Full Hypothesis Test Examples Tests on Means 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. Set up the Hypothesis Test: Since the problem is about a mean, this is a test of a single population mean . H 0 : μ = 16.43 H a : μ < 16.43 For Jeffrey to swim faster, his time will be less than 16.43 seconds. The "<" tells you this is left-tailed. 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) X ¯ ~ N ( μ , σ X n ) Therefore, X ¯ ~ N ( 16.43 , 0.8 15 ) μ = 16.43 comes from H 0 and not the data. σ = 0.8, and n = 15. Calculate the p -value using the normal distribution for a mean: p -value = P ( x ¯ < 16) = 0.0187 where the sample mean in the problem is given as 16. p -value = 0.0187 (This is called the actual level of significance .) The p -value is the area to the left of the sample mean is given as 16. Graph: μ = 16.43 comes from H 0 . Our assumption is μ = 16.43. Interpretation of the p -value: If H 0 is true , there is a 0.0187 probability (1.87%)that Jeffrey's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% 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 Make a decision: Since α > p -value, reject H 0 . This indicates that you reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds. Conclusion: At the 5% significance level, there is sufficient evidence that Jeffrey's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Thus, based on the sample data, we conclude that Jeffrey swims faster using the new goggles. 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 at least 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 free-style, on average, in less than 16.43 seconds. (Do not reject the null hypothesis when the null hypothesis is false.) Try It The mean throwing distance of a football for Marco, a high school 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. Since the problem is about a mean, this is a test of a single population mean. H 0 : μ = 40 H a : μ > 40 p = 0.0062 Because p < α , we reject the null hypothesis. There is sufficient evidence to suggest that the change in grip improved Marco’s throwing distance. Jasmine has just begun her new job on the sales force of a very competitive company. In a sample of 16 sales calls it was found that she closed the contract for an average value of 108 dollars with a standard deviation of 12 dollars. Company policy requires that new members of the sales force must exceed an average of $100 per contract during the trial employment period. Can we conclude that Jasmine has met this requirement at the confidence level of 95%? H 0 : µ ≤ 100 H a : µ > 100 The null and alternative hypothesis are for the parameter µ because the number of dollars of the contracts is a continuous random variable. Also, this is a one-tailed test because the company has only an interested if the number of dollars per contact is below a particular number not "too high" a number. This can be thought of as making a claim that the requirement is being met and thus the claim is in the alternative hypothesis. Test statistic: t c = x ¯ − µ 0 s n = 108 − 100 ( 12 16 ) = 2.67 Critical value: t a = 1.753 with n-1 degrees of freedom= 15 The test statistic is a Student's t because the sample size is below 30; therefore, we cannot use the normal distribution. Comparing the calculated value of the test statistic and the critical value of t ( t a ) at a 5% significance level, we see that the calculated value is in the tail of the distribution. Thus, we conclude that 108 dollars per contract is significantly larger than the hypothesized value of 100 and thus we cannot accept the null hypothesis. There is evidence that supports Jasmine's performance meets company standards. Try It 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 is recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, state your conclusion, and identify the Type I errors. H 0 : μ = 5 H a : μ < 5 p = 0.0082 Because p < α , we reject the null hypothesis. There is sufficient evidence to suggest that the stock price of the company grows at a rate less than $5 a week. Type I Error: To conclude that the stock price is growing slower than $5 a week when, in fact, the stock price is growing at $5 a week (reject the null hypothesis when the null hypothesis is true). Type II Error: To conclude that the stock price is growing at a rate of $5 a week when, in fact, the stock price is growing slower than $5 a week (do not reject the null hypothesis when the null hypothesis is false). A manufacturer of salad dressings uses machines to dispense liquid ingredients into bottles that move along a filling line. The machine that dispenses salad dressings is working properly when 8 ounces are dispensed. Suppose that the average amount dispensed in a particular sample of 35 bottles is 7.91 ounces with a variance of 0.03 ounces squared, s 2 . Is there evidence that the machine should be stopped and production wait for repairs? The lost production from a shutdown is potentially so great that management feels that the level of confidence in the analysis should be 99%. Again we will follow the steps in our analysis of this problem. STEP 1 : Set the Null and Alternative Hypothesis. The random variable is the quantity of fluid placed in the bottles. This is a continuous random variable and the parameter we are interested in is the mean. Our hypothesis therefore is about the mean. In this case we are concerned that the machine is not filling properly. From what we are told it does not matter if the machine is over-filling or under-filling, both seem to be an equally bad error. This tells us that this is a two-tailed test: if the machine is malfunctioning it will be shutdown regardless if it is from over-filling or under-filling. The null and alternative hypotheses are thus: H 0 : μ = 8 H a : μ ≠ 8 STEP 2 : Decide the level of confidence and draw the graph showing the critical value. This problem has already set the level of confidence at 99%. The decision seems an appropriate one and shows the thought process when setting the confidence level at 1%. Management wants to be very certain, as certain as probability will allow, that they are not shutting down a machine that is not in need of repair. To draw the distribution and the critical value, we need to know which distribution to use. Because this is a continuous random variable and we are interested in the mean, and the sample size is greater than 30, the appropriate distribution is the normal distribution and the relevant critical value is 2.575 from the normal table. We draw the graph and mark these points. STEP 3 : Calculate sample parameters and the test statistic. The sample parameters are provided, the sample mean is 7.91 and the sample variance is .03 and the sample size is 35. We need to note that the sample variance was provided not the sample standard deviation, which is what we need for the formula. Remembering that the standard deviation is simply the square root of the variance, we therefore know the sample standard deviation, s, is 0.173. With this information we calculate the test statistic as -3.07, and mark it on the graph. Z c = x ¯ - μ 0 s n = 7.91 - 8 .173 35 = -3.07 STEP 4 : Compare test statistic and the critical values Now we compare the test statistic and the critical value by placing the test statistic on the graph. We see that the test statistic is in the tail, decidedly greater than the critical value of 2.575. We note that even the very small difference between the hypothesized value and the sample value is still a large number of standard deviations. The sample mean is only 0.08 ounces different from the required level of 8 ounces, but it is 3 plus standard deviations away and thus we cannot accept the null hypothesis. STEP 5 : Reach a Conclusion Three standard deviations of a test statistic will guarantee that the test will fail. The probability that anything is within three standard deviations is almost zero. Actually it is 0.0026 on the normal distribution, which is certainly almost zero in a practical sense. Our formal conclusion would be “ At a 99% level of confidence we cannot accept the hypothesis that the sample mean came from a distribution with a mean of 8 ounces” Or less formally, and getting to the point, “At a 99% level of confidence we conclude that the machine is under filling the bottles and is in need of repair”. Try It A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2). Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475 . Set up the Hypothesis Test: H 0 : μ = 475 H 0 : μ > 475 This is a right-tailed test. The random variable X is the mean time of employees working in a day. Distribution for the test: σ is known. X ~ N 475 , 45 20 x ¯ = 475 . 5 minutes (from the data). σ = 45 minutes. We assume μ = 475 minutes unless our data shows otherwise. The p -value is calculated: p -value = P ( x ¯ > 475 . 5 ) = 0 . 5044 Interpretation of p -value: If H 0 is true, then there is a 0.5044, or 50.44%, chance that the mean time of employees working is 475.5 minutes or more. Since a 50.44% chance is large enough, the mean time of 475.5 is not a rare event. Compare α and p values: α = 0 . 025 and p = 0 . 5044 Make a decision: Since α < p value, do not reject H 0 . Conclusion: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean time of employees is more than 475 minutes. Hypothesis Test for Proportions Just as there were confidence intervals for proportions, or more formally, the population parameter p of the binomial distribution, there is the ability to test hypotheses concerning p . The population parameter for the binomial is p . The estimated value (point estimate) for p is p′ where p′ = x/n , x is the number of successes in the sample and n is the sample size. When you perform a hypothesis test of a population proportion p , you take a simple random sample from the population. The conditions for a binomial distribution must be met, which are: there are a certain number n of independent trials meaning random sampling, the outcomes of any trial are binary, success or failure, and each trial has the same probability of a success p . The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np′ and nq′ must both be greater than five ( np′ > 5 and nq′ > 5). In this case the binomial distribution of a sample (estimated) proportion can be approximated by the normal distribution with μ = np and σ = np . Remember that q = 1 – p . There is no distribution that can correct for this small sample bias and thus if these conditions are not met we simply cannot test the hypothesis with the data available at that time. We met this condition when we first were estimating confidence intervals for p . Again, we begin with the standardizing formula modified because this is the distribution of a binomial. Z = p' - p pq n Substituting p 0 , the hypothesized value of p , we have: Z c = p' - p 0 p 0 q 0 n This is the test statistic for testing hypothesized values of p , where the null and alternative hypotheses take one of the following forms: Two-tailed test One-tailed test One-tailed test H 0 : p = p 0 H 0 : p ≤ p 0 H 0 : p ≥ p 0 H a : p ≠ p 0 H a : p > p 0 H a : p < p 0 The decision rule stated above applies here also: if the calculated value of Z c shows that the sample proportion is "too many" standard deviations from the hypothesized proportion, the null hypothesis cannot be accepted. The decision as to what is "too many" is pre-determined by the analyst depending on the level of significance required in the test. The mortgage department of a large bank is interested in the nature of loans of first-time borrowers. This information will be used to tailor their marketing strategy. They believe that 50% of first-time borrowers take out smaller loans than other borrowers. They perform a hypothesis test to determine if the percentage is the same or different from 50% . They sample 100 first-time borrowers and find 53 of these loans are smaller that the other borrowers. For the hypothesis test, they choose a 5% level of significance. STEP 1 : Set the null and alternative hypothesis. H 0 : p = 0.50 H a : p ≠ 0.50 The words "is the same or different from" tell you this is a two-tailed test. The Type I and Type II errors are as follows: The Type I error is to conclude that the proportion of borrowers is different from 50% when, in fact, the proportion is actually 50%. (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 borrowers differs from 50% when, in fact, the proportion does differ from 50%. (You fail to reject the null hypothesis when the null hypothesis is false.) STEP 2 : Decide the level of significance and draw the graph showing the critical value The level of significance has been set by the problem at the 5% level. Because this is two-tailed test one-half of the alpha value will be in the upper tail and one-half in the lower tail as shown on the graph. The critical value for the normal distribution at the 95% level of confidence is 1.96. This can easily be found on the student’s t-table at the very bottom at infinite degrees of freedom remembering that at infinity the t-distribution is the normal distribution. Of course the value can also be found on the normal table but you have go looking for one-half of 95 (0.475) inside the body of the table and then read out to the sides and top for the number of standard deviations. STEP 3 : Calculate the sample parameters and critical value of the test statistic. The test statistic is a normal distribution, Z, for testing proportions and is: Z = p' - p 0 p 0 q 0 n = .53 - .50 .5 ( .5 ) 100 = 0.60 For this case, the sample of 100 found 53 of these loans were smaller than those of other borrowers. The sample proportion, p′ = 53/100= 0.53 The test question, therefore, is : “Is 0.53 significantly different from .50?” Putting these values into the formula for the test statistic we find that 0.53 is only 0.60 standard deviations away from .50. This is barely off of the mean of the standard normal distribution of zero. There is virtually no difference from the sample proportion and the hypothesized proportion in terms of standard deviations. STEP 4 : Compare the test statistic and the critical value. The calculated value is well within the critical values of ± 1.96 standard deviations and thus we cannot reject the null hypothesis. To reject the null hypothesis we need significant evident of difference between the hypothesized value and the sample value. In this case the sample value is very nearly the same as the hypothesized value measured in terms of standard deviations. STEP 5 : Reach a conclusion The formal conclusion would be “At a 5% level of significance we cannot reject the null hypothesis that 50% of first-time borrowers take out smaller loans than other borrowers.” Notice the length to which the conclusion goes to include all of the conditions that are attached to the conclusion. Statisticians, for all the criticism they receive, are careful to be very specific even when this seems trivial. Statisticians cannot say more than they know, and the data constrain the conclusion to be within the metes and bounds of the data. Try It A teacher believes that 85% of students in the class will want to go on a field trip to the local zoo. The teacher performs a hypothesis test to determine if the percentage is the same or different from 85%. The teacher samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance. Since the problem is about percentages, this is a test of single population proportions. H 0 : p = 0.85 H a : p ≠ 0.85 p = 0.7554 Because p > α , we fail to reject the null hypothesis. There is not sufficient evidence to suggest that the proportion of students that want to go to the zoo is not 85%. Suppose a consumer group suspects that the proportion of households that have three or more cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three or more cell phones. Here is an abbreviate version of the system to solve hypothesis tests applied to a test on a proportions. H 0 : p = 0.3 H a : p ≠ 0.3 n = 150 p' = x n = 43 150 = 0.287 Z c = p' - p 0 p 0 q 0 n = 0.287 - 0.3 .3 ( .7 ) 150 = 0.347 Try It Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which, 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypothesis, find the p -value, state your conclusion, and identify the Type I and Type II errors. 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; .98; .98; 1.02; .95; .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. Let’s follow a four-step process to answer this statistical question. 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 H 0 : μ ≤ 1 H a : μ > 1 Plan : We are testing a sample mean without a known population standard deviation with less than 30 observations. Therefore, we need to use a Student's-t distribution. Assume the underlying population is normal. Do the calculations and draw the graph . State the Conclusions : We cannot accept the null hypothesis. It is reasonable to state that the data supports the claim that the average conductivity level is greater than one. Try It The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows: 205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205 Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal. The hypotheses, at a significance level of 0.1, will be as follows: H 0 : μ ≤ 200 H 0 : μ > 200 Since the standard deviation is not known, Student’s t -distribution is to be used. The data is input into the TI-83 as follows: Since the p -value is less than alpha value, we will reject the null hypothesis. It is reasonable to state that the data supports the claim that the average boiling point is greater than 200. 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 non-cell phone users is 0.0340%). 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. We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be H 0 : p ≤ 0.00034 H a : p > 0.00034 If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer. 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. Try It In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users (the rate of skin diseases for non-moisturizer users is 0.041%). 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. The test hypotheses are as follows: H 0 : p ≤ 0 . 00041 H 0 : p > 0 . 00041 If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes disease-causing environments, we want to minimize the chances of incorrectly identifying causes of diseases. The testing will be done with x = 138 and n = 390,000. The sample is sufficiently large because we have np = 390,000(0.00041) = 159.9, nq = 390,000(0.99959) = 389,840.1, two independent outcomes, and a fixed probability of success p = 0.00041. Thus, we will be able to generalize our results to the population. The associated TI results are: Since the p -value = 0.0832 is greater than alpha value = 0.005, we cannot reject the null. Therefore, we conclude that there is not enough evidence to support the claim of higher skin diseases rates for moisturizer users. Chapter Review The hypothesis test itself has an established process. This can be summarized as follows: Determine H 0 and H a . Remember, they are contradictory. Determine the random variable. Determine the distribution for the test. Draw a graph and calculate the test statistic. Compare the calculated test statistic with the Z critical value determined by the level of significance required by the test and make a decision (cannot reject H 0 or cannot accept H 0 ), and write a clear conclusion using English sentences. Formula Review Test statistic for a hypothesis test of proportions: Z c = p ' – p 0 p 0 ( 1 – p 0 ) n Assume H 0 : μ = 9 and H a : μ < 9. Is this a left-tailed, right-tailed, or two-tailed test? This is a left-tailed test. Assume H 0 : μ ≤ 6 and H a : μ > 6. Is this a left-tailed, right-tailed, or two-tailed test? Assume H 0 : p = 0.25 and H a : p ≠ 0.25. Is this a left-tailed, right-tailed, or two-tailed test? This is a two-tailed test. Draw the general graph of a left-tailed test. Draw the graph of a two-tailed test. 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? Your friend claims that their mean golf score is 63. You want to show that it is higher than that. What type of test would you use? a right-tailed test 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? You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use? a left-tailed test If the alternative hypothesis has a not equals ( ≠ ) symbol, you know to use which type of test? Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test? This is a left-tailed test. Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test? 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? This is a two-tailed test. Homework 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, is the data highly inconsistent with the claim? H 0 : μ ≥ 50,000 H a : μ < 50,000 Let X ¯ = the average lifespan of a brand of tires. normal distribution z = -2.315 p -value = 0.0103 Answers may vary. alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is less than 0.05. Conclusion: There is sufficient evidence to conclude that the mean lifespan of the tires is less than 50,000 miles. (43,537, 49,463) From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant of around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level? 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% level? H 0 : μ = $1.00 H a : μ ≠ $1.00 Let X ¯ = the average cost of a daily newspaper. normal distribution z = –0.866 p -value = 0.3865 Answers may vary. Alpha: 0.01 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.01. Conclusion: There is sufficient evidence to support the claim that the mean cost of daily papers is $1. The mean cost could be $1. ($0.84, $1.06) 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% level? The mean number of sick days an employee takes per year is believed to be about ten. 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 ten? H 0 : μ = 10 H a : μ ≠ 10 Let X ¯ the mean number of sick days an employee takes per year. Student’s t -distribution t = –1.13 p -value = 0.1299 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.05. Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean number of sick days is not ten. (4.9443, 11.806) 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. 81 women were surveyed with the following results. The sample mean was 83; the sample standard deviation was ten. Does it appear that the mean work week has increased for women at the 5% level? Your statistics instructor claims that 60% of the students who take their Elementary Statistics class go through life feeling more enriched. For some reason that the instructor can't quite figure out, most people don't believe them. You decide to check this out on your own. You randomly survey 64 of their past Elementary Statistics students and find that 34 feel more enriched as a result of taking the class. Now, what do you think? H 0 : p ≥ 0.6 H a : p < 0.6 Let P′ = the proportion of students who feel more enriched as a result of taking Elementary Statistics. normal for a single proportion 1.13 p -value = 0.1299 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for decision: The p -value is greater than 0.05. Conclusion: There is insufficient evidence to conclude that less than 60 percent of the students feel more enriched. Confidence Interval: (0.409, 0.654) The “plus-4s” confidence interval is (0.411, 0.648) 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. Refer to Exercise 9.63 . 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. H 0 : μ = 4 H a : μ ≠ 4 Let X ¯ the average I.Q. of a set of brown trout. two-tailed Student's t-test t = 1.95 p -value = 0.076 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is greater than 0.05 Conclusion: There is insufficient evidence to conclude that the average IQ of brown trout is not four. (3.8865,5.9468) According to an article in Newsweek , the birth ratio of females to males is 100:105. In China, the birth ratio is 100: 114 (46.7% females). Suppose you want to explore the reported figures of the percent of females born in China. You conduct a study. In this study, you count the number of females and males born in 150 randomly chosen recent births. There are 60 females and 90 males born of the 150. Based on your study, do you believe that the percent of females born in China is 46.7? A poll done for Newsweek found that 13% of Americans have seen or sensed the presence of an angel. A contingent doubts that the percent is really that high. It conducts its own survey. Out of 76 Americans surveyed, only two had seen or sensed the presence of an angel. As a result of the contingent’s survey, would you agree with the Newsweek poll? In complete sentences, also give three reasons why the two polls might give different results. H 0 : p ≥ 0.13 H a : p < 0.13 Let P′ = the proportion of Americans who have seen or sensed angels normal for a single proportion –2.688 p -value = 0.0036 Answers may vary. alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is less than 0.05. Conclusion: There is sufficient evidence to conclude that the percentage of Americans who have seen or sensed an angel is less than 13%. (0, 0.0623). The“plus-4s” confidence interval is (0.0022, 0.0978) 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. They ask ten engineering friends in start-ups for the lengths of their mean work weeks. Based on the results that follow, should they 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; 50; 55. Sixty-eight percent of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% 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% represents California. NOTE: For more accurate results, use more California community colleges and this past year's data. According to New York City's Community Health Survey, the most recent adult smoking rate is 12%. Suppose that a survey is conducted to determine this year’s rate. Nine out of 70 randomly chosen N.Y. City residents reply that they smoke. Conduct a hypothesis test to determine if the rate is still 12% or if it has decreased. H 0 : p = 0.12 H a : p < 0.12 Let P ' = the proportion of NYC residents that smoke. Normal for a single proportion 0.2207 p -value = 0.5873 Answers may vary alpha 0.05 Decision: Do not reject the null hypothesis. Reason for Decision: The p -value is greater than 0.05. At the 5% significance level, there is insufficient evidence to conclude that the proportion of NYC residents who smoke is less than 0.14. Confidence interval: Confidence Interval: (0.0502, 0.2070): The “plus-4s” confidence interval (see Confidence Intervals ) is (0.0676, 0.2297). 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. They randomly survey 56 online students and find that the sample mean is 29.4 with a standard deviation of 2.1. Conduct a hypothesis test. 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. H 0 : μ = 69,110 H a : μ > 69,110 Let X ¯ = the mean salary in dollars for California registered nurses. Student's t -distribution t = 1.719 p -value: 0.0466 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: The p -value is less than 0.05. Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean salary of California registered nurses exceeds $69,110. ($68,757, $73,485) 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 U.S. is less than four years old. Over the past few decades, public health officials have examined the link between weight concerns and teen girls' smoking. Researchers surveyed a group of 273 randomly selected teen girls living in Massachusetts (between 12 and 15 years old). After four years the girls were surveyed again. Sixty-three said they smoked to stay thin. Is there good evidence that more than 30% of the teen girls smoke to stay thin? After conducting the test, your decision and conclusion are Reject H 0 : There is sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin. Do not reject H 0 : There is not sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin. Do not reject H 0 : There is not sufficient evidence to conclude that more than 30% of teen girls smoke to stay thin. Reject H 0 : There is sufficient evidence to conclude that less than 30% of teen girls smoke to stay thin. c A statistics instructor believes that fewer than 20% of Evergreen Valley College (EVC) students attended the opening night midnight showing of the latest Marvel Universe movie. The instructor surveys 84 students and finds that 11 of them attended the midnight showing. At a 1% level of significance, an appropriate conclusion is: There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of a Marvel Universe movie is less than 20%. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of a Marvel Universe movie is more than 20%. There is sufficient evidence to conclude that the percent of EVC students who attended the midnight showing of a Marvel Universe movie is less than 20%. There is insufficient evidence to conclude that the percent of EVC students who attended the midnight showing of a Marvel Universe movie is at least 20%. 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? There is enough evidence to conclude that the mean number of hours is more than 4.75 There is enough evidence to conclude that the mean number of hours is more than 4.5 There is not enough evidence to conclude that the mean number of hours is more than 4.5 There is not enough evidence to conclude that the mean number of hours is more than 4.75 c Instructions: For the following ten exercises, Hypothesis testing: For the following ten exercises, answer each question. State the null and alternate hypothesis. State the p -value. State alpha. What is your decision? Write a conclusion. Answer any other questions asked in the problem. According to the Center for Disease Control website, in 2022 at least 2% of high school students reported they had smoked a cigarette. An Introduction to Statistics class conducted a hypothesis test at the local high school (a medium sized–approximately 1,200 students–small city demographic) to determine if the local high school’s percentage was lower. One hundred fifty students were chosen at random and surveyed. Of the 150 students surveyed, 82 have smoked. Use a significance level of 0.05 and using appropriate statistical evidence, conduct a hypothesis test and state the conclusions. A recent survey in the N.Y. Times Almanac indicated that 48.8% of families own stock. A broker wanted to determine if this survey could be valid. They 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? H 0 : p = 0.488 H a : p ≠ 0.488 p -value = 0.0114 alpha = 0.05 Reject the null hypothesis. There is enough evidence to conclude that the percent of families owning stocks is not 48.8%. The survey does not appear to be accurate. Driver error can be listed as the cause of approximately 54% 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? The US Department of Energy reported that 51.7% 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 in Kentucky? Are the results applicable across the country? Why? H 0 : p = 0.517 H a : p ≠ 0.517 p -value = 0.9203. alpha = 0.05. Do not reject the null hypothesis. At the 5% significance level, there is not enough evidence to conclude that the proportion of homes in Kentucky that are heated by natural gas is different from the proportion of homes heated by natural gas across the country. 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. For Americans using library services, the American Library Association claims that at most 67% of patrons borrow books. The library director in Owensboro, Kentucky feels this is not true, so they 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 that do borrow books from the Owensboro Library? The Weather Underground reported that the mean amount of summer rainfall for the northeastern US 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. H 0 : µ ≥ 11.52 H a : µ < 11.52 p -value = 0.000002 which is almost 0. alpha = 0.05. Reject the null hypothesis. At the 5% 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. We would make the same conclusion if alpha was 1% because the p -value is almost 0. A survey 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 US cities? 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? H 0 : µ ≤ 5.8 H a : µ > 5.8 p -value = 0.9987 alpha = 0.05 Do not reject the null hypothesis. At the 5% 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. According to the U.S. Census Bureau , the mean family size in the U.S. is 3.13. 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 level, is the class’ mean family size greater than the national average? Does the census result remain valid? Why? The student academic group on a college campus claims that first-year 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 first-year 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? H 0 : µ ≥ 150 H a : µ < 150 p -value = 0.0622 alpha = 0.01 Do not reject the null hypothesis. At the 1% significance level, there is not enough evidence to conclude that first-year students study less than 2.5 hours per day, on average. The student academic group’s claim appears to be correct. References Data from Amit Schitai. Director of Instructional Technology and Distance Learning. LBCC. Data from Bloomberg Businessweek . Available online at http://www.businessweek.com/news/2011- 09-15/nyc-smoking-rate-falls-to-record-low-of-14-bloomberg-says.html. Data from energy.gov. Available online at http://energy.gov (accessed June 27. 2013). Data from Gallup®. Available online at www.gallup.com (accessed June 27, 2013). Data from Growing by Degrees by Allen and Seaman. Data from La Leche League International. Available online at http://www.lalecheleague.org/Law/BAFeb01.html. Data from the American Automobile Association. Available online at www.aaa.com (accessed June 27, 2013). Data from the American Library Association. Available online at www.ala.org (accessed June 27, 2013). Data from the Bureau of Labor Statistics. Available online at http://www.bls.gov/oes/current/oes291111.htm. Data from the Centers for Disease Control and Prevention. Available online at www.cdc.gov (accessed June 27, 2013) Data from the U.S. Census Bureau, available online at http://quickfacts.census.gov/qfd/states/00000.html (accessed June 27, 2013). Data from the United States Census Bureau. Available online at http://www.census.gov/hhes/socdemo/language/. Data from Toastmasters International. Available online at http://toastmasters.org/artisan/detail.asp?CategoryID=1&SubCategoryID=10&ArticleID=429&Page=1. Data from Weather Underground. Available online at www.wunderground.com (accessed June 27, 2013). Federal Bureau of Investigations. “Uniform Crime Reports and Index of Crime in Daviess in the State of Kentucky enforced by Daviess County from 1985 to 2005.” Available online at http://www.disastercenter.com/kentucky/crime/3868.htm (accessed June 27, 2013). “Foothill-De Anza Community College District.” De Anza College, Winter 2006. Available online at http://research.fhda.edu/factbook/DAdemofs/Fact_sheet_da_2006w.pdf. Johansen, C., J. Boice, Jr., J. McLaughlin, J. Olsen. “Cellular Telephones and Cancer—a Nationwide Cohort Study in Denmark.” Institute of Cancer Epidemiology and the Danish Cancer Society, 93(3):203-7. Available online at http://www.ncbi.nlm.nih.gov/pubmed/11158188 (accessed June 27, 2013). Central Limit Theorem Given a random variable (RV) with known mean μ and known standard deviation σ. We are sampling with size n and we are interested in two new RVs - the sample mean, X ¯ . If the size n of the sample is sufficiently large, then X ¯ ~ N ( μ , σ n ) . If the size n of the sample is sufficiently large, then the distribution of the sample means will approximate a normal distribution regardless of the shape of the population. The expected value of the mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, σ n , is called the standard error of the mean.

Business


Introduction 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: modification of work “scrambled egg, toast and home-made berry jam (no added sugar) for breakfast” by Chloe Lim/ Flickr, CC BY 2.0) Studies often compare two groups. For example, researchers are interested in the effect aspirin has in preventing heart 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 SAT or GRE preparatory courses really help raise their scores. Many business applications require comparing two groups. It may be the investment returns of two different investment strategies, or the differences in production efficiency of different management styles. To compare two means or two proportions, you work with two groups. The groups are classified either as independent 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 of each group. Independent Groups two samples that are selected from two populations, and the values from one population are not related in any way to the values from the other population. Matched Pairs two samples that are dependent. Differences between a before and after scenario are tested by testing one population mean of differences.

Business


Comparing Two Independent Population Means The comparison of two independent population means is very common and provides a way to test the hypothesis that the two groups differ from each other. Is the night shift less productive than the day shift, are the rates of return from fixed asset investments different from those from common stock investments, and so on? An observed difference between two sample means depends on both the means and the sample standard deviations. Very different means can occur by chance if there is great variation among the individual samples. The test statistic will have to account for this fact. 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 we will see later was developed by Aspin-Welch. When we developed the hypothesis test for the mean and proportions we began with the Central Limit Theorem. We recognized that a sample mean came from a distribution of sample means, and sample proportions came from the sampling distribution of sample proportions. This made our sample parameters, the sample means and sample proportions, into random variables. It was important for us to know the distribution that these random variables came from. The Central Limit Theorem gave us the answer: the normal distribution. Our Z and t statistics came from this theorem. This provided us with the solution to our question of how to measure the probability that a sample mean came from a distribution with a particular hypothesized value of the mean or proportion. In both cases that was the question: what is the probability that the mean (or proportion) from our sample data came from a population distribution with the hypothesized value we are interested in? Now we are interested in whether or not two samples have the same mean. Our question has not changed: Do these two samples come from the same population distribution? To approach this problem we create a new random variable. We recognize that we have two sample means, one from each set of data, and thus we have two random variables coming from two unknown distributions. To solve the problem we create a new random variable, the difference between the sample means. This new random variable also has a distribution and, again, the Central Limit Theorem tells us that this new distribution is normally distributed, regardless of the underlying distributions of the original data. A graph may help to understand this concept. Pictured are two distributions of data, X 1 and X 2 , with unknown means and standard deviations. The second panel shows the sampling distribution of the newly created random variable ( X ¯ 1 - X ¯ 2 ). This distribution is the theoretical distribution of many sample means from population 1 minus sample means from population 2. The Central Limit Theorem tells us that this theoretical sampling distribution of differences in sample means is normally distributed, regardless of the distribution of the actual population data shown in the top panel. Because the sampling distribution is normally distributed, we can develop a standardizing formula and calculate probabilities from the standard normal distribution in the bottom panel, the Z distribution. We have seen this same analysis before in The Central Limit Theorem Figure 7.2 . The Central Limit Theorem, as before, provides us with the standard deviation of the sampling distribution, and further, that the expected value of the mean of the distribution of differences in sample means is equal to the differences in the population means. Mathematically this can be stated: E ( µ x ¯ 1 - µ x ¯ 2 ) = µ 1 - µ 2 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: ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 We remember that substituting the sample variance for the population variance when we did not have the population variance was the technique we used when building the confidence interval and the test statistic for the test of hypothesis for a single mean back in Confidence Intervals and Hypothesis Testing with One Sample . The test statistic ( t -score) is calculated as follows: t c = ( x – 1 – x – 2 ) – δ 0 ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 where: s 1 and s 2 , the sample standard deviations, are estimates of σ 1 and σ 2 , respectively and σ 1 and σ 2 are the unknown population standard deviations. x – 1 and x – 2 are the sample means. μ 1 and μ 2 are the unknown population means. δ = the hypothesized differences between the two populations means. Substituting the sample variances of the two sample groups when the populations variances are unknown and the sample size is large, the t -distribution will provide good estimates using degrees of freedom: v = n 1 + n 2 - 2 . Large in this case would be greater than 100 observations from the combined two groups ( n 1 + n 2 ) although other statisticians suggest that the definition of large is greater than 30 observations for each of the two groups. The number of degrees of freedom ( df ) requires a somewhat complicated calculation. The df are not always a whole number. The test statistic above is approximated by the Student's t -distribution with df as follows: Degrees of Freedom The standard error is: d f = ( ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ) 2 ( 1 n 1 – 1 ) ( ( s 1 ) 2 n 1 ) 2 + ( 1 n 2 – 1 ) ( ( s 2 ) 2 n 2 ) 2 When both sample sizes n 1 and n 2 are 30 or larger, the Student's t approximation is very good. If each sample has more than 30 observations then the degrees of freedom can be calculated as n 1 + n 2 − 2 . The format of the sampling distribution, differences in sample means, specifies that the format of the null and alternative hypothesis is: The test hypothesis for differences between two means H 0 : µ 1 - µ 2 = δ 0 H a : µ 1 - µ 2 ≠ δ 0 where δ 0 is the hypothesized difference between the two means. If the question is simply “is there any difference between the means?” then δ 0 = 0 and the null and alternative hypotheses becomes: H 0 : µ 1 = µ 2 H a : µ 1 ≠ µ 2 An example of when δ 0 might not be zero is when the comparison of the two groups requires a specific difference for the decision to be meaningful. Imagine that you are making a capital investment. You are considering changing from your current model machine to another. You measure the productivity of your machines by the speed they produce the product. It may be that a contender to replace the old model is faster in terms of product throughput, but is also more expensive. The second machine may also have more maintenance costs, setup costs, etc. The null hypothesis would be set up so that the new machine would have to be better than the old one by enough to cover these extra costs in terms of speed and cost of production. This form of the null and alternative hypothesis shows how valuable this particular hypothesis test can be. For most of our work we will be testing simple hypotheses asking if there is any difference between the two distribution means. Independent groups The Kona Iki Corporation produces coconut milk. They take coconuts and extract the milk inside by drilling a hole and pouring the milk into a vat for processing. They have both a day shift (called the B shift) and a night shift (called the G shift) to do this part of the process. They would like to know if the day shift and the night shift are equally efficient in processing the coconuts. A study is done sampling 9 shifts of the G shift and 16 shifts of the B shift. The results of the number of hours required to process 100 pounds of coconuts is presented in . A study is done and data are collected, resulting in the data in . Sample Size Average Number of Hours to Process 100 Pounds of Coconuts Sample Standard Deviation G Shift 9 2 0.866 B Shift 16 3.2 1.00 Is there a difference in the mean amount of time for each shift to process 100 pounds of coconuts? Test at the 5% level of significance. The population standard deviations are not known and cannot be assumed to equal each other. Let g be the subscript for the G Shift and b be the subscript for the B Shift. Then, μ g is the population mean for G Shift and μ b is the population mean for B Shift. This is a test of two independent groups , two population means . Random variable : X ¯ g − X ¯ b = difference in the sample mean amount of time between the G Shift and the B Shift takes to process the coconuts. H 0 : μ g = μ b H 0 : μ g – μ b = 0 H a : μ g ≠ μ b H a : μ g – μ b ≠ 0 The words "the same" tell you H 0 has an "=". Since there are no other words to indicate H a , is either faster or slower. This is a two tailed test. Distribution for the test: Use t df where df is calculated using the df formula for independent groups, two population means above. Using a calculator, df is approximately 18.8462. Graph: t c = ( X ¯ 1 − X ¯ 2 ) − δ 0 S 1 2 n 1 + S 2 2 n 2 = -3.01 We next find the critical value on the t-table using the degrees of freedom from above. The critical value, 2.093, is found in the .025 column, this is α/2, at 19 degrees of freedom. (The convention is to round up the degrees of freedom to make the conclusion more conservative.) Next we calculate the test statistic and mark this on the t-distribution graph. Make a decision: Since the calculated t-value is in the tail we cannot accept the null hypothesis that there is no difference between the two groups. The means are different. The graph has included the sampling distribution of the differences in the sample means to show how the t-distribution aligns with the sampling distribution data. We see in the top panel that the calculated difference in the two means is -1.2 and the bottom panel shows that this is 3.01 standard deviations from the mean. Typically we do not need to show the sampling distribution graph and can rely on the graph of the test statistic, the t-distribution in this case, to reach our conclusion. Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that the G Shift takes to process 100 pounds of coconuts is different from the B Shift (mean number of hours for the B Shift is greater than the mean number of hours for the G Shift). Try It Two samples are shown in . 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% level of significance. Sample Size Sample Mean Sample Standard Deviation Population A 25 5 1 Population B 16 4.7 1.2 NOTE When the sum of the sample sizes is larger than 30 ( n 1 + n 2 > 30) you can use the normal distribution to approximate the Student's t . Use the Student's t distribution, however, whenever possible. A study is done to determine if Company A retains its workers longer than Company B. It is believed that Company A has a higher retention than Company B. The study finds that in a sample of 11 workers at Company A their average time with the company is four years with a standard deviation of 1.5 years. A sample of 9 workers at Company B finds that the average time with the company was 3.5 years with a standard deviation of 1 year. Test this proposition at the 1% level of significance. a. Is this a test of two means or two proportions? a. two means because time is a continuous random variable. b. Are the populations standard deviations known or unknown? b. unknown c. Which distribution do you use to perform the test? c. Student's t d. What is the random variable? d. X ¯ A - X ¯ B e. What are the null and alternate hypotheses? e. H Ø : μ A ≤ μ B H a : μ A > μ B f. Is this test right-, left-, or two-tailed? f. right one-tailed test g. What is the value of the test statistic? t c = ( X ¯ 1 − X ¯ 2 ) − δ 0 S 1 2 n 1 + S 2 2 n 2 = 0.89 h. Can you accept/reject the null hypothesis? h. Cannot reject the null hypothesis that there is no difference between the two groups. Test statistic is not in the tail. The critical value of the t -distribution is 2.764 with 10 degrees of freedom. This example shows how difficult it is to reject a null hypothesis with a very small sample. The critical values require very large test statistics to reach the tail. i. Conclusion: i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the retention of workers at Company A is longer than Company B, on average. Try It 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 five 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. Are the population standard deviations known? Conduct an appropriate hypothesis test. At the 5% significance level, what is your conclusion? An interesting research question is the effect, if any, that different types of teaching formats have on the grade outcomes of students. To investigate this issue one sample of students' grades was taken from a hybrid class and another sample taken from a standard lecture format class. Both classes were for the same subject. The mean course grade in percent for the 35 hybrid students is 74 with a standard deviation of 16. The mean grades of the 40 students form the standard lecture class was 76 percent with a standard deviation of 9. Test at 5% to see if there is any significant difference in the population mean grades between standard lecture course and hybrid class. We begin by noting that we have two groups, students from a hybrid class and students from a standard lecture format class. We also note that the random variable, what we are interested in, is students' grades, a continuous random variable. We could have asked the research question in a different way and had a binary random variable. For example, we could have studied the percentage of students with a failing grade, or with an A grade. Both of these would be binary and thus a test of proportions and not a test of means as is the case here. Finally, there is no presumption as to which format might lead to higher grades so the hypothesis is stated as a two-tailed test. H 0 : µ 1 = µ 2 H a : µ 1 ≠ µ 2 As would virtually always be the case, we do not know the population variances of the two distributions and thus our test statistic is: t c = ( x ¯ 1 − x ¯ 2 ) − δ 0 s 2 n 1 + s 2 n 2 = ( 74 − 76 ) − 0 16 2 35 + 9 2 40 = −0.65 To determine the critical value of the Student's t we need the degrees of freedom. For this case we use: df = n1 + n2 - 2 = 35 + 40 -2 = 73. This is large enough to consider it the normal distribution thus t a/2 = 1.96. Again as always we determine if the calculated value is in the tail determined by the critical value. In this case we do not even need to look up the critical value: the calculated value of the difference in these two average grades is not even one standard deviation apart. Certainly not in the tail. Conclusion: Cannot reject the null at α=5%. Therefore, evidence does not exist to prove that the grades in hybrid and standard classes differ. Try It Two professors, A and B, teach classes on the same subjects. Scores of 10 students from each class are selected randomly. The final exam scores of the students are as follows: Professor A: 97 62 73 58 84 74 66 93 73 85 Professor B: 85 64 74 55 76 67 72 84 71 98 Professor A says that the mean score of their class is more than the mean of professor B’s class. Is professor A correct? Test at a 5% significance level. Answer the following questions: Is this a test of two means or two proportions? Are the population standard deviations known or unknown? Which distribution do you use to perform the test? What is the random variable? What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. Is this test right-, left-, or two-tailed? What is the p -value? Do you reject or not reject the null hypothesis? At the ___ level of significance, from the sample data, there ___ (is/is not) sufficient evidence to conclude that ___. two means unknown Student’s t X 1 - X 2 H 0 : μ 1 = μ 2 Null hypothesis: the means of final exam scores are equal for both professors. H 0 : μ 2 < μ 1 Alternative hypothesis: the mean of final exam scores for class of professor B is less than mean of final exam scores for class of professor A. left-tailed p -value = 0.03 Reject the null hypothesis. Professor A is correct. The evidence shows that the mean of final exam scores of the class of professor A is more than that of the class of professor B. At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of final exam scores of the class of professor A is more than that of the class of professor B. References Data from Graduating Engineer + Computer Careers. Available online at http://www.graduatingengineer.com Data from Microsoft Bookshelf . Data from the United States Senate website, available online at www.Senate.gov (accessed June 17, 2013). “List of current United States Senators by Age.” Wikipedia. Available online at http://en.wikipedia.org/wiki/List_of_current_United_States_Senators_by_age (accessed June 17, 2013). “Sectoring by Industry Groups.” Nasdaq. Available online at http://www.nasdaq.com/markets/barchart-sectors.aspx?page=sectors&base=industry (accessed June 17, 2013). “World Series History.” Baseball-Almanac, 2013. Available online at http://www.baseball-almanac.com/ws/wsmenu.shtml (accessed June 17, 2013). Chapter Review 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) Formula Review Standard error: SE = ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 Test statistic ( t -score): t c = ( x ¯ 1 − x ¯ 2 ) − δ 0 ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 Degrees of freedom: d f = ( ( s 1 ) 2 n 1 + ( s 2 ) 2 n 2 ) 2 ( 1 n 1 − 1 ) ( ( s 1 ) 2 n 1 ) 2 + ( 1 n 2 − 1 ) ( ( s 2 ) 2 n 2 ) 2 where: s 1 and s 2 are the sample standard deviations, and n 1 and n 2 are the sample sizes. x ¯ 1 and x ¯ 2 are the sample means. Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for independent group means, population standard deviations, and/or variances known independent group means, population standard deviations, and/or variances unknown matched or paired samples single mean two proportions single proportion It is believed that 70% of men pass their drivers test in the first attempt, while 65% of women pass the test in the first attempt. Of interest is whether the proportions are in fact equal. two proportions 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. 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. matched or paired samples 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 mid-level professionals are paid differently, on average. The average worker in Germany gets eight weeks of paid vacation. single mean According to a television commercial, 80% of dentists agree that Ultrafresh toothpaste is the best on the market. It is believed that the average grade on an English essay in a particular school system for women is higher than for men. A random sample of 31 women had a mean score of 82 with a standard deviation of three, and a random sample of 25 men had a mean score of 76 with a standard deviation of four. independent group means, population standard deviations and/or variances unknown 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? 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? two proportions A new medicine is said to help improve sleep. Eight subjects are picked at random and given the medicine. The means hours slept for each person were recorded before starting the medication and after. 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. independent group means, population standard deviations and/or variances unknown Varsity athletes practice five times a week, on average. 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 of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions different? independent group means, population standard deviations and/or variances unknown 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? A high school principal claims that 30% of student athletes drive themselves to school, while 4% of non-athletes drive themselves to school. In a sample of 20 student athletes, 45% drive themselves to school. In a sample of 35 non-athlete students, 6% drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes? two proportions 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 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. Are standard deviations known or unknown? What is the random variable? The random variable is the difference between the mean amounts of sugar in the two soft drinks. Is this a one-tailed or two-tailed test? Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for White people born in 1900 and 33.0 years for non-White people. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 White people, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 non-White people, 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 White people and non-White people. Is this a test of means or proportions? means State the null and alternative hypotheses. H 0 : __________ H a : __________ Is this a right-tailed, left-tailed, or two-tailed test? two-tailed In symbols, what is the random variable of interest for this test? In words, define the random variable of interest for this test. the difference between the mean life spans of White and non-White people Which distribution (normal or Student's t ) would you use for this hypothesis test? Explain why you chose the distribution you did for . This is a comparison of two population means with unknown population standard deviations. Calculate the test statistic. 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. Answers may vary. At a pre-conceived α = 0.05, what is your: Decision: Reason for the decision: Conclusion (write out in a complete sentence): Cannot accept the null hypothesis p -value < 0.05 At the 5% level of significance, the evidence supports the claim that life expectancy in the 1900s was different between White people and non-White people. Does it appear that the means are the same? Why or why not? Homework The mean number of English courses taken in a two–year time period by men and women college students is believed to be about the same. An experiment is conducted and data are collected from 29 men and 16 women. The men took an average of three English courses with a standard deviation of 0.8. The women took an average of four English courses with a standard deviation of 1.0. Are the means statistically the same? 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. Subscripts: 1: two-year colleges; 2: four-year colleges H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2 X ¯ 1 – X ¯ 2 is the difference between the mean enrollments of the two-year colleges and the four-year colleges. Student’s- t test statistic: -0.2480 p -value: 0.4019 Answers may vary. Alpha: 0.05 Decision: Cannot reject Reason for Decision: p -value > alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean enrollment at four-year colleges is higher than at two-year colleges. At Rachel’s 11 th birthday party, eight guests were timed to see how long (in seconds) they could hold their breath in a relaxed position. After a two-minute rest, they timed themselves while jumping. The children thought that the mean difference between their jumping and relaxed times would be zero. Test their hypothesis. Relaxed time (seconds) Jumping time (seconds) 26 21 47 40 30 28 22 21 23 25 45 43 37 35 29 32 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 actually 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. Subscripts: 1: mechanical engineering; 2: electrical engineering H 0 : μ 1 ≥ μ 2 H a : μ 1 < μ 2 X ¯ 1 − X ¯ 2 is the difference between the mean entry level salaries of mechanical engineers and electrical engineers. t 108 test statistic: t = –0.82 p -value: 0.2061 Answers may vary. Alpha: 0.05 Decision: Cannot reject the null hypothesis. Reason for Decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the mean entry-level salaries of mechanical engineers is lower than that of electrical engineers. Marketing companies have collected data implying that teenage girls use more social media apps on their smartphones than teenage boys do. In one particular study of 40 randomly chosen teenage girls and boys (20 of each) with smartphones, the mean number of social media apps 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 Data Sets in Appendix A: Statistical Tables to answer the next four exercises. 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. H 0 : μ 1 = μ 2 H a : μ 1 ≠ μ 2 X ¯ 1 − X ¯ 2 is the difference between the mean times for completing a lap in races and in practices. t 20.32 test statistic: –4.70 p -value: 0.0001 Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. Repeat the test in , but use Lap 5 data this time. Repeat the test in , but this time combine the data from Laps 1 and 5. H 0 : μ 1 = μ 2 H a : μ 1 ≠ μ 2 is the difference between the mean times for completing a lap in races and in practices. t 40.94 test statistic: –5.08 p -value: zero Answers may vary. Alpha: 0.05 Decision: Cannot accept the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean time for completing a lap in races is different from that in practices. 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. Western Eastern Los Angeles 9 D.C. 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 Conduct a hypothesis test to answer the next two exercises. The exact distribution for the hypothesis test is: the normal distribution the Student's t -distribution the uniform distribution the exponential distribution If the level of significance is 0.05, the conclusion is: There is sufficient evidence to conclude that the W Division teams score fewer goals, on average, than the E teams There is insufficient evidence to conclude that the W Division teams score more goals, on average, than the E teams. There is insufficient evidence to conclude that the W teams score fewer goals, on average, than the E teams score. Unable to determine c 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. They take 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. A concluding statement is: 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. 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. 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. 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. Elijah wants to know whether textbook costs are different for different courses of study. He selects a random sample of 33 sociology textbooks offered on a popular online site. The mean price of his sample is $74.64 with a standard deviation of $49.36. He then selects a random sample of 33 math and science textbooks from the same site. The mean price of this sample is $111.56 with a standard deviation of $66.90. Is the mean price of a sociology textbook lower than the mean price of a math or science textbook? Test at a 1% significance level. Test: two independent sample means, population standard deviations unknown. μ 1 = the mean price of a sociology text on the selected site μ 2 = the mean price of a math/science text on the selected site Random variable: X 1 - X 2 = the difference in the sample mean textbook price between sociology texts and math/science texts. Hypotheses: H 0 : μ 1 - μ 2 = 0 , H a : μ 1 - μ 2 < 0 which can be expressed as H 0 s : μ 1 = μ 2 , H a : μ 1 < μ 2 Distribution for the test: Use tdf ; because each sample has more than 30 observations, d f = n 1 + n 2 - 2 = 33 + 33 - 2 = 64 . Estimate the critical value on the t -table using the nearest available degrees of freedom, 60. The critical value, 2.660, is found in the .0005 column. Calculate the test statistic: t c = ( X 1 - X 2 ) - 0 s 1 2 n 1 + s 2 2 n 2 = ( 74 . 64 - 111 . 56 ) - 0 49 . 36 2 33 + 66 . 90 2 33 = - 2 . 55 Using a calculator with t c = - 2 . 55 and d f = 64 , the left-tailed p -value: 0.0066. Decision: Reject H 0 . Conclusion: At the 1% level of significance, from the sample data, there is sufficient evidence to conclude that the mean price of sociology textbooks is less than the mean price of textbooks for math/science. 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. 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. They take 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: μ day > μ night μ day < μ night μ day = μ night μ day ≠ μ night d Cohen’s d a measure of effect size based on the differences between two means. If d is between 0 and 0.2 then the effect is small. If d approaches is 0.5, then the effect is medium, and if d approaches 0.8, then it is a large effect. Pooled Variance a weighted average of two variances that can then be used when calculating standard error.

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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 United States 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. Cohen's Standard Effect Sizes Size of effect d Small 0.2 Medium 0.5 Large 0.8 Cohen's d is the measure of the difference between two means divided by the pooled standard deviation: d = x ¯ 1 – x ¯ 2 s p o o l e d where s p o o l e d = ( n 1 – 1 ) s 1 2 + ( n 2 – 1 ) s 2 2 n 1 + n 2 – 2 It is important to note that Cohen's d does not provide a level of confidence as to the magnitude of the size of the effect comparable to the other tests of hypothesis we have studied. The sizes of the effects are simply indicative. 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. x̅ 1 = 4 s 1 = 1.5 n 1 = 11 x̅ 2 = 3.5 s 2 = 1 n 2 = 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 companies is small indicating that there is not a significant difference between them. Try It Calculate Cohen’s d for Try It 10.4. Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem. μ 1 = 5 s 1 = 1 . 2 n 1 = 15 μ 2 = 4 . 5 s 2 = 0 . 8 n 2 = 20 d = 0.49 The effect is medium because 0.49 is very close to Cohen’s value of 0.5 for medium effect size. The size of the differences of the means for the two companies is small, indicating that there is not a significant difference between them. Chapter Review Cohen's d is a measure of “effect size” based on the differences between two means. It is important to note that Cohen's d does not provide a level of confidence as to the magnitude of the size of the effect comparable to the other tests of hypothesis we have studied. The sizes of the effects are simply indicative. Formula Review Cohen’s d is the measure of effect size: d = x ¯ 1 − x ¯ 2 s p o o l e d where s p o o l e d = ( n 1 − 1 ) s 1 2 + ( n 2 − 1 ) s 2 2 n 1 + n 2 − 2

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Test for Differences in Means: Assuming Equal Population Variances Typically we can never expect to know any of the population parameters, mean, proportion, or standard deviation. When testing hypotheses concerning differences in means we are faced with the difficulty of two unknown variances that play a critical role in the test statistic. We have been substituting the sample variances just as we did when testing hypotheses for a single mean. And as we did before, we used a Student's t to compensate for this lack of information on the population variance. There may be situations, however, when we do not know the population variances, but we can assume that the two populations have the same variance. If this is true then the pooled sample variance will be smaller than the individual sample variances. This will give more precise estimates and reduce the probability of discarding a good null. The null and alternative hypotheses remain the same, but the test statistic changes to: t c = ( x – 1 − x – 2 ) − δ 0 S p 2 ( 1 n 1 + 1 n 2 ) where S p 2 is the pooled variance given by the formula: s p 2 = ( n 1 − 1 ) s 1 2 + ( n 2 − 1 ) s 2 2 n 1 + n 2 − 2 A drug trial is attempted using a real drug and a pill made of just sugar. 18 people are given the real drug in hopes of increasing the production of endorphins. The increase in endorphins is found to be on average 8 micrograms per person, and the sample standard deviation is 5.4 micrograms. 11 people are given the sugar pill, and their average endorphin increase is 4 micrograms with a standard deviation of 2.4. From previous research on endorphins it is determined that it can be assumed that the variances within the two samples can be assumed to be the same. Test at 5% to see if the population mean for the real drug had a significantly greater impact on the endorphins than the population mean with the sugar pill. First we begin by designating one of the two groups Group 1 and the other Group 2. This will be needed to keep track of the null and alternative hypotheses. Let's set Group 1 as those who received the actual new medicine being tested and therefore Group 2 is those who received the sugar pill. We can now set up the null and alternative hypothesis as: H 0 : µ 1 ≤ µ 2 H 1 : µ 1 > µ 2 This is set up as a one-tailed test with the claim in the alternative hypothesis that the medicine will produce more endorphins than the sugar pill. We now calculate the test statistic which requires us to calculate the pooled variance, S p 2 using the formula above. t c = ( x ¯ 1 − x ¯ 2 ) − δ 0 S p 2 ( 1 n 1 + 1 n 2 ) = ( 8 − 4 ) − 0 20.4933 ( 1 18 + 1 11 ) = 2.31 t α , allows us to compare the test statistic and the critical value. t α = 1.703 at d f = n 1 + n 2 − 2 = 18 + 11 − 2 = 27 The test statistic is clearly in the tail, 2.31 is larger than the critical value of 1.703, and therefore we cannot maintain the null hypothesis. Thus, we conclude that there is significant evidence at the 95% level of confidence that the new medicine produces the effect desired. Try It 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 and , respectively. Northeast 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 West 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 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% significance level. Answer the following questions: Is this a test of two means or two proportions? Are the population standard deviations known or unknown? Which distribution do you use to perform the test? What is the random variable? What are the null and alternative hypotheses? Write the null and alternative hypotheses in words and in symbols. Is this test right, left, or two tailed? What is the p -value? Do you reject or not reject the null hypothesis? At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ______. Calculate Cohen’s d and interpret it. Chapter Review In situations when we do not know the population variances but assume the variances are the same, the pooled sample variance will be smaller than the individual sample variances. This will give more precise estimates and reduce the probability of discarding a good null. Formula Review t c = ( x ¯ 1 − x ¯ 2 ) − δ 0 S p 2 ( 1 n 1 + 1 n 2 ) where S p 2 is the pooled variance given by the formula: S p 2 = ( n 1 − 1 ) s 1 2 + ( n 2 − 1 ) s 2 2 n 1 + n 2 − 2

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