software is released and sold to the public. Ironically, I work for a company that sells products with known problems. Unfortunately, most of the problems are difficult to create, which makes them difficult to fix. I usually use the test program X, which tests the product, to try to create a specific problem. When the test program is run to make an error occur, the likelihood of generating an error is 1 percent. So, armed with this knowledge, I wrote a new test program Y that will generate the same error that test program X creates, but more often. To find out if my test program is better than the original, so that I can convince the management that I'm right, I ran my test program to find out how often I can generate the same error. When I ran my test program 50 times, I generated the error twice. While this may not seem much better, I think that I can convince the management to use my test program instead of the original test program. Am I right? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 567 94. "Japanese Girls’ Names" by Kumi Furuichi It used to be very typical for Japanese girls’ names to end with “ko.” The trend might have started around my grandmothers’ generation and its peak might have been around my mother’s generation. “Ko” means “child” in Chinese characters. Parents would name their daughters with “ko” attaching to other Chinese characters that have meanings that they want their daughters to become, such as Sachiko—happy child, Yoshiko—a good child, Yasuko—a healthy child, and so on. However, I noticed recently that only two out of nine of my Japanese girlfriends at this school have names that end with “ko.” More and more, parents seem to have become creative, modernized, and, |
sometimes, westernized in naming their children. I have a feeling that, while 70 percent or more of my mother’s generation would have names with “ko” at the end, the proportion has dropped among my peers. I wrote down all my Japanese friends’, ex-classmates’, coworkers’, and acquaintances’ names that I could remember. Following are the names. Some are repeats. Test to see if the proportion has dropped for this generation. Ai, Akemi, Akiko, Ayumi, Chiaki, Chie, Eiko, Eri, Eriko, Fumiko, Harumi, Hitomi, Hiroko, Hiroko, Hidemi, Hisako, Hinako, Izumi, Izumi, Junko, Junko, Kana, Kanako, Kanayo, Kayo, Kayoko, Kazumi, Keiko, Keiko, Kei, Kumi, Kumiko, Kyoko, Kyoko, Madoka, Maho, Mai, Maiko, Maki, Miki, Miki, Mikiko, Mina, Minako, Miyako, Momoko, Nana, Naoko, Naoko, Naoko, Noriko, Rieko, Rika, Rika, Rumiko, Rei, Reiko, Reiko, Sachiko, Sachiko, Sachiyo, Saki, Sayaka, Sayoko, Sayuri, Seiko, Shiho, Shizuka, Sumiko, Takako, Takako, Tomoe, Tomoe, Tomoko, Touko, Yasuko, Yasuko, Yasuyo, Yoko, Yoko, Yoko, Yoshiko, Yoshiko, Yoshiko, Yuka, Yuki, Yuki, Yukiko, Yuko, Yuko. 568 Chapter 9 | Hypothesis Testing with One Sample 95. "Phillip’s Wish," by Suzanne Osorio My nephew likes to play Chasing the girls makes his day. He asked his mother If it is okay To get his ear pierced. She said, “No way!” To poke a hole through your ear, Is not what I want for you, dear. He argued his point quite well, Says even my macho pal, Mel, Has gotten this done. It’s all just for fun. C’mon please, mom, please, what the hell. Again Phillip complained to his mother, Saying half his |
report they fear public speaking. Conduct a hypothesis test to determine if the percentage at her school is less than 40. 98. Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68 percent also represents California’s percent for full-time faculty teaching the online classes, Long Beach City College (LBCC) in California was randomly selected for comparison. In the same year, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test to determine if 68 percent represents California. Note: For more accurate results, use more California community colleges and this past year's data. 99. According to an article in a local poll, a city found that 14 percent of its residents walk for exercise. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen city residents replied that they walk for exercise. Conduct a hypothesis test to determine if the rate is still 14 percent or if it has decreased. 570 Chapter 9 | Hypothesis Testing with One Sample 100. The mean age of De Anza College students in a previous term was 26.6 years old. An instructor thinks the mean age for online students is older than 26.6. She randomly surveys 56 online students and finds that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test. 101. Registered nurses earned an average annual salary of $69,110. For that same year, a survey was conducted of 41 California registered nurses to determine if the annual salary is higher than $69,110 for California nurses. The sample average was $71,121 with a sample standard deviation of $7,489. Conduct a hypothesis test. 102. La Leche League International reports that the mean age of weaning a child from breastfeeding is age four to five worldwide. In America, most nursing mothers wean their children much earlier. Suppose a random survey is conducted of 21 U.S. mothers who recently weaned their children. The mean weaning age was nine months (3/4 year) with a standard deviation of 4 months. Conduct a hypothesis test to determine if the mean weaning age in the United States is less than four years old. 103. Harley Davidson motorcycles are the largest selling motorcycle in the United States, with 14 percent of all motorcycles sold in 2012. Interestingly, a random sample of 1,945 stolen motorcycles was selected, and it was found that |
just 8 percent of them were Harleys. Is there good evidence that the proportion of Harleys among stolen motorcycles is significantly less than their share of all motorcycles? After conducting the test, what decision and conclusion would you make? a. Reject H0: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles b. Do not reject H0: There is not sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles c. Do not reject H0: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly more than their share of all motorcycles d. Reject H0: There is sufficient evidence to conclude that the proportion of Harleys stolen is significantly more than their share of all motorcycles 104. A statistics instructor believes that fewer than 20 percent of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Harry Potter movie. She surveys 84 of her students and finds that 11 of them attended the midnight showing. At a 1 percent level of significance, what is an appropriate conclusion? a. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20 percent. b. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is more than 20 percent. c. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is less than 20 percent. d. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of Harry Potter is at least 20 percent. 105. Previously, an organization reported that teenagers spent 4.5 hours per week, on average, on the phone. The organization thinks that, currently, the mean is higher. Fifteen randomly chosen teenagers were asked how many hours per week they spend on the phone. The sample mean was 4.75 hours with a sample standard deviation of 2.0. Conduct a hypothesis test. At a significance level of a = 0.05, what is the correct conclusion? a. There is enough evidence to conclude that the mean number of hours is more than 4.75. b. There is enough evidence to conclude that the mean number of hours is more than 4.5. c. There is not enough evidence to conclude that the mean number of hours is more than 4.5. d. There is not enough evidence to |
conclude that the mean number of hours is more than 4.75. Hypothesis testing: For the following 10 exercises, answer each question. a. State the null and alternate hypotheses. b. State the p-value. c. State alpha. d. What is your decision? e. Write a conclusion. f. Answer any other questions asked in the problem. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 571 106. A research group is studying a particular infectious disease. In 2011 at least 18 percent of nursing home residents had the disease. An Introduction to Statistics class in Daviess County, KY, conducted a hypothesis test at the nursing home (approximately 1,200 residents) to determine if the local nursing home's incidence was lower. One hundred fifty residents were chosen at random and surveyed. Of the 150 residents surveyed, 82 have the disease. Use a significance level of 0.05 and, using appropriate statistical evidence, conduct a hypothesis test and state the conclusions. 107. A recent survey in the New York Times Almanac indicated that 48.8 percent of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of 250 families and found that 142 owned some type of stock. At the 0.05 significance level, can the survey be considered to be accurate? 108. Driver error can be listed as the cause of approximately 54 percent of all fatal auto accidents, according to the American Automobile Association. Thirty randomly selected fatal accidents are examined, and it is determined that 14 were caused by driver error. Using α = 0.05, is the AAA proportion accurate? 109. The U.S. Department of Energy reported that 51.7 percent of homes were heated by natural gas. A random sample of 221 homes in Kentucky found that 115 were heated by natural gas. Does the evidence support the claim for Kentucky at the α = 0.05 level? Are the results applicable across the country? Why? 110. For Americans using library services, the American Library Association claims that at most 67 percent of patrons borrow books. The library director in Owensboro, KY, feels this is not true, so she asked a local college statistic class to conduct a survey. The class randomly selected 100 patrons and found that 82 borrowed books. Did the class demonstrate that the percentage was higher in Owensboro, KY? Use α = 0.01 level of |
significance. What is the possible proportion of patrons who do borrow books from the Owensboro Library? 111. The Weather Underground reported that the mean amount of summer rainfall for the northeastern United States is at least 11.52 inches. Ten cities in the northeast are randomly selected and the mean rainfall amount is calculated to be 7.42 inches with a standard deviation of 1.3 inches. At the α = 0.05 level, can it be concluded that the mean rainfall was below the reported average? What if α = 0.01? Assume the amount of summer rainfall follows a normal distribution. 112. A survey in the New York Times Almanac finds the mean commute time (one way) is 25.4 minutes for the 15 largest US cities. The Austin, TX, chamber of commerce feels that Austin’s commute time is less and wants to publicize this fact. The mean for 25 randomly selected commuters is 22.1 minutes with a standard deviation of 5.3 minutes. At the α = 0.10 level, is the Austin, TX, commute significantly less than the mean commute time for the 15 largest U.S. cities? 113. A report by the Gallup Poll found that a woman visits her doctor, on average, at most 5.8 times each year. A random sample of 20 women results in these yearly visit totals: 3; 2; 1; 3; 7; 2; 9; 4; 6; 6; 8; 0; 5; 6; 4; 2; 1; 3; 4; 1 At the α = 0.05 level, can it be concluded that the sample mean is higher than 5.8 visits per year? 114. According to the New York Times Almanac the mean family size in the United States is 3.18. A sample of a college math class resulted in the following family sizes: 5; 4; 5; 4; 4; 3; 6; 4; 3; 3; 5; 5; 6; 3; 3; 2; 7; 4; 5; 2; 2; 2; 3; 2 At α = 0.05, is the class’s mean family size greater than the national average? Does the Almanac result remain valid? Why? 115. The student academic group on a college campus claims that freshman students study at least 2.5 hours per day, on average. One Introduction to Statistics class was skeptical. The class took a random sample of 30 freshman students and found a mean study time of 137 minutes with a |
standard deviation of 45 minutes. At α = 0.01 level, is the student academic group’s claim correct? REFERENCES 9.1 Null and Alternative Hypotheses Centers for Disease Control and Prevention. (n.d.). Physical activity facts. Retrieved from http://www.cdc.gov/ healthyschools/physicalactivity/facts.htm National Institute of Mental Health. (n.d.). Publications about depression. Retrieved from http://www.nimh.nih.gov/ publicat/depression.cfm 572 Chapter 9 | Hypothesis Testing with One Sample 9.5 Additional Information and Full Hypothesis Test Examples American Automobile Association. (n.d.). Retrieved from www.aaa.com American Library Association. (n.d.). Retrieved from www.ala.org Amit Schitai. (n.d.). Data. Bureau of Labor Statistics. (n.d.). Occupational employment statistics. Retrieved from http://www.bls.gov/oes/current/ oes291111.htm Centers for Disease Control and Prevention. (n.d.). Retrieved from www.cdc.gov De Anza College. (2006). Foothill-De Anza Community College District. Retrieved from http://research.fhda.edu/factbook/ DAdemofs/Fact_sheet_da_2006w.pdf Federal Bureau of Investigation. (n.d.). Uniform crime reports and index of crime in Daviess in the state of Kentucky enforced by Daviess County from 1985 to 2005. Retrieved from http://www.disastercenter.com/kentucky/crime/3868.htm Gallup. (n.d.). Retrieved from www.gallup.com Johansen, C., Boice, Jr. J., McLaughlin, J., & Olsen, J. (2001, Feb. 7). Cellular telephones and cancer—a nationwide cohort study in Denmark. Journal of National Cancer Institute, 93(3), 203–7. Retrieved from http://www.ncbi.nlm.nih.gov/ pubmed/11158188 La Leche League International. (n.d.). Retrieved from http://www.lalecheleague.org/Law/BAFeb01.html Online Learning Consortium. (2005 Nov.). Growing by degrees: Online education in the United States, 2005. Newburyport, MA: Allen, I. E., & Seaman, J |
. Available at http://files.eric.ed.gov/fulltext/ED530062.pdf Toastmasters detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1 International. Retrieved (n.d.). from http://toastmasters.org/artisan/ U.S. Census Bureau. (n.d.). Language use. Retrieved from https://www.census.gov/topics/population/language-use.html U.S. Census Bureau. (n.d.). QuickFacts. Retrieved from https://www.census.gov/quickfacts/table/PST045216/00 U.S. Department of Energy. (n.d.). Retrieved from http://energy.gov Weather Underground. (n.d.). Retrieved from www.wunderground.com SOLUTIONS 1 The random variable is the mean Internet speed in megabits per second. 3 The random variable is the mean number of children an American family has. 5 The random variable is the proportion of people picked at random in Times Square visiting the city. 7 a. H0: p = 0.42 b. Ha: p < 0.42 9 a. H0: μ = 15 b. Ha: μ ≠ 15 11 Type I: The mean price of mid-sized cars is $32,000, but we conclude that it is not $32,000. Type II: The mean price of mid-sized cars is not $32,000, but we conclude that it is $32,000. 13 α = the probability that you think the bag cannot withstand –15 degrees F, when, in fact, it can. β = the probability that you think the bag can withstand –15 degrees F, when, in fact, it cannot. 15 Type I: The procedure will go well, but the doctors think it will not. Type II: The procedure will not go well, but the doctors think it will. 17 0.019 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 573 19 0.998 21 A normal distribution or a Student’s t-distribution 23 Use a Student’s t-distribution 25 a normal distribution for a single population mean 27 It must be approximately normally distributed. 29 They must both be greater than five. 31 bin |
omial distribution 33 The outcome of winning is very unlikely. 35 H0: μ > = 73 Ha: μ < 73 The p-value is almost zero, which means there is sufficient data to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches at the 5 percent level. The data do support the claim. 37 The shaded region shows a low p-value. 39 Do not reject H0. 41 means 43 the mean time spent on homework for 26 students 45 a. 3 b. 1.5 c. 1.8 d. 26 ¯ 47 X ~ N ⎛ ⎝2.5, 1.5 26 ⎞ ⎠ 49 This is a left-tailed test. 51 This is a two-tailed test. 53 Figure 9.23 55 a right-tailed test 57 a left-tailed test 59 This is a left-tailed test. 61 This is a two-tailed test. Chapter 9 | Hypothesis Testing with One Sample 574 62 a. H0: μ = 34; Ha: μ ≠ 34 b. H0: p ≤ 0.60; Ha: p > 0.60 c. H0: μ ≥ 100,000; Ha: μ < 100,000 d. H0: p = 0.29; Ha: p ≠ 0.29 e. H0: p = 0.05; Ha: p < 0.05 f. H0: μ ≤ 10; Ha: μ > 10 g. H0: p = 0.50; Ha: p ≠ 0.50 h. H0: μ = 6; Ha: μ ≠ 6 i. H0: p ≥ 0.11; Ha: p < 0.11 j. H0: μ ≤ 20,000; Ha: μ > 20,000 64 c 66 a. Type I error: We conclude that the mean is not 34 years, when it really is 34 years. Type II error: We conclude that the mean is 34 years, when in fact it really is not 34 years. b. Type I error: We conclude that more than 60 percent of Americans vote in presidential elections, when the actual percentage is at most 60 percent.Type II error: We conclude that at most 60 percent of Americans vote in presidential elections when, in fact, more than 60 percent do. c. Type I error: We conclude that the mean starting salary is less than $100,000 |
, when it really is at least $100,000. Type II error: We conclude that the mean starting salary is at least $100,000 when, in fact, it is less than $100,000. d. Type I error: We conclude that the proportion of high school seniors who take physical education daily is not 29%, when it really is 29%. Type II error: We conclude that the proportion of high school seniors who take physical education daily is 29% when, in fact, it is not 29%. e. Type I error: We conclude that fewer than 5 percent of adults ride the bus to work in Los Angeles, when the percentage that do is really 29%. Type II error: We conclude that 29%. or more adults ride the bus to work in Los Angeles when, in fact, fewer that 29% do. f. Type I error: We conclude that the mean number of cars a person owns in his or her lifetime is more than 10, when in reality it is not more than 10. Type II error: We conclude that the mean number of cars a person owns in his or her lifetime is not more than 10 when, in fact, it is more than 10. g. Type I error: We conclude that the proportion of Americans who prefer to live away from cities is not about half, though the actual proportion is about half. Type II error: We conclude that the proportion of Americans who prefer to live away from cities is half when, in fact, it is not half. h. Type I error: We conclude that the duration of paid vacations each year for Europeans is not six weeks, when in fact it is six weeks. Type II error: We conclude that the duration of paid vacations each year for Europeans is six weeks when, in fact, it is not. i. Type I error: We conclude that the proportion is less than 11 percent, when it is really at least 11 percent. Type II error: We conclude that the proportion of women who develop breast cancer is at least 11 percent, when in fact it is less than 11 percent. j. Type I error: We conclude that the average tuition cost at private universities is more than $20,000, though in reality it is at most $20,000. Type II error: We conclude that the average tuition cost at private universities is at most $20,000 when, in fact, it is more than $20,000. 68 b 70 d 72 d 74 a. H0: μ ≥ 50,000 This Open |
Stax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 575 b. Ha: μ < 50,000 c. Let X ¯ = the average lifespan of a brand of tires. d. normal distribution e. z = -2.315 f. p-value = 0.0103 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles. i. (43,537, 49,463) 75 a. H0: μ ≥ 35.5 b. Ha: μ < 35.5 c. Let x¯ = the average mpg for the sample of cars and trucks in the fleet d. normal distribution e. z = -0.648 f. p-value = 0.2578 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is sufficient evidence to support the claim that the manufacturer’s fleet meets the fuel economy standards in the 2016 policy. i. (31.88 mpg, 37.32 mpg) 76 a. H0: μ = $1.00 b. Ha: μ ≠ $1.00 c. Let x¯ = the average cost of a daily newspaper. d. normal distribution e. z = –0.866 f. p-value = 0.3865 g. Check student’s solution. h. i. Alpha: 0.01 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.01. iv. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1. i. ($0.84, $1.06) Chapter 9 | Hypothesis Testing with One Sample 576 78 a. H0: μ = 10 b. Ha: μ ≠ 10 c. Let X ¯ = the mean number of sick days an employee takes per year. d |
. Student’s t-distribution e. t = –1.12 f. p-value = 0.300 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the mean number of sick days is not 10. i. (4.9443, 11.806) 80 a. H0: p ≥ 0.6 b. Ha: p < 0.6 c. Let P′ = the proportion of students who feel more enriched as a result of taking elementary statistics. d. normal for a single proportion e. 1.12 f. p-value = 0.1308 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is insufficient evidence to conclude that less than 60 percent of her students feel more enriched. i. Confidence interval: (0.409, 0.654) The “plus-4s” confidence interval is (0.411, 0.648) 82 a. H0: μ = 4 b. Ha: μ ≠ 4 ¯ c. Let X the average I.Q. of a set of brown trout. d. e. two-tailed Student's t-test t = 1.95 f. p-value = 0.076 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 577 iv. Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four. i. (3.8865, 5.9468) 84 a. H0: p ≥ 0.13 b. Ha: p < 0.13 c. Let P′ = the proportion of Americans who have the disease d. normal for a single proportion e. –2.688 f. p-value = |
0.0036 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have been diagnosed with the disease is less than 13 percent. i. (0, 0.0623). The plus-4s confidence interval is (0.0022, 0.0978) 86 a. H0: μ ≥ 129 b. Ha: μ < 129 c. Let X ¯ = the average time in seconds that Terri finishes Lap 4. d. Student's t-distribution e. t = 1.209 f. 0.8792 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is insufficient evidence to conclude that Terri’s mean lap time is less than 129 seconds. i. (128.63, 130.37) 88 a. H0: p = 0.60 b. Ha: p < 0.60 c. Let P′ = the proportion of family members who shed tears at a reunion. d. normal for a single proportion e. –1.71 f. 0.0438 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. 578 Chapter 9 | Hypothesis Testing with One Sample iii. Reason for decision: p-value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of family members who shed tears at a reunion is less than 0.60. However, the test is weak because the p-value and alpha are quite close, so other tests should be done. i. We are 95 percent confident that between 38.29 percent and 61.71 percent of family members will shed tears at a family reunion. (0.3829, 0.6171). The plus-4s confidence interval (see chapter 8) is (0.3861, 0.6139) Note that here the large-sample 1 – PropZTest provides the approximate p-value of 0.0438. Whenever a p-value based on a normal approximation is close |
to the level of significance, the exact p-value based on binomial probabilities should be calculated whenever possible. This is beyond the scope of this course. 89 a. H0: μ ≥ 22 b. Ha: μ < 22 c. Let X ¯ = the mean number of bubbles per blow. d. Student's t-distribution e. –2.667 f. p-value = 0.00486 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: There is sufficient evidence to conclude that the mean number of bubbles per blow is less than 22. i. (18.501, 21.499) 91 a. H0: μ ≤ 1 b. Ha: μ > 1 c. Let X ¯ = the mean cost in dollars of macaroni and cheese in a certain town. d. Student's t-distribution e. t = 0.340 f. p-value = 0.36756 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05 iv. Conclusion: The mean cost could be $1, or less. At the 5 percent significance level, there is insufficient evidence to conclude that the mean price of a box of macaroni and cheese is more than $1. i. (0.8291, 1.241) 93 a. H0: p = 0.01 b. Ha: p > 0.01 c. Let P′ = the proportion of errors generated d. Normal for a single proportion This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 579 e. 2.13 f. 0.0165 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of errors generated is more than 0.01. i. Confidence interval: (0, 0.094). The plus |
-4s confidence interval is (0.004, 0.144). 95 a. H0: p = 0.50 b. Ha: p < 0.50 c. Let P′ = the proportion of friends that has a pierced ear. d. normal for a single proportion e. –1.70 f. p-value = 0.0448 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. (However, they are very close.) iv. Conclusion: There is sufficient evidence to support the claim that less than 50 percent of his friends have pierced ears. i. Confidence interval: (0.245, 0.515): The plus-4s confidence interval is (0.259, 0.519). 97 a. H0: p = 0.40 b. Ha: p < 0.40 c. Let P′ = the proportion of schoolmates who fear public speaking. d. normal for a single proportion e. –1.01 f. p-value = 0.1563 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. Conclusion: There is insufficient evidence to support the claim that less than 40 percent of students at the school fear public speaking. i. Confidence interval: (0.3241, 0.4240): The plus-4s confidence interval is (0.3257, 0.4250). 99 a. H0: p = 0.14 b. Ha: p < 0.14 580 Chapter 9 | Hypothesis Testing with One Sample c. Let P′ = the proportion of nursing home residents that have the disease. d. normal for a single proportion e. –0.2756 f. p-value = 0.3914 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject the null hypothesis. iii. Reason for decision: The p-value is greater than 0.05. iv. At the 5 percent significance level, there is insufficient evidence to conclude that the proportion of nursing home residents that have the disease is less than 0.14. i. Confidence interval: (0.050 |
2, 0.2070): The plus-4s confidence interval (see chapter 8) is (0.0676, 0.2297). 101 a. H0: μ = 69,110 b. Ha: μ > 69,110 c. Let X ¯ = the mean salary in dollars for California registered nurses. d. Student's t-distribution e. t = 1.719 f. p-value: 0.0466 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: The p-value is less than 0.05. iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110. i. ($68,757, $73,485) 103 a. H0: p ≥ 0.14, Ha: p < 0.14 b. p-value < 0.0002 c. Alpha: 0.05 d. Reject the null hypothesis. e. At the 5 percent significance level, there is sufficient evidence to conclude that the proportion of Harleys stolen is significantly less than their share of all motorcycles. (conclusion a) 105 c 107 a. H0: p = 0.488 Ha: p ≠ 0.488 b. p-value = 0.0114 c. alpha = 0.05 d. Reject the null hypothesis. e. At the 5 percent level of significance, there is enough evidence to conclude that 48.8 percent of families own stocks. f. The survey does not appear to be accurate. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 9 | Hypothesis Testing with One Sample 581 109 a. H0: p = 0.517 Ha: p ≠ 0.517 b. p-value = 0.9203. c. alpha = 0.05. d. Do not reject the null hypothesis. e. At the 5 percent significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is 0.517. f. However, we cannot generalize this result to the entire nation. First, the sample’s population is only the state of Kentucky. Second, it is reasonable to assume that homes in the extreme north and south will have extreme |
high usage and low usage, respectively. We would need to expand our sample base to include these possibilities if we wanted to generalize this claim to the entire nation. 111 a. H0: µ ≥ 11.52 Ha: µ < 11.52 b. p-value = 0.000002 which is almost 0. c. alpha = 0.05. d. Reject the null hypothesis. e. At the 5 percent significance level, there is enough evidence to conclude that the mean amount of summer rain in the northeaster US is less than 11.52 inches, on average. f. We would make the same conclusion if alpha was 1 percent because the p-value is almost 0. 113 a. H0: µ ≤ 5.8 Ha: µ > 5.8 b. p-value = 0.9987 c. alpha = 0.05 d. Do not reject the null hypothesis. e. At the 5 percent level of significance, there is not enough evidence to conclude that a woman visits her doctor, on average, more than 5.8 times a year. 115 a. H0: µ ≥ 150 Ha: µ < 150 b. p-value = 0.0622 c. alpha = 0.01 d. Do not reject the null hypothesis. e. At the 1 percent significance level, there is not enough evidence to conclude that freshmen students study less than 2.5 hours per day, on average. f. The student academic group’s claim appears to be correct. 582 Chapter 9 | Hypothesis Testing with One Sample This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 583 10 | HYPOTHESIS TESTING WITH TWO SAMPLES Figure 10.1 If you want to test a claim that involves two groups (the types of breakfasts eaten east and west of the Mississippi River), you can use a slightly different technique when conducting a hypothesis test. (credit: Chloe Lim) Introduction By the end of this chapter, the student should be able to do the following: Chapter Objectives • Classify hypothesis tests by type • Conduct and interpret hypothesis tests for two population means, population standard deviations known • Conduct and interpret hypothesis tests for two population means, population standard deviations unknown • Conduct and interpret hypothesis tests for two population proportions • Conduct and interpret hypothesis tests for matched or paired samples Studies often compare two groups. For |
example, researchers are interested in the effect aspirin has in preventing heart 584 Chapter 10 | Hypothesis Testing with Two Samples attacks. Over the last few years, newspapers and magazines have reported various aspirin studies involving two groups. Typically, one group is given aspirin and the other group is given a placebo. Then, the heart attack rate is studied over several years. There are other situations that deal with the comparison of two groups. For example, studies compare various diet and exercise programs. Politicians compare the proportion of individuals from different income brackets who might vote for them. Students are interested in whether the SAT or GRE preparatory courses really help raise their scores. You have learned to conduct hypothesis tests on single means and single proportions. You will expand upon that in this chapter. You will compare two means or two proportions to each other. The general procedure is the same, just expanded. To compare two means or two proportions, you work with two groups. The groups are classified as independent groups or matched pairs. Independent groups consist of two samples that are independent, that is, sample values selected from one population are not related in any way to sample values selected from the other population. Matched pairs consist of two samples that are dependent. The parameter tested using matched pairs is the population mean. The parameters tested using independent groups are either population means or population proportions. NOTE This chapter relies on either a calculator or a computer to calculate the degrees of freedom, the test statistics, and p values. TI-83+ and TI-84 instructions are included, as well as the test statistic formulas. When using a TI-83+ or TI-84 calculator, we do not need to separate two population means, independent groups, or population variances unknown into large and small sample sizes. However, most statistical computer software has the ability to differentiate these tests. This chapter deals with the following hypothesis tests: • Independent groups (samples are independent) ◦ Test of two population means ◦ Test of two population proportions • Matched or paired samples (samples are dependent) ◦ Test of the two population proportions by testing one population mean of differences 10.1 | Two Population Means with Unknown Standard Deviations 1. The two independent samples are simple random samples from two distinct populations. 2. For the two distinct populations ◦ ◦ if the sample sizes are small, the distributions are important (should be normal), and if the sample sizes are large, the distributions are not important (need not be normal) The test comparing two independent population means |
with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch. The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual ¯ samples. To account for the variation, we take the difference of the sample means, X error to standardize the difference. The result is a t-score test statistic. ¯ 1 − X 2, and divide by the standard Because we do not know the population standard deviations, we estimate them using the two sample standard deviations from our independent samples. For the hypothesis test, we calculate the estimated standard deviation, or standard error, ¯ of the difference in sample means, X ¯ 1 − X 2. The standard error is calculated as follows: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 585 (s1 )2 n1 (s2 )2 n2 + The test statistic (t-score) is calculated as follows: ( x¯ 1 – x¯ 2 ) – (μ1 – μ2 ) (s2 )2 n2 + (s1 )2 n1 where • s1 and s2, the sample standard deviations, are estimates of σ1 and σ2, respectively, • σ1 and σ1 are the unknown population standard deviations, • x¯ 1 and x¯ 2 are the sample means, and • μ1 and μ2 are the population means. The number of degrees of freedom (df) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student’s t-distribution with df as follows: Degrees of freedom d f = 2 ⎛ ⎜ ⎝ (s1)2 n1 (s2)2 n2 ⎞ ⎟ ⎠ + 2 ⎛ ⎝ 1 n1 – 1 ⎛ ⎞ ⎜ ⎠ ⎝ (s1)2 n1 ⎞ ⎟ ⎠ + ⎛ ⎝ 1 n2 – 1 ⎛ ⎞ |
⎜ ⎠ ⎝ (s2)2 n2 ⎞ ⎟ ⎠ 2 When both sample sizes n1 and n2 are five or larger, the Student’s t approximation is very good. Notice that the sample variances (s1)2 and (s2)2 are not pooled. (If the question comes up, do not pool the variances.) It is not necessary to compute this by hand. A calculator or computer easily computes it. Example 10.1 Independent groups The average amount of time boys and girls aged 7 to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in Table 10.1. Each populations has a normal distribution. Sample Size Average Number of Hours Playing Sports per Day Sample Standard Deviation Girls 9 Boys 16 Table 10.1 2 3.2 0.866 1.00 Is there a difference in the mean amount of time boys and girls aged 7 to 11 play sports each day? Test at the 5 percent level of significance. Solution 10.1 The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μg is the population mean for girls and μb is the population mean for boys. This is a test of two independent groups, two population means. 586 Chapter 10 | Hypothesis Testing with Two Samples b ¯ g − X = difference in the sample mean amount of time girls and boys play sports each ¯ Random variable: X day. H0: μg = μb Ha: μg ≠ μb The words the same tell you H0 has an "=". Since there are no other words to indicate Ha, assume it says is different. This is a two-tailed test. H0: μg – μb = 0 Ha: μg – μb ≠ 0 Distribution for the test: Use tdf where df is calculated using the df formula for independent groups, two population means. Using a calculator, df is approximately 18.8462. Do not pool the variances. Calculate the p-value using a Student’s t-distribution: p-value = 0.0054 Graph: Figure 10.2 sg = 0.866 sb = 1 So, x¯ Half the p-value is below –1.2, and half is above 1.2. = 2 – 3.2 = –1.2 |
g – x¯ b Make a decision: Since α > p-value, reject H0. This means you reject μg = μb. The means are different. Press STAT. Arrow over to TESTS and press 4:2-SampTTest. Arrow over to Stats and press ENTER. Arrow down and enter 2 for the first sample mean, 0.866 for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER. Arrow down to Pooled: and No. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is –3.14. Do the procedure again, but instead of Calculate do Draw. Conclusion—: At the 5 percent level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged 7 to 11 play sports per day is different (mean number of hours boys aged 7 to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged 7 to 11 play sports per day is greater than the mean number of hours played by boys). This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 587 10.1 Two samples are shown in Table 10.2. Both have normal distributions. The means for the two populations are thought to be the same. Is there a difference in the means? Test at the 5 percent level of significance. Sample Size Sample Mean Sample Standard Deviation Population A 25 Population B 16 Table 10.2 5 4.7 1 1.2 NOTE When the sum of the sample sizes is larger than 30 (n1 + n2 > 30), you can use the normal distribution to approximate the Student’s t. Example 10.2 A study is done by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is 4 math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of 1 math class. The community group believes that |
a student who graduates from College A has taken more math classes, on average. Both populations have a normal distribution. Test at a 1 percent significance level. Answer the following questions: a. Is this a test of two means or two proportions? Solution 10.2 a. two means b. Are the populations standard deviations known or unknown? Solution 10.2 b. unknown c. Which distribution do you use to perform the test? Solution 10.2 c. Student’s t d. What is the random variable? Solution 10.2 ¯ d. X ¯ A - X B e. What are the null and alternate hypotheses? Write the null and alternate hypotheses in symbols. 588 Chapter 10 | Hypothesis Testing with Two Samples Solution 10.2 e. Ho : μ A ≤ μB Ha : μ A > μB f. Is this test right-, left-, or two-tailed? Solution 10.2 f. Figure 10.3 right g. What is the p-value? Solution 10.2 g. 0.1928 h. Do you reject or not reject the null hypothesis? Solution 10.2 h. do not reject i. Conclusion: Solution 10.2 i. At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from College A has taken more math classes, on average, than a student who graduates from College B. 10.2 A study is done to determine if Company A retains its workers longer than Company B. Company A samples 15 workers, and their average time with the company is 5 years with a standard deviation of 1.2. Company B samples 20 workers, and their average time with the company is 4.5 years with a standard deviation of 0.8. The populations are normally distributed. a. Are the population standard deviations known? b. Conduct an appropriate hypothesis test. At the 5 percent significance level, what is your conclusion? This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 589 Example 10.3 A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to- |
face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in Table 10.3 and Table 10.4. 67.6 41.2 85.3 55.9 82.4 91.2 73.5 94.1 64.7 64.7 70.6 38.2 61.8 88.2 70.6 58.8 91.2 73.5 82.4 35.5 94.1 88.2 64.7 55.9 88.2 97.1 85.3 61.8 79.4 79.4 Table 10.3 Online Class 77.9 95.3 81.2 74.1 98.8 88.2 85.9 92.9 87.1 88.2 69.4 57.6 69.4 67.1 97.6 85.9 88.2 91.8 78.8 71.8 98.8 61.2 92.9 90.6 97.6 100 95.3 83.5 92.9 89.4 Table 10.4 Face-to-Face Class Is the mean of the final exam scores of the online class lower than the mean of the final exam scores of the faceto-face class? Test at a 5 percent significance level. Answer the following questions: a. Is this a test of two means or two proportions? b. Are the population standard deviations known or unknown? c. Which distribution do you use to perform the test? d. What is the random variable? e. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. f. Is this test right-, left-, or two-tailed? g. What is the p-value? h. Do you reject or not reject the null hypothesis? i. At the _____ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. (See the conclusion in Example 10.2, and write yours in a similar fashion.) First put the data for each group into two lists (such as L1 and L2). Press STAT. Arrow over to TESTS and press 4:2SampTTest. Make sure Data is highlighted and press ENTER. Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to μ1: and arrow to ≠ μ2 (does not equal). Press ENTER. Arrow down to Pooled: No. Press |
ENTER. Arrow down to Calculate and press ENTER. NOTE Be careful not to mix up the information for Group 1 and Group 2! 590 Chapter 10 | Hypothesis Testing with Two Samples Solution 10.3 two means a. b. unknown c. Student’s t ¯ d. X ¯ 1 – X 2 e. 1. H0: μ1 = μ2 Null hypothesis: The means of the final exam scores are equal for the online and face-to- face statistics classes. 2. Ha: μ1 < μ2 Alternative hypothesis: The mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class. f. left-tailed g. p-value = 0.0011 Figure 10.4 h. Reject the null hypothesis. i. The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class. At the 5 percent level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class. Cohen’s Standards for Small, Medium, and Large Effect Sizes Cohen’s d is a measure of effect size based on the differences between two means. Cohen’s d, named for U.S. statistician Jacob Cohen, measures the relative strength of the differences between the means of two populations based on sample data. The calculated value of effect size is then compared to Cohen’s standards of small, medium, and large effect sizes. Size of Effect d Small medium Large 0.2 0.5 0.8 Table 10.5 Cohen’s Standard Effect Sizes Cohen’s d is the measure of the difference between two means divided by the pooled standard deviation: d = x¯ 1 – x¯ s pooled 2 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 591 where s pooled = (n1 – 1)s1 2 2 + (n2 – 1)s2 n1 + n2 – 2. Example 10.4 Calculate Cohen’s d for Example 10.2. Is the size of the effect small, |
medium, or large? Explain what the size of the effect means for this problem. Solution 10.4 μ1 = 4 s1 = 1.5 n1 = 11 μ2 = 3.5 s2 = 1 n2 = 9 d = 0.384 The effect is small because 0.384 is between Cohen’s value of 0.2 for small effect size and 0.5 for medium effect size. The size of the differences of the means for the two colleges is small, indicating that there is not a significant difference between them. Example 10.5 Calculate Cohen’s d for Example 10.3. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem. Solution 10.5 d = 0.834; large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the final exam scores of online students and students in a face-to-face class is large, indicating a significant difference. 10.5 Weighted alpha is a measure of risk-adjusted performance of stocks over a period of a year. A high positive weighted alpha signifies a stock whose price has risen, while a small positive weighted alpha indicates an unchanged stock price during the time period. Weighted alpha is used to identify companies with strong upward or downward trends. The weighted alpha for the top 30 stocks of banks in the Northeast and in the West as identified by Nasdaq on May 24, 2013 are listed in Table 10.6 and Table 10.7, respectively. 94.2 75.2 69.6 52.0 48.0 41.9 36.4 33.4 31.5 27.6 77.3 71.9 67.5 50.6 46.2 38.4 35.2 33.0 28.7 26.5 76.3 71.7 56.3 48.7 43.2 37.6 33.7 31.8 28.5 26.0 Table 10.6 Northeast 126.0 70.6 65.2 51.4 45.5 37.0 33.0 29.6 23.7 22.6 116.1 70.6 58.2 51.2 43.2 36.0 31.4 28.7 23.5 21.6 78.2 68.2 55.6 50.3 39.0 34.1 31.0 25.3 23 |
.4 21.5 Table 10.7 West Is there a difference in the weighted alpha of the top 30 stocks of banks in the Northeast and in the West? Test at a 5 percent significance level. Answer the following questions: 592 Chapter 10 | Hypothesis Testing with Two Samples a. Is this a test of two means or two proportions? b. Are the population standard deviations known or unknown? c. Which distribution do you use to perform the test? d. What is the random variable? e. What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. f. Is this test right-, left-, or two-tailed? g. What is the p-value? h. Do you reject or not reject the null hypothesis? i. At the _____ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. j. Calculate Cohen’s d and interpret it. 10.2 | Two Population Means with Known Standard Deviations Even though this situation is not likely (knowing the population standard deviations), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the ¯ difference between the means is normal, and both populations must be normal. The random variable is X normal distribution has the following format: Normal distribution is ¯ 1 – X 2. The ¯ μ1 – μ2, ⎣ (σ1)2 n1 + (σ2)2 n2 ⎤ ⎥. ⎦ The standard deviation is (σ1)2 n1 (σ2)2 n2. + The test statistic (z-score) is z = Example 10.6 ( x¯ 1 – x¯ 2) – (μ1 – μ2) (σ2)2 n2 + (σ1)2 n1. Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax. Both populations have a normal distribution. The data are recorded in Table 10.8. Wax Sample Mean Number of Months Floor Wax Lasts Population Standard Deviation 1 2 3 2.9 Table 10.8 0.33 0.36 Does the data indicate that Wax 1 is more effective than Wax 2? Test at a 5 percent level of significance. This OpenStax book is available for free at http://cnx.org/content |
/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 593 Solution 10.6 This is a test of two independent groups, two population means, population standard deviations known. ¯ Random Variable: X ¯ 1 – X 2 = difference in the mean number of months the competing floor waxes last. H0: μ1 ≤ μ2 Ha: μ1 > μ2 The words is more effective says that Wax 1 lasts longer than Wax 2, on average. Longer is a > symbol and goes into Ha. Therefore, this is a right-tailed test. Distribution for the test: The population standard deviations are known, so the distribution is normal. Using the formula, the distribution is ¯ 0, 0.332 20 + 0.362 20 ⎞ ⎠. Since μ1 ≤ μ2, then μ1 – μ2 ≤ 0 and the mean for the normal distribution is zero. Calculate the p value using the normal distribution: p value = 0.1799 Graph: Figure 10..9 = 0.1 Compare α and the p value: α = 0.05 and p value = 0.1799. Therefore, α < p value. Make a decision: Since α < p value, do not reject H0. Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time Wax 1 lasts is longer (Wax 1 is more effective) than the mean time Wax 2 lasts. Press STAT. Arrow over to TESTS and press 3:2-SampZTest. Arrow over to Stats and press ENTER. Arrow down and enter.33 for sigma1,.36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to μ1: and arrow to > μ2. Press ENTER. Arrow down to Calculate and press ENTER. The p value is p = 0.1799, and the test statistic is 0.9157. Do the procedure again, but instead of Calculate do Draw. 594 Chapter 10 | Hypothesis Testing with Two Samples 10.6 The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines are randomly assigned to be tested. Both populations have normal distributions. Table 10.9 shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1 |
? Test at a 5 percent level of significance. Engine Sample Mean Number of RPM Population Standard Deviation 1 2 1,500 1,600 Table 10.9 50 60 Example 10.7 An interested citizen wanted to know if Democratic U.S. senators are older than Republican U.S. senators, on average. On May 26, 2013, the mean age of 30 randomly selected Republican senators was 61 years 247 days (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days (61.704 years) with a standard deviation of 9.55 years. Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5 percent level of significance. Solution 10.7 This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators; 2: Republican senators ¯ Random variable: X ¯ 1 – X 2 = difference in the mean age of Democratic and Republican U.S. senators. H0: µ1 ≤ µ2 H0: µ1 – µ2 ≤ 0 Ha: µ1 > µ2 Ha: µ1 – µ2 > 0 The words older than translates as a > symbol and goes into Ha. Therefore, this is a right-tailed test. Distribution for the test: The distribution is the normal approximation to the Student’s t for means, independent groups. Using the formula, the distribution is ¯ () X ¯ 1 – X 2 ∼ N[0, (9.55)2 30 + (10.17)2 30 ] Since µ1 ≤ µ2, µ1 – µ2 ≤ 0 and the mean for the normal distribution is zero. Calculating the p value using the normal distribution gives p value = 0.4040. Graph: This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 595 Figure 10.6 Compare α and the p value: α = 0.05 and p value = 0.4040. Therefore, α < p value. Make a decision: Since α < p value, do not reject H0. Conclusion: At the 5 percent level of |
significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators. 10.3 | Comparing Two Independent Population Proportions When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present: 1. The two independent samples are simple random samples that are independent. 2. The number of successes is at least five, and the number of failures is at least five, for each of the samples. 3. Growing literature states that the population must be at least 10 or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results. Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions. The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H0: pA = pB. To conduct the test, we use a pooled proportion, pc. The pooled proportion is calculated as follows: The distribution for the differences is pc = x A + xB n A + nB. P′ A − P′B ~ N[0, pc(1 − pc)( 1 n A + 1 nB )]. The test statistic (z-score) is z = Example 10.8 (p′ A − p′B) − (p A − pB) ) pc(1 − pc)( 1 n A + 1 nB. Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given Medication A still had hives 30 minutes 596 Chapter 10 | Hypothesis Testing with Two Samples after taking the medication. Twelve out of another random sample of 200 adults given Medication B still had hives 30 minutes after taking the medication. Test at a 1 percent level of significance. Solution 10.8 The problem asks for a difference in proportions, making it a test of two proportions. Let A and B be the subscripts for Medication A and Medication B, respectively. Then, pA and pB are the desired population proportions. Random Variable: P′A – P′B = |
difference in the proportions of adult patients who did not react after 30 minutes to Medication A and to Medication B. H0: pA = pB pA – pB = 0 Ha: pA ≠ pB pA – pB ≠ 0 The words is a difference tell you the test is two-tailed. Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal: pc = x A + xB n A + nB = 20 + 12 200 + 200 = 0.08 1 – pc = 0.92 P′ A – P′B ~ N ⎡ ⎣0, (0.08)(0.92)( 1 200 + 1 200 ⎤ ) ⎦ P′A – P′B follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.1404. Estimated proportion for group A: p′ A = Estimated proportion for group B: p′B = x A n A xB nB = 20 200 = 12 200 = 0.1 = 0.06 Graph: Figure 10.7 P′A – P′B = 0.1 – 0.06 = 0.04. Half the p-value is below –0.04, and half is above 0.04. Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α < p-value. Make a decision: Since α < p-value, do not reject H0. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 597 Conclusion: At a 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to Medication A and Medication B. Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1: and arrow to not equal p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1404, and the test statistic is 1.47. Do the procedure again, |
but instead of Calculate do Draw. 10.8 Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5 percent level of significance. Example 10.9 A research study was conducted about gender differences in texting. The researcher believed that the proportion of girls involved in texting is less than the proportion of boys involved. The data collected in spring 2010 among a random sample of middle and high school students in a large school district in the southern United States is summarized in Table 10.9. Is the proportion of girls sending texts less than the proportion of boys texting? Test at a 1 percent level of significance. Males Females Sent texts 183 Total number surveyed 2231 156 2169 Table 10.10 Solution 10.9 This is a test of two population proportions. Let M and F be the subscripts for males and females. Then, pM and pF are the desired population proportions. Random variable: p′F − p′M = difference in the proportions of males and females who sent texts. H0: pF = pM H0: pF – pM = 0 Ha: pF < pM Ha: pF – pM < 0 The words less than tell you the test is left-tailed. Distribution for the test: Since this is a test of two population proportions, the distribution is normal: 598 Chapter 10 | Hypothesis Testing with Two Samples = 156 + 183 2169 + 2231 = 0.077 pc = xF + x M nF + n M 1 − pc = 0.923 Therefore, ⎛ ⎝0, + 1 1 2169 p′F – p′ M ∼ N ⎛ (0.077)(0.923) ⎝ ⎞ ⎞ ⎠ ⎠ 2231 p′F – p′M follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.1045 Estimated proportion for females: 0.0719 Estimated proportion for males: 0.082 Graph: Figure 10.8 Decision: Since α < p-value, do not reject H0. Conclusion: At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that |
the proportion of girls sending texts is less than the proportion of boys sending texts. Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1045 and the test statistic is z = –1.256. Example 10.10 Researchers conducted a study of smartphone use (Phone A versus Phone B) among adults. A cell phone company claimed that Phone B smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5 percent own Phone B. Of the 1,343 white cell phone owners randomly sampled, 10 percent own Phone B. Test at the 5 percent level of significance. Is the proportion of white Phone B owners greater than the proportion of African American Phone B owners? Solution 10.10 This is a test of two population proportions. Let W and A be the subscripts for the whites and African Americans. Then, pW and pA are the desired population proportions. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 599 Random variable: p′W – p′A = difference in the proportions of Phone A and Phone B users. H0: pW = pA H0: pW – pA = 0 Ha: pW > pA Ha: pW – pA > 0 The words more popular indicate that the test is right-tailed. Distribution for the test: The distribution is approximately normal. pc = xW + x A nW + n A = 134 + 12 1343 + 232 = 0.0927 1 − pc = 0.9073 Therefore, p′W – p′ A ∽ N ⎛ ⎝0, ⎛ (0.0927)(0.9073) ⎝ 1 1343 + 1 232 ⎞ ⎞ ⎠ ⎠ p′W – p′ A follows an approximate normal distribution. Calculate the p-value using the normal distribution: p-value = 0.0077 Estimated proportion |
for group A: 0.10 Estimated proportion for group B: 0.05 Graph: Figure 10.9 Decision: Since α > p-value, reject the H0. Conclusion: At the 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of white cell phone owners use Phone B than African Americans. TI-83+ and TI-84: Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 135 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to greater than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.0092, and the test statistic is z = 2.33. 600 Chapter 10 | Hypothesis Testing with Two Samples 10.10 A group of citizens wanted to know if the proportion of homeowners in their small city was different in 2011 than in 2010. Their research showed that of the 113,231 available homes in their city in 2010, 7,622 of them were owned by the families who live there. In 2011, 7,439 of the 104,873 of the available homes were owned by city residents. Test at a 5 percent significance level. Answer the following questions: a. Is this a test of two means or two proportions? b. Which distribution do you use to perform the test? c. What is the random variable? d. What are the null and alternative hypotheses? Write the null and alternative hypotheses in symbols. e. Is this test right-, left-, or two-tailed? f. What is the p-value? g. Do you reject or not reject the null hypothesis? h. At the ______ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. 10.4 | Matched or Paired Samples (Optional) When using a hypothesis test for matched or paired samples, the following characteristics should be present: 1. Simple random sampling is used. 2. Sample sizes are often small. 3. Two measurements (samples) are drawn from the same pair of individuals or objects. 4. Differences are calculated from the matched or paired samples. 5. The differences form the sample that is used for the hypothesis test. 6. Either the matched pairs have differences that come from a population that is normal or the number |
of differences is sufficiently large so that distribution of the sample mean of differences is approximately normal. In a hypothesis test for matched or paired samples, subjects are matched in pairs and differences are calculated. The differences are the data. The population mean for the differences, μd, is then tested using a Student’s-t test for a single population mean with n – 1 degrees of freedom, where n is the number of differences. The test statistic (t-score) is x¯ t = d − μd sd ⎞ ⎛ ⎠ ⎝ n. Example 10.11 A study was conducted to investigate the effectiveness of pain-reducing medication. Results for randomly selected subjects are shown in Table 10.10. A lower score indicates less pain. The before value is matched to an after value, and the differences are calculated. The differences have a normal distribution. Are the sensory measurements, on average, lower after the medication? Test at a 5 percent significance level. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 601 Subject: A B C D E F G H Before 6.6 6.5 9.0 10.3 11.3 8.1 6.3 11.6 After 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2.0 Table 10.11 Solution 10.11 Corresponding before and after values form matched pairs. (Calculate after – before.) After Data Before Data Difference 6.8 2.4 7.4 8.5 8.1 6.1 3.4 2 Table 10.12 6.6 6.5 9 10.3 11.3 8.1 6.3 11.6 0.2 –4.1 –1.6 –1.8 –3.2 –2 –2.9 –9.6 The data for the test are the differences: {0.2, –4.1, –1.6, –1.8, –3.2, –2, –2.9, –9.6} The sample mean and sample standard deviation of the differences are: xd = –3.13 and sd = 2.91 Verify these values. Let μd be the population mean for the differences. We use the subscript d to denote differences. ¯ Random variable: X d = |
the mean difference of the sensory measurements. H0: μd ≥ 0 The null hypothesis is zero or positive, meaning that there is the same or more pain felt after taking the medication. That means the subject shows no improvement. μd is the population mean of the differences. Ha: μd < 0 The alternative hypothesis is negative, meaning there is less pain felt after taking the medication. That means the subject shows improvement. The score should be lower after taking the medication, so the difference ought to be negative to indicate improvement. Distribution for the test: The distribution is a Student’s t with df = n – 1 = 8 – 1 = 7. Use t7. Note —that the test is for a single population mean. Calculate the p-value using the Student’s-t distribution: p-value = 0.0095 Graph: 602 Chapter 10 | Hypothesis Testing with Two Samples Figure 10.10 ¯ X d is the random variable for the differences. The sample mean and sample standard deviation of the differences are as follows: x¯ s¯ d d = –3.13 = 2.91 Compare α and the p-value: α = 0.05 and p-value = 0.0095. α > p-value. Make a decision: Since α > p-value, reject H0. This means that μd < 0 and there is improvement. Conclusion: At a 5 percent level of significance, from the sample data, there is sufficient evidence to conclude that the sensory measurements, on average, are lower after taking the medication. The medication appears to be effective in reducing pain. NOTE For the TI-83+ and TI-84 calculators, you can either calculate the differences ahead of time (after before) and put the differences into a list or you can put the after data into a first list and the before data into a second list. Then, go to a third list and arrow up to the name. Enter 1st list name – 2nd list name. The calculator will do the subtraction, and you will have the differences in the third list. Use your list of differences as the data. Press STAT and arrow over to TESTS. Press 2:T-Test. Arrow over to Data and press ENTER. Arrow down and enter 0 for μ0, the name of the list where you put the data, and 1 for Freq:. Arrow down to μ: and arrow over to < μ0. Press ENTER. Arrow |
down to Calculate and press ENTER. The p-value is 0.0094, and the test statistic is –3.04. Do these instructions again except, arrow to Draw instead of Calculate. Press ENTER. 10.11 A study was conducted to investigate how effective a new diet was in lowering cholesterol. Results for the randomly selected subjects are shown in the table. The differences have a normal distribution. Are the subjects’ cholesterol levels lower on average after the diet? Test at the 5 percent level. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 603 Subject A B C D E F G H I Before 209 210 205 198 216 217 238 240 222 After 199 207 189 209 217 202 211 223 201 Table 10.13 Example 10.12 A college football coach was interested in whether the college’s strength development class increased his players’ maximum lift (in pounds) on the bench press exercise. He asked four of his players to participate in a study. The amount of weight they could each lift was recorded before they took the strength development class. After completing the class, the amount of weight they could each lift was again measured. The data are as follows: Weight (in pounds) Player 1 Player 2 Player 3 Player 4 Amount of weight lifted prior to the class 205 Amount of weight lifted after the class 295 241 252 338 330 368 360 Table 10.14 The coach wants to know if the strength development class makes his players stronger, on average. Record the differences data. Calculate the differences by subtracting the amount of weight lifted prior to the class from the weight lifted after completing the class. The data for the differences are: {90, 11, -8, -8}. Assume the differences have a normal distribution. Using the differences data, calculate the sample mean and the sample standard deviation. x¯ d = 21.3, sd = 46.7 NOTE The data given here would indicate that the distribution is right-skewed. The difference 90 may be an extreme outlier. It is pulling the sample mean to be 21.3 (positive). The means of the other three data values are negative. Using the difference data, this becomes a test of a single __________. ¯ Define the random variable: X d is the mean difference in the maximum lift per player. The distribution for the hypothesis test is t3. H0: |
μd ≤ 0, Ha: μd > 0 Graph: 604 Chapter 10 | Hypothesis Testing with Two Samples Figure 10.11 Calculate the p-value: The p-value is 0.2150. Decision: If the level of significance is 5 percent, the decision is not to reject the null hypothesis, because α < p-value. What is the conclusion? At a 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the strength development class helped make the players stronger, on average. 10.12 A new prep class was designed to improve SAT test scores. Five students were selected at random. Their scores on two practice exams were recorded, one before the class and one after. The data are recorded in Table 10.15. Are the scores, on average, higher after the class? Test at a 5 percent level. SAT Scores Student 1 Student 2 Student 3 Student 4 Score before class 1840 Score after class 1920 1960 2160 1920 2200 2150 2100 Table 10.15 Example 10.13 Seven eighth-graders at Kennedy Middle School measured how far they could push the shot put with their dominant (writing) hand and their weaker (nonwriting) hand. They thought that they could push equal distances with both hands. The data are collected and recorded in Table 10.16. Distance (in feet) using Student 1 Student 2 Student 3 Student 4 Student 5 Student 6 Student 7 Dominant Hand Weaker Hand 30 28 26 14 34 27 17 18 19 17 26 26 20 16 Table 10.16 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 605 Conduct a hypothesis test to determine whether the mean difference in distances between the children’s dominant versus weaker hands is significant. Record the differences data. Calculate the differences by subtracting the distances with the weaker hand from the distances with the dominant hand. The data for the differences are: {2, 12, 7, –1, 2, 0, 4}. The differences have a normal distribution. Using the differences data, calculate the sample mean and the sample standard deviation. x¯ = 3.71, sd d = 4.5. ¯ Random variable: X d = mean difference in the distances between the hands. Distribution for the hypothesis test: t6 H0: μd = 0 Ha: μd ≠ 0 Graph: Figure 10. |
12 Calculate the p-value: The p-value is 0.0716 (using the data directly). Test statistic = 2.18. p-value = 0.0719 using ⎛ ⎝ x¯ d = 3.71, sd = 4.5 ⎞ ⎠. Decision: Assume α = 0.05. Since α < p-value, do not reject H0. Conclusion: At the 5 percent level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the children’s weaker and dominant hands to push the shot put. 10.13 Five ball players think they can throw the same distance with their dominant hand (throwing) and off-hand (catching hand). The data were collected and recorded in Table 10.17. Conduct a hypothesis test to determine whether the mean 5 difference in distances between the dominant and off-hand is significant. Test at the 5 percent level. Player 1 Player 2 Player 3 Player 4 Player 5 Dominant Hand 120 Off-Hand 105 111 109 135 98 140 111 125 99 Table 10.17 606 Chapter 10 | Hypothesis Testing with Two Samples 10.5 | Hypothesis Testing for Two Means and Two Proportions This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 607 10.1 Hypothesis Testing for Two Means and Two Proportions Student Learning Outcomes • The student will select the appropriate distributions to use in each case. • The student will conduct hypothesis tests and interpret the results. Supplies: • The business section from two consecutive days’ newspapers • Three small packages of multicolored chocolates • Five small packages of peanut butter candies Increasing Stocks Survey Look at yesterday’s newspaper business section. Conduct a hypothesis test to determine if the proportion of New York Stock Exchange (NYSE) stocks that increased is greater than the proportion of NASDAQ stocks that increased. As randomly as possible, choose 40 NYSE stocks and 32 NASDAQ stocks and complete the following statements. 1. H0: _________ 2. Ha: _________ 3. In words, define the random variable. 4. The distribution to use for the test is _____________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of |
significance. a. Graph: Figure 10.13 b. Calculate the p value. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. 608 Chapter 10 | Hypothesis Testing with Two Samples Decreasing Stocks Survey Randomly pick eight stocks from the newspaper. Using two consecutive days’ business sections, test whether the stocks went down, on average, for the second day. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. The distribution to use for the test is _____________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph Figure 10.14 b. Calculate the p value: 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Candy Survey Buy three small packages of multicolored chocolates and five small packages of peanut butter candies (same net weight as the multicolored chocolates). Test whether the mean number of candy pieces per package is the same for the two brands. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. What distribution should be used for this test? 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 609 Figure 10.15 b. Calculate the p value. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. Shoe Survey Test whether women have, on average, more pairs of shoes than men. Include all forms of sneakers, shoes, sandals, and boots. Use your class as the sample. 1. H0: ________ 2. Ha: ________ 3. In words, define the random variable. 4. The distribution to use for the test is ________________. 5. Calculate the test statistic using your data. 6. Draw a graph and label it appropriately. Shade the actual level of significance. a. Graph Figure 10.16 b. Calculate the p value |
. 7. Do you reject or not reject the null hypothesis? Why? 8. Write a clear conclusion using a complete sentence. 610 Chapter 10 | Hypothesis Testing with Two Samples KEY TERMS degrees of freedom (df) the number of objects in a sample that are free to vary pooled proportion estimate of the common value of p1 and p2 standard deviation a number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation variable (random variable) a characteristic of interest in a population being studied. Common notation for variables are uppercase Latin letters X, Y, Z,... Common notation for a specific value from the domain (set of all possible values of a variable) are lowercase Latin letters x, y, z,.... For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3,.... Variables in statistics differ from variables in intermediate algebra in two ways: • The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color, then the domain is {black, blond, gray, green, orange}. • We can tell what specific value x of the random variable X takes only after performing the experiment. CHAPTER REVIEW 10.1 Two Population Means with Unknown Standard Deviations Two population means from independent samples where the population standard deviations are not known ¯ • Random variable: X ¯ 1 − X 2 = the difference of the sampling means • Distribution: Student’s t-distribution with degrees of freedom (variances not pooled) 10.2 Two Population Means with Known Standard Deviations A hypothesis test of two population means from independent samples where the population standard deviations are known (typically approximated with the sample standard deviations) will have these characteristics: ¯ • Random variable: X ¯ 1 − X 2 = the difference of the means • Distribution: normal distribution 10.3 Comparing Two Independent Population Proportions Test of two population proportions from independent samples • Random variable: p^ A – p^ B = difference between the two estimated proportions • Distribution: normal distribution 10.4 Matched or Paired Samples (Optional) A hypothesis test for matched or paired samples (t-test) has these characteristics: • Test the differences by subtracting one measurement from the other measurement • Random variable: x¯ |
d = mean of the differences. • Distribution: Student’s t distribution with n – 1 degrees of freedom. • If the number of differences is small (less than 30), the differences must follow a normal distribution. • Two samples are drawn from the same set of objects. • Samples are dependent. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 611 FORMULA REVIEW 10.1 Two Population Means with Unknown Standard Deviations Standard error: SE = (s1)2 n1 (s2)2 n2 + Test statistic (t-score): t = ( x¯ Degrees of freedom: 1 − x¯ 2) − (μ1 − μ2) (s2)2 n2 + (s1)2 n1 ⎛ ⎜ ⎝ (s1)2 n1 + 2 (s2)2 n2 ⎞ ⎟ ⎠ 2 ⎛ ⎝ 1 n1 − 1 ⎛ ⎞ ⎜ ⎠ ⎝ (s1)2 n1 ⎞ ⎟ ⎠ + ⎛ ⎝ 1 n2 − 1 ⎛ ⎞ ⎜ ⎠ ⎝ (s2)2 n2 ⎞ ⎟ ⎠ 2 d f = where: Generally, µ1 - µ2 = 0. where σ1 and σ2 are the known population standard deviations, n1 and n2 are the sample sizes, x¯ are the sample means, and μ1 and μ2 are the population means. and x¯ 1 2 10.3 Comparing Two Independent Population Proportions Pooled proportion: pc = xF + x M nF + n M Distribution for the differences: ⎡ ⎛ ⎣0, pc(1 − pc) ⎝ p′ A − p′B ∼ N 1 n A + 1 nB ⎤ ⎞ ⎦ ⎠ where the null hypothesis is H0: pA = pB = 0 or H0: pA – pB s1 and s2 are the sample standard deviations, and n1 and n2 are the sample sizes. Test statistic (z-score): z = (p′ A − p′ B |
) ⎛ pc(1 − pc) ⎝ 1 n A + 1 nB ⎞ ⎠ x¯ 1 and x¯ 2 are the sample means. Cohen’s d is the measure of effect size: d = 1 − x¯ x¯ s pooled 2 where s pooled = (n1 − 1)s1 2 2 + (n2 − 1)s2 n1 + n2 − 2. 10.2 Two Population Means with Known Standard Deviations Normal distributionμ1 − μ2, ⎣ (σ1)2 n1 + (σ2)2 n2 ⎤ ⎥. ⎦ Generally, µ1 – µ2 = 0. Test statistic (z-score): where the null hypothesis is H0: pA = pB = 0 or H0: pA − pB and where p′A and p′B are the sample proportions, pA and pB are the population proportions, Pc is the pooled proportion, and nA and nB are the sample sizes. 10.4 Matched or Paired Samples (Optional) Test statistic (t-score): t = x¯ d − μd sd ⎞ ⎛ ⎠ ⎝ n where: d is the mean of the sample differences, μd is the mean x¯ of the population differences, sd is the sample standard deviation of the differences, and n is the sample size. z = ( x¯ 1 − x¯ 2) − (μ1 − μ2) (σ2)2 n2 + (σ1)2 n1 PRACTICE 10.1 Two Population Means with Unknown Standard Deviations Use the following information to answer the next 15 exercises. Indicate if the hypothesis test is for a. independent group means, population standard deviations, and/or variances known, 612 Chapter 10 | Hypothesis Testing with Two Samples b. independent group means, population standard deviations, and/or variances unknown, c. matched or paired samples, d. e. f. single mean, two proportions, or single proportion. 1. It is believed that 70 percent of males pass their drivers test in the first attempt, while 65 percent of females pass the test in the first attempt. Of interest is whether the proportions are equal. 2. A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. |
A study is done to test this. 3. A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted. 4. The known standard deviation in salary for all mid-level professionals in the financial industry is $11,000. Company A and Company B are in the financial industry. Suppose samples are taken of mid-level professionals from Company A and from Company B. The sample mean salary for mid-level professionals in Company A is $80,000. The sample mean salary for mid-level professionals in Company B is $96,000. Company A and Company B management want to know if their midlevel professionals are paid differently, on average. 5. The average worker in Germany gets eight weeks of paid vacation. 6. According to a television commercial, 80% of dentists agree that a brand of fluoridated toothpaste is the best on the market. 7. It is believed that the average grade on an English essay in a particular school system is higher for females than for males. A random sample of 31 females had a mean score of 82 with a standard deviation of 3, and a random sample of 25 males had a mean score of 76 with a standard deviation of 4. 8. The league mean batting average is 0.280 with a known standard deviation of 0.06. The Rattlers and the Vikings belong to the league. The mean batting average for a sample of eight Rattlers is 0.210, and the mean batting average for a sample of eight Vikings is 0.260. There are 24 players on the Rattlers and 19 players on the Vikings. Are the batting averages of the Rattlers and Vikings statistically different? 9. In a random sample of 100 forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico? 10. A new medicine is said to help improve sleep. Eight subjects are picked at random and given the medicine. The mean hours slept for each person were recorded before starting the medication and after. 11. It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and |
a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6. 12. Varsity athletes practice five times a week, on average. 13. A sample of 12 in-state graduate school programs at School A has a mean tuition of $64,000 with a standard deviation of $8,000. At School B, a sample of 16 in-state graduate programs has a mean tuition of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions different? 14. A new WiFi range booster is being offered to consumers. A researcher tests the native range of 12 different routers under the same conditions. The ranges are recorded. Then, the researcher uses the new WiFi range booster and records the new ranges. Does the new WiFi range booster do a better job? 15. A high school principal claims that 30 percent of student athletes drive themselves to school, while 4 percent of nonathletes drive themselves to school. In a sample of 20 student athletes, 45 percent drive themselves to school. In a sample of 35 nonathlete students, 6 percent drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes? Use the following information to answer the next three exercises: A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in Beverage B is 38 grams with This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 613 a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal distributions. 16. Are standard deviations known or unknown? 17. What is the random variable? 18. Is this a one-tailed or two-tailed test? Use the following information to answer the next 12 exercises. The U.S. Centers for Disease Control reports that the mean life expectancy was 47.6 years for whites born in 1900 and 33.0 years for nonwhites. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the |
124 whites, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 nonwhites, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for whites and nonwhites. 19. Is this a test of means or proportions? 20. State the null and alternative hypotheses. a. H0: __________ b. Ha: __________ 21. Is this a right-tailed, left-tailed, or two-tailed test? 22. In symbols, what is the random variable of interest for this test? 23. In words, define the random variable of interest for this test. 24. Which distribution (normal or Student’s t) would you use for this hypothesis test? 25. Explain why you chose the distribution you did for Exercise 10.24. 26. Calculate the test statistic and p-value. 27. Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value. 28. Find the p-value. 29. At a preconceived α = 0.05, write the following: a. Your decision: b. The reason for your decision: c. Your conclusion (write out in a complete sentence): 30. Does it appear that the means are the same? Why or why not? 10.2 Two Population Means with Known Standard Deviations Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. Table 10.18 shows the result. Scouters believe that Rodriguez pitches a speedier fastball. Pitcher Sample Mean Speed of Pitches (mph) Population Standard Deviation 3 7 Wesley 86 Rodriguez 91 Table 10.18 31. What is the random variable? 32. State the null and alternative hypotheses. 33. What is the test statistic? 34. What is the p value? 35. At the 1 percent significance level, what is your conclusion? Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of 614 Chapter 10 | Hypothesis Testing with Two |
Samples the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. Plant Group Sample Mean Height of Plants (inches) Population Standard Deviation Food No food Table 10.19 16 14 2.5 1.5 36. Is the population standard deviation known or unknown? 37. State the null and alternative hypotheses. 38. What is the p value? 39. Draw the graph of the p value. 40. At the 1 percent significance level, what is your conclusion? Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. Fifteen pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. Sample Mean Melting Temperatures (°F) Population Standard Deviation Alloy Gamma 800 Alloy Zeta 900 Table 10.20 95 105 41. State the null and alternative hypotheses. 42. Is this a right-, left-, or two-tailed test? 43. What is the p value? 44. Draw the graph of the p value. 45. At the 1 percent significance level, what is your conclusion? 10.3 Comparing Two Independent Population Proportions Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1. 46. Is this a test of means or proportions? 47. What is the random variable? 48. State the null and alternative hypotheses. 49. What is the p-value? 50. What can you conclude about the two operating systems? Use the following information to answer the next 12 exercises. In the recent U.S. Census, 3 percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only 9 people reported being of two |
or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 615 a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota. 51. Is this a test of means or proportions? 52. State the null and alternative hypotheses. a. H0: _________ b. Ha: _________ 53. Is this a right-tailed, left-tailed, or two-tailed test? How do you know? 54. What is the random variable of interest for this test? 55. In words, define the random variable for this test. 56. Which distribution (normal or Student’s t) would you use for this hypothesis test? 57. Explain why you chose the distribution you did for the Exercise 10.56. 58. Calculate the test statistic. 59. Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value. Figure 10.17 60. Find the p-value. 61. At a preconceived α = 0.05, write the following: a. Your decision: b. The reason for your decision: c. Your conclusion (write out in a complete sentence): 62. Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not? 10.4 Matched or Paired Samples (Optional) Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table 10.21. The before value is matched to an after value, and the differences are calculated. The differences have a normal distribution. Test at the 1 percent significance level. Installation A B C D E F G H Before After Table 10.21 63. What is the random variable? 64. State the null and alternative hypotheses. 65. What is the p-value? 66. Draw the graph of the p-value. 67. What conclusion can you draw about the software patch? Use the following information to answer next five exercises. A |
study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal 616 Chapter 10 | Hypothesis Testing with Two Samples distribution. Test at the 1 percent significance level. Subject A B C D E F Before After Table 10.22 68. State the null and alternative hypotheses. 69. What is the p-value? 70. What is the sample mean difference? 71. Draw the graph of the p-value. 72. What conclusion can you draw about the juggling class? Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1 percent significance level. Patient A B C D E F Before 161 162 165 162 166 171 After 158 159 166 160 167 169 Table 10.23 73. State the null and alternative hypotheses. 74. What is the test statistic? 75. What is the p-value? 76. What is the sample mean difference? 77. What is the conclusion? HOMEWORK 10.1 Two Population Means with Unknown Standard Deviations DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the.doc or the.pdf files. NOTE If you are using a Student’s t-distribution for a homework problem in what follows, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption.) This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 617 78. The mean number of English courses taken in a two-year period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 29 males and 16 females. The males took an average of 3 English |
courses with a standard deviation of 0.8. The females took an average of 4 English courses with a standard deviation of 1.0. Are the means statistically the same? 79. A student at a four-year college claims that mean enrollment at four-year colleges is higher than at two-year colleges in the United States. Two surveys are conducted. Of the 35 two-year colleges surveyed, the mean enrollment was 5,068 with a standard deviation of 4,777. Of the 35 four-year colleges surveyed, the mean enrollment was 5,466 with a standard deviation of 8,191. 80. At Rachel’s eleventh birthday party, eight girls were timed to see how long (in seconds) they could sit perfectly still in a relaxed position. After a two-minute rest, they timed themselves while jumping. The girls thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis. Relaxed time (seconds) Jumping time (seconds) 26 47 30 22 23 45 37 29 Table 10.24 21 40 28 21 25 43 35 32 81. Mean entry-level salaries for college graduates with mechanical engineering degrees and electrical engineering degrees are believed to be approximately the same. A recruiting office thinks that the mean mechanical engineering salary is lower than the mean electrical engineering salary. The recruiting office randomly surveys 50 entry-level mechanical engineers and 60 entry-level electrical engineers. Their mean salaries were $46,100 and $46,700, respectively. Their standard deviations were $3,450 and $4,210, respectively. Conduct a hypothesis test to determine if you agree that the mean entry-level mechanical engineering salary is lower than the mean entry-level electrical engineering salary. 82. Marketing companies have collected data implying that teenage girls use more ringtones on their smartphones than teenage boys do. In one study of 40 randomly chosen teenage girls and boys (20 of each) with smartphones, the mean number of ringtones for the girls was 3.2 with a standard deviation of 1.5. The mean for the boys was 1.7 with a standard deviation of 0.8. Conduct a hypothesis test to determine if the means are approximately the same or if the girls’ mean is higher than the boys’ mean. Use the information from Appendix C to answer the next four exercises. 83. Using the data from Lap 1 only, conduct a hypothesis test to determine if the mean time for completing a lap in races is the same as it is in practices. 84. Repeat |
the test in Exercise 10.83, but use Lap 5 data this time. 85. Repeat the test in Exercise 10.83, but this time combine the data from Laps 1 and 5. 86. In two to three complete sentences, explain in detail how you might use Terri Vogel’s data to answer the following question: Does Terri Vogel drive faster in races than she does in practices? Use the following information to answer the next two exercises. The Eastern and Western Major League Soccer conferences have a new Reserve Division that allows new players to develop their skills. Data for a randomly picked date showed the following annual goals. 618 Chapter 10 | Hypothesis Testing with Two Samples Western Eastern Los Angeles 9 D United 9 FC Dallas 3 Chicago 8 Chivas USA 4 Columbus 7 Real Salt Lake 3 New England 6 Colorado 4 MetroStars 5 San Jose 4 Kansas City 3 Table 10.25 Conduct a hypothesis test to answer the next two exercises. 87. The exact distribution for the hypothesis test is a. b. c. d. the normal distribution the Student’s t-distribution the uniform distribution the exponential distribution 88. If the level of significance is 0.05, the conclusion is: a. There is sufficient evidence to conclude that the W Division teams score fewer goals, on average, than the E teams. b. There is insufficient evidence to conclude that the W Division teams score more goals, on average, than the E teams. c. There is insufficient evidence to conclude that the W teams score fewer goals, on average, than the E teams. d. There is not sufficient evidence to determine a conclusion. 89. Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The day subscript refers to the statistics day students. The night subscript refers to the statistics night students. Which of the following is a concluding statement: a. There is sufficient evidence to conclude that statistics night students’ mean on Exam 2 is better than the statistics day students’ mean on Exam 2. b. There is insufficient evidence to conclude that the statistics day students’ mean on Exam 2 is better than the statistics night students’ mean on Exam 2. c. There |
is insufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. d. There is sufficient evidence to conclude that there is a significant difference between the means of the statistics day students and night students on Exam 2. 90. Researchers interviewed people in a certain industry in Canada and the United States. The mean age of the 100 Canadians upon entering this industry was 18 with a standard deviation of 6. The mean age of the 130 Americans upon entering this industry was 20 with a standard deviation of 8. Is the mean age of entering this industry in Canada lower than the mean age in the United States? Test at a 1 percent significance level. 91. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. The powder diet group had a mean weight loss of 42 pounds with a standard deviation of 12 pounds. The liquid diet group had a mean weight loss of 45 pounds with a standard deviation of 14 pounds. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 619 92. Suppose a statistics instructor believes that there is no significant difference between the mean class scores of statistics day students on Exam 2 and statistics night students on Exam 2. She takes random samples from each of the populations. The mean and standard deviation for 35 statistics day students were 75.86 and 16.91, respectively. The mean and standard deviation for 37 statistics night students were 75.41 and 19.73. The day subscript refers to the statistics day students. The night subscript refers to the statistics night students. An appropriate alternative hypothesis for the hypothesis test is a. μday > μnight b. μday < μnight c. μday = μnight d. μday ≠ μnight 10.2 Two Population Means with Known Standard Deviations DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the.doc or the.pdf files. NOTE If you are using a Student’s t-distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. |
(When using these tests in a real situation, you must first prove that assumption.) 93. A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3. 94. Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is known that the population standard deviation for each group is $180. Determine whether you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls. 95. A group of transfer-bound students wondered if they will spend the same mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were $947 and $1,011, respectively. The population standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to determine if the means are statistically the same. 96. Some manufacturers claim that nonhybrid sedan cars have a lower mean miles per gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of 7 mpg. Thirty-one nonhybrid sedans get a mean of 22 mpg with a standard deviation of 4 mpg. Suppose that the population standard deviations are known to be 6 and 3, respectively. Conduct a hypothesis test to evaluate the manufacturers’ claim. 97. A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, |
and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test. 620 Chapter 10 | Hypothesis Testing with Two Samples 98. One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from 1 (strongly agree) to 5 (strongly disagree). Table 10.26 contains 10 of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife). Wife’s Score Husband’s Score Table 10.26 10.3 Comparing Two Independent Population Proportions DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the.doc or the.pdf files. NOTE If you are using a Student’s t-distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption.) 99. A recent drug survey showed an increase in the use of prescription medication among local senior citizens as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of prescription medication use is higher locally or nationally. Locally, 65 senior citizens reported taking prescription medication within the past month, while 60 national seniors reported using them. larger + smaller dimension larger dimension 100. Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ⎛ ⎝ to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation?, was |
the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 ⎞ ⎠ 101. A year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic students out of a total of 2,441 students. In general, do you think that the percent of Hispanic students at the two colleges is basically the same or different? Use the following information to answer the next three exercises. Neuroinvasive West Nile virus is a severe disease that affects a person’s nervous system. It is spread by the Culex species of mosquito. In the United States in 2010, there were 629 reported cases of neuroinvasive West Nile virus out of a total of 1,021 reported cases, and there were 486 neuroinvasive reported cases out of a total of 712 cases reported in 2011. Is the 2011 proportion of neuroinvasive West Nile virus cases more than the 2010 proportion of neuroinvasive West Nile virus cases? Using a 1 percent level of significance, conduct an appropriate hypothesis test. • 2011 subscript: 2011 group. • 2010 subscript: 2010 group 102. This is a. a test of two proportions b. a test of two independent means c. a test of a single mean d. a test of matched pairs. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 621 103. An appropriate null hypothesis is a. p2011 ≤ p2010 b. p2011 ≥ p2010 c. μ2011 ≤ μ2010 d. p2011 > p2010 104. The p-value is 0.0022. At a 1 percent level of significance, what is the appropriate conclusion? a. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus. b. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus. c. There is insufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuro |
invasive West Nile virus is less than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus. d. There is sufficient evidence to conclude that the proportion of people in the United States in 2011 who contracted neuroinvasive West Nile virus is more than the proportion of people in the United States in 2010 who contracted neuroinvasive West Nile virus. 105. Researchers conducted a study to find out if there is a difference in the use of e-readers by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7 percent of the 628 surveyed use e-readers, while 11 percent of the 2,309 participants 30 years old and older use e-readers. 106. Adults aged 18 years and older were randomly selected for a survey about a specific disease. The researchers wanted to determine if the proportion of women who have the disease is less than the proportion of southern men who do. The results are shown in Table 10.27. Test at the 1 percent level of significance. Number diagnosed with disease Sample size Men 42,769 Women 67,169 Table 10.27 155,525 248,775 107. Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than of people age 30 and older. Table 10.28 details the number of tablet owners for each age group. Test at the 1 percent level of significance. 16–29 year olds 30 years and older Own a Tablet 69 Sample Size 628 Table 10.28 231 2,309 108. A group of friends debated whether more men use smartphones than women. They consulted a research study of smartphone use among adults. The results of the survey indicate that of the 973 men randomly sampled, 379 use smartphones. For women, 404 of the 1,304 who were randomly sampled use smartphones. Test at the 5 percent level of significance. 109. While her husband spent 2.5 hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who do. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey. 622 Chapter 10 | Hypothesis Testing with Two Samples 110. We are interested in whether children’s educational computer software costs less, on average, than children� |
�s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of $4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of $10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software. 111. A researcher recently claimed that the proportion of college-age males who wear at least one piece of jewelery is as high as the proportion of college-age females. She conducted a survey in her classes. Out of 107 males, 20 wear at least one piece of jewelery. Out of 92 females, 47 wear at least one piece of jewelery. Do you believe that the proportion of males has reached the proportion of females? 112. Use the data sets found in Appendix C to answer this exercise. Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of practice laps she completes slower than 135 seconds? 113. To Breakfast or Not to Breakfast? by Richard Ayore In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what he or she has achieved for the past year and also focuses ahead for more to come. If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine. Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!” And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we |
worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in Table 10.29, solve our problem. Work hours with breakfast Work hours without breakfast 8 7 9 5 9 8 10 Table 10.29 10.4 Matched or Paired Samples (Optional) DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E. Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the.doc or the.pdf files. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 623 NOTE If you are using a Student’s t-distribution for the homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption.) 114. Ten individuals went on a low-fat diet for 12 weeks to lower their cholesterol. The data are recorded in Table 10.30. Do you think that their cholesterol levels were significantly lowered? Starting cholesterol level Ending cholesterol level 140 220 110 240 200 180 190 360 280 260 Table 10.30 140 230 120 220 190 150 200 300 300 240 Use the following information to answer the next two exercises. A new preventative medication was tried on a group of 224 patients who had the same risk factors for a disease. 45 patients developed the disease after four years. In a control group of 224 patients, 68 developed the disease after four years. We want to test whether the method of treatment reduces the proportion of patients who develop the disease after four years. Let the subscript t = treated patient and ut = untreated patient. 115. The appropriate hypotheses are a. H0: pt < put and Ha: pt ≥ put b. H0: pt ≤ put and Ha: pt > put c. H0: pt = put and Ha: pt ≠ put d. H0: pt = put and Ha: pt < put 116. If the p-value is 0.0062, what is the conclusion? Use α = 0.05. a. The method has no effect. b. There is sufficient evidence to conclude that the method reduces the proportion of patients who develop the |
disease after four years. c. There is sufficient evidence to conclude that the method increases the proportion of patients who develop the disease after four years. d. There is insufficient evidence to conclude that the method reduces the proportion of patients who develop the disease after four years. Use the following information to answer the next two exercises. An experiment is conducted to show that blood pressure can be consciously reduced in people trained in a biofeedback exercise program. Six subjects were randomly selected, and blood pressure measurements were recorded before and after the training. The difference between blood pressures was calculated (after – before), producing the following results: x¯ pressure has decreased after the training. = −10.2 sd = 8.4. Using the data, test the hypothesis that the blood d 624 Chapter 10 | Hypothesis Testing with Two Samples 117. The distribution for the test is t5 t6 a. b. c. N(−10.2, 8.4) d. N(−10.2, 8.4 6 ) 118. If α = 0.05, the p-value and the conclusion are a. 0.0014; There is sufficient evidence to conclude that the blood pressure decreased after the training. b. 0.0014; There is sufficient evidence to conclude that the blood pressure increased after the training. c. 0.0155; There is sufficient evidence to conclude that the blood pressure decreased after the training. d. 0.0155; There is sufficient evidence to conclude that the blood pressure increased after the training. 119. A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and then after having taken her class. She conducts a hypothesis test. The data are as follows. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 Mean score after class 80 78 80 93 86 87 86 Table 10.31 The correct decision is reject H0. a. b. do not reject H0. 120. A local research group is studying a chronic disease. They believe the number of cases of the disease is higher in 2013 than in 2012 in the southern United States. The group compared the estimates of new cases by southern state in 2012 and 2013. The results are in Table 10.32. Southern States 2012 2013 Alabama Arkansas Florida Georgia Kentucky Louisiana 3,450 3,720 2,150 2,280 15,540 15,710 6,970 |
7,310 3,160 3,300 3,320 3,630 Mississippi 1,990 2,080 North Carolina 7,090 7,430 Oklahoma 2,630 2,690 South Carolina 3,570 3,580 Tennessee 4,680 5,070 Texas Virginia Table 10.32 15,050 14,980 6,190 6,280 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 625 121. A traveler wanted to know if the prices of hotels are different in the 10 cities that he visits the most often. The list of the cities with the corresponding prices for his two favorite hotel chains is in Table 10.33. Test at the 1 percent level of significance. Hyatt Regency prices in dollars Hilton prices in dollars Cities Atlanta Boston Chicago Dallas Denver Indianapolis Los Angeles New York City Philadelphia 107 358 209 209 167 179 179 625 179 Washington, DC 245 Table 10.33 169 289 299 198 169 214 169 459 159 239 122. A politician asked his staff to determine whether the underemployment rate in the Northeast decreased from 2011 to 2012. The results are in Table 10.34. Northeastern States 2011 2012 Connecticut Delaware Maine Maryland Massachusetts New Hampshire New Jersey New York Ohio Pennsylvania Rhode Island Vermont West Virginia Table 10.34 17.3 17.4 19.3 16.0 17.6 15.4 19.2 18.5 18.2 16.5 20.7 14.7 15.5 16.4 13.7 16.1 15.5 18.2 13.5 18.7 18.7 18.8 16.9 22.4 12.3 17.3 BRINGING IT TOGETHER: HOMEWORK Use the following information to answer the next 10 exercises. Indicate which of the following choices best identifies the hypothesis test. 626 Chapter 10 | Hypothesis Testing with Two Samples A. Independent group means, population standard deviations and/or variances known B. Independent group means, population standard deviations and/or variances unknown C. Matched or paired samples D. Single mean E. Two proportions F. Single proportion 123. A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. The population standard deviations are two pounds and three pounds, respectively. Of interest is whether the liquid diet yields a higher mean weight loss than the powder diet. 124. A new |
chocolate bar is taste-tested on consumers. Of interest is whether the proportion of children who like the new chocolate bar is greater than the proportion of adults who like it. 125. The mean number of English courses taken in a two-year time period by male and female college students is believed to be about the same. An experiment is conducted and data are collected from 9 males and 16 females. 126. A football league reported that the mean number of touchdowns per game was five. A study is done to determine if the mean number of touchdowns has decreased. 127. A study is done to determine if students in the California state university system take longer to graduate than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. From years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. 128. According to a doctor’s magazine, 75 percent of senior citizens think that yearly checkups are very important. A study is done to verify this. 129. According to a recent study, U.S. companies have a mean maternity leave of six weeks. 130. A recent survey showed an increase in use of prescription medication among local senior citizens as compared to the national percent. Suppose that a survey of 100 local senior citizens and 100 national senior citizens is conducted to see if the proportion of prescription medication use is higher locally than nationally. 131. A new SAT study course is tested on 12 individuals. Pre-course and post-course scores are recorded. Of interest is the mean increase in SAT scores. The following data are collected: Pre-course score Post-course score 1 960 1010 840 1100 1250 860 1330 790 990 1110 740 Table 10.35 300 920 1100 880 1070 1320 860 1370 770 1040 1200 850 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 627 132. According to a statistics college professor, 68 percent of his students pass the final exam. A graduate researcher designs a study to determine if this claim is true. 133. Lesley E. Tan investigated the relationship between left-handedness versus right-handedness and motor competence in preschool children. Random samples of 41 left-handed preschool children and 41 right-handed preschool children were given several tests of motor skills to determine if there is evidence of a difference between the |
children based on this experiment. The experiment produced the means and standard deviations shown in Table 10.36. Determine the appropriate test and best distribution to use for that test. Left-handed Right-handed Sample size Sample mean 41 97.5 Sample standard deviation 17.5 41 98.1 19.2 Table 10.36 a. Two independent means, normal distribution b. Two independent means, Student’s t-distribution c. Matched or paired samples, Student’s t-distribution d. Two population proportions, normal distribution 134. A golf instructor is interested in determining if her new technique for improving players’ golf scores is effective. She takes four new students. She records their 18-hole scores before learning the technique and after having taken her class. She conducts a hypothesis test. The data are shown in Table 10.37. Player 1 Player 2 Player 3 Player 4 Mean score before class 83 Mean score after class 80 78 80 93 86 87 86 Table 10.37 This is a. a test of two independent means. b. a test of two proportions. c. a test of a single mean. d. a test of a single proportion. REFERENCES 10.1 Two Population Means with Unknown Standard Deviations Baseball-Almanac. (2013). World series history. Retrieved from http://www.baseball-almanac.com/ws/wsmenu.shtml Graduating Engineer + Computer Careers. (n.d.). Retrieved from http://www.graduatingengineer.com Microsoft Bookshelf. (n.d.). Nasdaq. (n.d.). Sectoring by industry groups. Retrieved from http://www.nasdaq.com/markets/barchart-sectors.aspx Prostitution Research and Education. (2013). Strip clubs: Where prostitution and trafficking happen. Retrieved from www.prostitutionresearch.com/ProsViolPosttrauStress.html U.S. Senate. (n.d.). Retrieved from www.senate.gov Wikipedia. List_of_current_United_States_Senators_by_age (n.d.). List of current United States Senators by age. Retrieved from http://en.wikipedia.org/wiki/ 628 Chapter 10 | Hypothesis Testing with Two Samples 10.2 Two Population Means with Known Standard Deviations Centers for Disease Control and Prevention. (2008, July 18). State-specific prevalence of obesity among adults—United States, 2007. |
MMWR, 57(28), 765–768. Retrieved from http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm Federal Bureau of Investigation. (n.d.). Texas Crime Rates 1960–1012. Available at http://www.disastercenter.com/crime/ txcrime.htm Hinduja, S. http://cyberbullying.us/blog/sexting-research-and-gender-differences/ (2013). Sexting Research and Gender Differences. Cyberbulling Research Center. Retrieved from Humes, K. R., Jones, N. A., & Ramirez, R. R. (2011 March). Overview of race and Hispanic origin: 2010 (2010 Census Briefs). Washington, DC: U.S. Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/ c2010br-02.pdf Smith, A. http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (2011, July 11). 35% of American adults own a smartphone. Pew Internet. Available online at Visually. (2013). Smart phone users, by the numbers. Retrieved from http://visual.ly/smart-phone-users-numbers 10.3 Comparing Two Independent Population Proportions American Cancer Society. (n.d.). Retrieved from http://www.cancer.org/index Centers for Disease Control and Prevention. (n.d.). West Nile virus. Retrieved from http://www.cdc.gov/ncidod/dvbid/ westnile/index.htm Chancellor’s Office, California Community Colleges. (1994, Nov.). Educational Resources. (n.d.). Gallup. (2013). State of the states. Retrieved from http://www.gallup.com/poll/125066/State-States.aspx?ref=interactive Hilton Hotels. (n.d.). Retrieved from http://www.hilton.com Hyatt Hotels. Retrieved from http://hyatt.com San Jose Museum of Art. (n.d.). Whitney exhibit (on loan). U.S. Department of Health and Human Services. (n.d). Statistics. Retrieved from https://www.hhs.gov/ SOLUTIONS 1 two proportions 3 matched or paired samples 5 single mean |
7 independent group means, population standard deviations and/or variances unknown 9 two proportions 11 independent group means, population standard deviations and/or variances unknown 13 independent group means, population standard deviations and/or variances unknown 15 two proportions 17 The random variable is the difference between the mean amounts of sugar in the two soft drinks. 19 means 21 two-tailed 23 the difference between the mean life spans of whites and nonwhites 25 This is a comparison of two population means with unknown population standard deviations. 27 Check student’s solution. This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 629 29 a. Reject the null hypothesis. b. p-value < 0.05 c. There is not enough evidence at the 5 percent level of significance to support the claim that life expectancy in the 1900s is different between whites and nonwhites. 31 the difference in mean speeds of the fastball pitches of the two pitchers 33 –2.46 35 At the 1 percent significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s. 37 Subscripts: 1 = Food, 2 = No Food H0: μ1 ≤ μ2 Ha: μ1 > μ2 39 Figure 10.18 41 Subscripts: 1 = Gamma, 2 = Zeta H0: μ1 = μ2 Ha: μ1 ≠ μ2 43 0.0062 45 There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma. 47 P′OS1 – P′OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS1 and OS2. 49 0.1018 51 proportions 53 right-tailed 55 The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota. 57 Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test. 59 Check student’s solution. 61 a. Reject the null hypothesis. b. p-value < alpha c. At the 5 percent significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more |
races in Nevada is statistically higher than that in North Dakota. 630 Chapter 10 | Hypothesis Testing with Two Samples 63 the mean difference of the system failures 65 0.0067 67 With a p-value 0.0067, we can reject the null hypothesis. There is enough evidence to support that the software patch is effective in reducing the number of system failures. 69 0.0021 71 Figure 10.19 73 H0: μd ≥ 0 Ha: μd < 0 75 0.0699 77 We decline to reject the null hypothesis. There is not sufficient evidence to support that the medication is effective. 79 Subscripts: 1: two-year colleges, 2: four-year colleges a. H0: μ1 ≥ μ2 b. Ha: μ1 < μ2 ¯ c. X ¯ 1 – X 2 d. Student’s t is the difference between the mean enrollments of the two-year colleges and the four-year colleges. e. test statistic: -0.2480 f. p-value: 0.4019 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Do not reject. iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges. 81 Subscripts: 1: mechanical engineering, 2: electrical engineering a. H0: µ1 ≥ µ2 b. Ha: µ1 < µ2 ¯ c. X ¯ 1 − X 2 is the difference between the mean entry-level salaries of mechanical engineers and electrical engineers. d. e. t108 test statistic: t = –0.82 f. p-value: 0.2061 g. Check student’s solution. h. i. Alpha: 0.05 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 631 ii. Decision: Do not reject the null hypothesis. iii. Reason for Decision: p-value > alpha iv. Conclusion: At the 5 percent significance level, there is insufficient evidence to conclude that the mean entry- level salaries of mechanical engineers is lower than that of electrical engineers. 83 a. H0: µ1 = µ2 b. Ha: µ1 ≠ µ2 ¯ c |
. X ¯ 1 − X 2 is the difference between the mean times for completing a lap in races and in practices. d. e. t20.32 test statistic: –4.70 f. p-value: 0.0001 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. 85 a. H0: µ1 = µ2 b. Ha: µ1 ≠ µ2 c. is the difference between the mean times for completing a lap in races and in practices. d. e. t40.94 test statistic: –5.08 f. p-value: zero g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p-value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. 88 c ¯ 90 Test: two independent sample means, population standard deviations unknown. Random variable: X Distribution: H0: μ1 = μ2, Ha: μ1 < μ2 The mean age of entering the industry in Canada is lower than the mean age in the United States. ¯ 1 − X 2 632 Chapter 10 | Hypothesis Testing with Two Samples Figure 10.20 Graph: left-tailed p-value : 0.0151 Decision: Do not reject H0. Conclusion: At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of entering the industry in Canada is lower than the mean age in the United States. 92 d 94 Subscripts: 1 = boys, 2 = girls a. H0: µ1 ≤ µ2 b. Ha: µ1 > µ2 c. The random variable is the difference in the mean auto insurance costs for boys and girls. d. normal e. test statistic: z = 2.50 f. p value: 0.0062 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for Decision: p value < alpha |
iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls. 96 Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans a. H0: µ1 ≥ µ2 b. Ha: µ1 < µ2 c. The random variable is the difference in the mean miles per gallon of nonhybrid sedans and hybrid sedans. d. normal e. test statistic: 6.36 f. p-value: 0 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p value < alpha iv. Conclusion: At the 5 percent significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans. 98 a. H0: µd = 0 b. Ha: µd < 0 This OpenStax book is available for free at http://cnx.org/content/col30309/1.8 Chapter 10 | Hypothesis Testing with Two Samples 633 c. The random variable Xd is the average difference between husband’s and wife’s satisfaction level. d. t9 test statistic: t = –1.86 e. f. p value: 0.0479 g. Check student’s solution h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis, but run another test. iii. Reason for Decision: p value < alpha iv. Conclusion: This is a weak test because alpha and the p value are close. However, there is insufficient evidence to conclude that the mean difference is negative. 101 Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College a. H0: p1 = p2 b. Ha: p1 ≠ p2 c. The random variable is the difference between the proportions of Hispanic students at Cabrillo College and Lake Tahoe College. d. normal for two proportions e. test statistic: 4.29 f. p-value: 0.00002 g. Check student’s solution. h. i. Alpha: 0.05 ii. Decision: Reject the null hypothesis. iii. Reason for decision: p-value < alpha iv. Conclusion: There is sufficient evidence to conclude that the proportions |
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