Two Population Means with Unknown Standard Deviations The two independent samples are simple random samples from two distinct populations. For the two distinct populations: if the sample sizes are small, the distributions are important (should be normal) if the sample sizes are large, the distributions are not important (need not be normal) NOTE The test comparing two independent population means with unknown and possibly unequal population standard deviations is called the Aspin-Welch t-test. The degrees of freedom formula was developed by Aspin-Welch. The comparison of two population means is very common. A difference between the two samples depends on both the means and the standard deviations. Very different means can occur by chance if there is great variation among the individual samples. In order to account for the variation, we take the difference of the sample means, X ¯ 1 – X ¯ 2 , and divide by the standard error in order to standardize the difference. The result is a t-score test statistic. 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 The test statistic ( t -score) is calculated as follows: ( x ¯ 1 – x ¯ 2 ) – ( μ 1 – μ 2 ) ( 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. 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 population means. The number of degrees of freedom ( df ) requires a somewhat complicated calculation. However, a computer or calculator calculates it easily. The df are not always a whole number. The test statistic calculated previously is approximated by the Student's t -distribution with df as follows: Degrees of freedom d f = ( ( 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 five or larger, the Student's t approximation is very good. Notice that the sample variances ( s 1 ) 2 and ( s 2 ) 2 are not pooled. (If the question comes up, do not pool the variances.) NOTE It is not necessary to compute this by hand. A calculator or computer easily computes it. Independent groups The average amount of time boys and girls aged seven to 11 spend playing sports each day is believed to be the same. A study is done and data are collected, resulting in the data in . Each populations has a normal distribution. Sample Size Average Number of Hours Playing Sports Per Day Sample Standard Deviation Girls 9 2 0.866 Boys 16 3.2 1.00 Is there a difference in the mean amount of time boys and girls aged seven to 11 play sports each day? Test at the 5% level of significance. The population standard deviations are not known. Let g be the subscript for girls and b be the subscript for boys. Then, μ g is the population mean for girls and μ b is the population mean for boys. 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 girls and boys play sports each day. 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 , assume it says "is different." 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. Using a calculator, df is approximately 18.8462. Do not pool the variances. Calculate the p -value using a Student's t -distribution: p -value = 0.0054 Graph: s g = 0.866 s b = 1 So, x ¯ g – x ¯ b = 2 – 3.2 = –1.2 Half the p -value is below –1.2 and half is above 1.2. Make a decision: Since α > p -value, reject H 0 . This means you reject μ g = μ b . The means are different. Press STAT . Arrow over to TESTS and press 4:2-SampTTest . Arrow over to Stats and press ENTER . Arrow down and enter 2 for the first sample mean, 0.866 for Sx1, 9 for n1, 3.2 for the second sample mean, 1 for Sx2, and 16 for n2. Arrow down to μ1: and arrow to does not equal μ2. Press ENTER . Arrow down to Pooled: and No . Press ENTER . Arrow down to Calculate and press ENTER . The p -value is p = 0.0054, the dfs are approximately 18.8462, and the test statistic is -3.14. Do the procedure again but instead of Calculate do Draw. Conclusion: At the 5% level of significance, the sample data show there is sufficient evidence to conclude that the mean number of hours that girls and boys aged seven to 11 play sports per day is different (mean number of hours boys aged seven to 11 play sports per day is greater than the mean number of hours played by girls OR the mean number of hours girls aged seven to 11 play sports per day is greater than the mean number of hours played by boys). 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 by a community group in two neighboring colleges to determine which one graduates students with more math classes. College A samples 11 graduates. Their average is four math classes with a standard deviation of 1.5 math classes. College B samples nine graduates. Their average is 3.5 math classes with a standard deviation of one math class. The community group believes that a student who graduates from college A has taken more math classes, on the average. Both populations have a normal distribution. Test at a 1% significance level. Answer the following questions. a. Is this a test of two means or two proportions? a. two means 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? Write the null and alternate hypotheses in words and in symbols. e. H o : μ A ≤ μ B H a : μ A > μ B f. Is this test right-, left-, or two-tailed? f. right g. What is the p -value? g. 0.1928 h. Do you reject or not reject the null hypothesis? h. Do not reject. i. Conclusion: i. At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a student who graduates from college A has taken more math classes, on the average, than a student who graduates from college B. 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? A professor at a large community college wanted to determine whether there is a difference in the means of final exam scores between students who took his statistics course online and the students who took his face-to-face statistics class. He believed that the mean of the final exam scores for the online class would be lower than that of the face-to-face class. Was the professor correct? The randomly selected 30 final exam scores from each group are listed in and . Online Class 67.6 41.2 85.3 55.9 82.4 91.2 73.5 94.1 64.7 64.7 70.6 38.2 61.8 88.2 70.6 58.8 91.2 73.5 82.4 35.5 94.1 88.2 64.7 55.9 88.2 97.1 85.3 61.8 79.4 79.4 Face-to-face Class 77.9 95.3 81.2 74.1 98.8 88.2 85.9 92.9 87.1 88.2 69.4 57.6 69.4 67.1 97.6 85.9 88.2 91.8 78.8 71.8 98.8 61.2 92.9 90.6 97.6 100 95.3 83.5 92.9 89.4 Is the mean of the Final Exam scores of the online class lower than the mean of the Final Exam scores of the face-to-face class? 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 ______. (See the conclusion in , and write yours in a similar fashion) two means unknown Student's t X ¯ 1 – X ¯ 2 H 0 : μ 1 = μ 2 Null hypothesis: the means of the final exam scores are equal for the online and face-to-face statistics classes. H a : μ 1 < μ 2 Alternative hypothesis: the mean of the final exam scores of the online class is less than the mean of the final exam scores of the face-to-face class. left-tailed p -value = 0.0011 Reject the null hypothesis The professor was correct. The evidence shows that the mean of the final exam scores for the online class is lower than that of the face-to-face class. At the 5% level of significance, from the sample data, there is (is/is not) sufficient evidence to conclude that the mean of the final exam scores for the online class is less than the mean of final exam scores of the face-to-face class. First put the data for each group into two lists (such as L1 and L2). Press STAT . Arrow over to TESTS and press 4:2SampTTest . Make sure Data is highlighted and press ENTER . Arrow down and enter L1 for the first list and L2 for the second list. Arrow down to μ 1 : and arrow to < μ 2 (less than). Press ENTER . Arrow down to Pooled: No. Press ENTER . Arrow down to Calculate and press ENTER . NOTE Be careful not to mix up the information for Group 1 and Group 2! 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. 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 Calculate Cohen’s d for . Is the size of the effect small, medium, or large? Explain what the size of the effect means for this problem. μ 1 = 4 s 1 = 1.5 n 1 = 11 μ 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 colleges is small indicating that there is not a significant difference between them. Try It Calculate Cohen’s d for . 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. Calculate Cohen’s d for . Is the size of the effect small, medium or large? Explain what the size of the effect means for this problem. d = 0.834; Large, because 0.834 is greater than Cohen’s 0.8 for a large effect size. The size of the differences between the means of the Final Exam scores of online students and students in a face-to-face class is large indicating a significant difference. 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. 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 = ( x ¯ 1 − x ¯ 2 ) − ( μ 1 − μ 2 ) ( 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. 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 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 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 and p -value. 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. Find the p -value. At a pre-conceived α = 0.05, what is your: Decision: Reason for the decision: Conclusion (write out in a complete sentence): Reject 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 DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in Appendix E . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t -distribution for a homework problem in what follows, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) 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: Do not 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: Do not 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 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: Reject 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 Exercise 10.83 , but use Lap 5 data this time. Repeat the test in Exercise 10.83 , 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: Reject 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 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 1 ¯ = 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 < μ 2 which can be expressed as H 0 s : μ 1 - μ 2 , Ha μ 1 < μ 2 . Distribution for the test: Use t d f ; 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 2 + 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: 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 Degrees of Freedom ( df ) the number of objects in a sample that are free to vary. Standard Deviation A number that is equal to the square root of the variance and measures how far data values are from their mean; notation: s for sample standard deviation and σ for population standard deviation. Variable (Random Variable) a characteristic of interest in a population being studied. Common notation for variables are upper-case Latin letters X , Y , Z ,... Common notation for a specific value from the domain (set of all possible values of a variable) are lower-case Latin letters x , y , z ,.... For example, if X is the number of children in a family, then x represents a specific integer 0, 1, 2, 3, .... Variables in statistics differ from variables in intermediate algebra in the two following ways. The domain of the random variable (RV) is not necessarily a numerical set; the domain may be expressed in words; for example, if X = hair color, then the domain is {black, blond, gray, green, orange}. We can tell what specific value x of the random variable X takes only after performing the experiment.

Math


Two Population Means with Known Standard Deviations Even though this situation is not likely (knowing the population standard deviations is not likely), the following example illustrates hypothesis testing for independent means, known population standard deviations. The sampling distribution for the difference between the means is normal and both populations must be normal. The random variable is X 1 ¯ – X 2 ¯ . The normal distribution has the following format: Normal distribution is: X ¯ 1 – X ¯ 2 ~ N [ μ 1 – μ 2 , ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 ] The standard deviation is: ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 The test statistic ( z -score) is: z = ( x ¯ 1 – x ¯ 2 ) – ( μ 1 – μ 2 ) ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 Independent groups, population standard deviations known: The mean lasting time of two competing floor waxes is to be compared. Twenty floors are randomly assigned to test each wax . Both populations have a normal distributions. The data are recorded in . Wax Sample Mean Number of Months Floor Wax Lasts Population Standard Deviation 1 3 0.33 2 2.9 0.36 Does the data indicate that wax 1 is more effective than wax 2 ? Test at a 5% level of significance. This is a test of two independent groups, two population means, population standard deviations known. Random Variable : X ¯ 1 – X ¯ 2 = difference in the mean number of months the competing floor waxes last. H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 The words "is more effective" says that wax 1 lasts longer than wax 2 , on average. "Longer" is a “>” symbol and goes into H a . Therefore, this is a right-tailed test. Distribution for the test: The population standard deviations are known so the distribution is normal. Using the formula, the distribution is: X ¯ 1 – X ¯ 2 ~ N ( 0 , 0.33 2 20 + 0.36 2 20 ) Since μ 1 ≤ μ 2 then μ 1 – μ 2 ≤ 0 and the mean for the normal distribution is zero. Calculate the p -value using the normal distribution: p -value = 0.1799 Graph: X ¯ 1 – X ¯ 2 = 3 – 2.9 = 0.1 Compare α and the p -value: α = 0.05 and p -value = 0.1799. Therefore, α < p -value. Make a decision: Since α < p -value, do not reject H 0 . Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean time wax 1 lasts is longer (wax 1 is more effective) than the mean time wax 2 lasts. Press STAT . Arrow over to TESTS and press 3:2-SampZTest . Arrow over to Stats and press ENTER . Arrow down and enter .33 for sigma1, .36 for sigma2, 3 for the first sample mean, 20 for n1, 2.9 for the second sample mean, and 20 for n2. Arrow down to μ 1: and arrow to > μ 2 . Press ENTER . Arrow down to Calculate and press ENTER . The p -value is p = 0.1799 and the test statistic is 0.9157. Do the procedure again, but instead of Calculate do Draw . Try It The means of the number of revolutions per minute of two competing engines are to be compared. Thirty engines of each type are randomly assigned to be tested. Both populations have normal distributions. shows the result. Do the data indicate that Engine 2 has higher RPM than Engine 1? Test at a 5% level of significance. Engine Sample Mean Number of RPM Population Standard Deviation 1 1,500 50 2 1,600 60 An interested citizen wanted to know if Democratic U. S. senators are older than Republican U.S. senators, on average. During a certain year, the mean age of 30 randomly selected Republican Senators was 61 years 247 days old (61.675 years) with a standard deviation of 10.17 years. The mean age of 30 randomly selected Democratic senators was 61 years 257 days old (61.704 years) with a standard deviation of 9.55 years. Do the data indicate that Democratic senators are older than Republican senators, on average? Test at a 5% level of significance. This is a test of two independent groups, two population means. The population standard deviations are unknown, but the sum of the sample sizes is 30 + 30 = 60, which is greater than 30, so we can use the normal approximation to the Student’s-t distribution. Subscripts: 1: Democratic senators 2: Republican senators Random variable: X ¯ 1 – X ¯ 2 = difference in the mean age of Democratic and Republican U.S. senators. H 0 : µ 1 ≤ µ 2 H 0 : µ 1 – µ 2 ≤ 0 H a : µ 1 > µ 2 H a : µ 1 – µ 2 > 0 The words "older than" translates as a “>” symbol and goes into H a . Therefore, this is a right-tailed test. Distribution for the test: The distribution is the normal approximation to the Student’s t for means, independent groups. Using the formula, the distribution is: X ¯ 1 – X ¯ 2 ∼ N [ 0 , ( 9.55 ) 2 30 + ( 10.17 ) 2 30 ] Since µ 1 ≤ µ 2 , µ 1 – µ 2 ≤ 0 and the mean for the normal distribution is zero. (Calculating the p -value using the normal distribution gives p -value = 0.4955) Graph: Compare α and the p -value: α = 0.05 and p -value = 0.4955. Therefore, α < p -value. Make a decision: Since α < p -value, do not reject H 0 . Conclusion: At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of Democratic senators is greater than the mean age of the Republican senators. Try It The average age of 10 professors selected randomly in university A is 46.672 with a standard deviation of 8.53. The average age of 10 professors selected randomly in university B is 47.531 with a standard deviation of 7.83. Does the data indicate that university A has older professors than university B, on average? Test at a 5% level of significance. The total sample size is 10 + 10 = 20., which is greater than 10, so we can use the Student’s t -distribution. The random variable is X 1 - X 2 = difference between the mean age of universities A and B. H 0 : μ 1 ≤ μ 2 H 0 : μ 1 - μ 2 ≤ 0 H 0 : μ 1 > μ 2 H 0 : μ 1 - μ 2 > 0 The phrase “older than” translates as a “>” symbol and goes into H 0 . Therefore, this is a right-tailed test. The distribution is: X 1 - X 2 ~ N 0 , ( 8 . 53 ) 2 10 + ( 7 . 83 ) 2 10 The mean for the normal distribution is 0 since μ 1 ≤ μ 2 and μ 1 - μ 2 ≤ 0 . The p -value is obtained as 0.4258. Since alpha is less than the p -value, do not reject the H 0 . At the 5% level of significance, from the sample data, there is not sufficient evidence to conclude that the mean age of professors in university A is more than the mean age of professors in university B. References Data from the United States Census Bureau. Available online at http://www.census.gov/prod/cen2010/briefs/c2010br-02.pdf Hinduja, Sameer. “Sexting Research and Gender Differences.” Cyberbulling Research Center, 2013. Available online at http://cyberbullying.us/blog/sexting-research-and-gender-differences/ (accessed June 17, 2013). “Smart Phone Users, By the Numbers.” Visually, 2013. Available online at http://visual.ly/smart-phone-users-numbers (accessed June 17, 2013). Smith, Aaron. “35% of American adults own a Smartphone.” Pew Internet, 2013. Available online at http://www.pewinternet.org/~/media/Files/Reports/2011/PIP_Smartphones.pdf (accessed June 17, 2013). “State-Specific Prevalence of Obesity AmongAdults—Unites States, 2007.” MMWR, CDC. Available online at http://www.cdc.gov/mmwr/preview/mmwrhtml/mm5728a1.htm (accessed June 17, 2013). “Texas Crime Rates 1960–1012.” FBI, Uniform Crime Reports, 2013. Available online at: http://www.disastercenter.com/crime/txcrime.htm (accessed June 17, 2013). Chapter Review A hypothesis test of two population means from independent samples where the population standard deviations are known will have these characteristics: Random variable: X ¯ 1 − X ¯ 2 = the difference of the means Distribution: normal distribution Formula Review Normal Distribution: X ¯ 1 − X ¯ 2 ∼ N [ μ 1 − μ 2 , ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 ] . Generally µ 1 – µ 2 = 0. Test Statistic ( z -score): z = ( x ¯ 1 − x ¯ 2 ) − ( μ 1 − μ 2 ) ( σ 1 ) 2 n 1 + ( σ 2 ) 2 n 2 Generally µ 1 - µ 2 = 0. where: σ 1 and σ 2 are the known population standard deviations. n 1 and n 2 are the sample sizes. x ¯ 1 and x ¯ 2 are the sample means. μ 1 and μ 2 are the population means. Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. shows the result. Scouters believe that Rodriguez pitches a speedier fastball. Pitcher Sample Mean Speed of Pitches (mph) Population Standard Deviation Wesley 86 3 Rodriguez 91 7 What is the random variable? The difference in mean speeds of the fastball pitches of the two pitchers State the null and alternative hypotheses. What is the test statistic? –2.46 What is the p -value? At the 1% significance level, what is your conclusion? At the 1% significance level, we can reject the null hypothesis. There is sufficient data to conclude that the mean speed of Rodriguez’s fastball is faster than Wesley’s. Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller. Plant Group Sample Mean Height of Plants (inches) Population Standard Deviation Food 16 2.5 No food 14 1.5 Is the population standard deviation known or unknown? State the null and alternative hypotheses. Subscripts: 1 = Food, 2 = No Food H 0 : μ 1 ≤ μ 2 H a : μ 1 > μ 2 What is the p -value? Draw the graph of the p -value. At the 1% significance level, what is your conclusion? Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point. Sample Mean Melting Temperatures (°F) Population Standard Deviation Alloy Gamma 800 95 Alloy Zeta 900 105 State the null and alternative hypotheses. Subscripts: 1 = Gamma, 2 = Zeta H 0 : μ 1 = μ 2 H a : μ 1 ≠ μ 2 Is this a right-, left-, or two-tailed test? What is the p -value? 0.0062 Draw the graph of the p -value. At the 1% significance level, what is your conclusion? There is sufficient evidence to reject the null hypothesis. The data support that the melting point for Alloy Zeta is different from the melting point of Alloy Gamma. Homework DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t -distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (When using these tests in a real situation, you must first prove that assumption, however.) A study is done to determine if students in the California state university system take longer to graduate, on average, than students enrolled in private universities. One hundred students from both the California state university system and private universities are surveyed. Suppose that from years of research, it is known that the population standard deviations are 1.5811 years and 1 year, respectively. The following data are collected. The California state university system students took on average 4.5 years with a standard deviation of 0.8. The private university students took on average 4.1 years with a standard deviation of 0.3. Parents of teenage boys often complain that auto insurance costs more, on average, for teenage boys than for teenage girls. A group of concerned parents examines a random sample of insurance bills. The mean annual cost for 36 teenage boys was $679. For 23 teenage girls, it was $559. From past years, it is known that the population standard deviation for each group is $180. Determine whether or not you believe that the mean cost for auto insurance for teenage boys is greater than that for teenage girls. Subscripts: 1 = boys, 2 = girls H 0 : µ 1 ≤ µ 2 H a : µ 1 > µ 2 The random variable is the difference in the mean auto insurance costs for boys and girls. normal test statistic: z = 2.50 p -value: 0.0062 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean cost of auto insurance for teenage boys is greater than that for girls. A group of transfer bound students wondered if they will spend the same mean amount on texts and supplies each year at their four-year university as they have at their community college. They conducted a random survey of 54 students at their community college and 66 students at their local four-year university. The sample means were $947 and $1,011, respectively. The population standard deviations are known to be $254 and $87, respectively. Conduct a hypothesis test to determine if the means are statistically the same. Some manufacturers claim that non-hybrid sedan cars have a lower mean miles-per-gallon (mpg) than hybrid ones. Suppose that consumers test 21 hybrid sedans and get a mean of 31 mpg with a standard deviation of seven mpg. Thirty-one non-hybrid sedans get a mean of 22 mpg with a standard deviation of four mpg. Suppose that the population standard deviations are known to be six and three, respectively. Conduct a hypothesis test to evaluate the manufacturers claim. Subscripts: 1 = non-hybrid sedans, 2 = hybrid sedans H 0 : µ 1 ≥ µ 2 H a : µ 1 < µ 2 The random variable is the difference in the mean miles per gallon of non-hybrid sedans and hybrid sedans. normal test statistic: 6.36 p -value: 0 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the mean miles per gallon of non-hybrid sedans is less than that of hybrid sedans. A baseball fan wanted to know if there is a difference between the number of games played in a World Series when the American League won the series versus when the National League won the series. From 1922 to 2012, the population standard deviation of games won by the American League was 1.14, and the population standard deviation of games won by the National League was 1.11. Of 19 randomly selected World Series games won by the American League, the mean number of games won was 5.76. The mean number of 17 randomly selected games won by the National League was 5.42. Conduct a hypothesis test. One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement “I’m pleased with the way we divide the responsibilities for childcare.” The ratings went from one (strongly agree) to five (strongly disagree). contains ten of the paired responses for husbands and wives. Conduct a hypothesis test to see if the mean difference in the husband’s versus the wife’s satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife). Wife’s Score 2 2 3 3 4 2 1 1 2 4 Husband’s Score 2 2 1 3 2 1 1 1 2 4 H 0 : µ d = 0 H a : µ d < 0 The random variable X d is the average difference between husband’s and wife’s satisfaction level. t 9 test statistic: t = –1.86 p -value: 0.0479 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis, but run another test. Reason for Decision: p -value < alpha Conclusion: This is a weak test because alpha and the p -value are close. However, there is insufficient evidence to conclude that the mean difference is negative.

Math


Comparing Two Independent Population Proportions When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present: The two independent samples are simple random samples that are independent. The number of successes is at least five, and the number of failures is at least five, for each of the samples. Growing literature states that the population must be at least ten or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results. Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions. Like the case of differences in sample means, we construct a sampling distribution for differences in sample proportions: p ' A = X A n A and p ' B = X B n B are the sample proportions for the two sets of data in question X A and X B . The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H 0 : p A = p B . To conduct the test, we use a pooled proportion , p c . The pooled proportion is calculated as follows: p c = x A + x B n A + n B The distribution for the differences is: P ′ A − P ′ B ~ N [ 0 , p c ( 1 − p c ) ( 1 n A + 1 n B ) ] The test statistic ( z -score) is: z = ( p ′ A − p ′ B ) − ( p A − p B ) p c ( 1 − p c ) ( 1 n A + 1 n B ) Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance. The problem asks for a difference in proportions, making it a test of two proportions. Let A and B be the subscripts for medication A and medication B, respectively. Then p A and p B are the desired population proportions. Random Variable: P′ A – P′ B = difference in the proportions of adult patients who did not react after 30 minutes to medication A and to medication B. H 0 : p A = p B p A – p B = 0 H a : p A ≠ p B p A – p B ≠ 0 The words "is a difference" tell you the test is two-tailed. Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal: p c = x A + x B n A + n B = 20 + 12 200 + 200 = 0.08 1 – p c = 0.92 P ′ A – P ′ B ~ N [ 0 , ( 0.08 ) ( 0.92 ) ( 1 200 + 1 200 ) ] P′ A – P′ B follows an approximate normal distribution. Calculate the p -value using the normal distribution: p -value = 0.1404. Estimated proportion for group A: p ′ A = x A n A = 20 200 = 0.1 Estimated proportion for group B: p ′ B = x B n B = 12 200 = 0.06 Graph: P′ A – P′ B = 0.1 – 0.06 = 0.04. Half the p -value is below –0.04, and half is above 0.04. Compare α and the p -value: α = 0.01 and the p -value = 0.1404. α < p -value. Make a decision: Since α < p -value, do not reject H 0 . Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to medication A and medication B . Press STAT . Arrow over to TESTS and press 6:2-PropZTest . Arrow down and enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1 : and arrow to not equal p2 . Press ENTER . Arrow down to Calculate and press ENTER . The p -value is p = 0.1404 and the test statistic is 1.47. Do the procedure again, but instead of Calculate do Draw . Try It Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5% level of significance. A research study was conducted about gender differences regarding the use of seat belts in motor vehicles. The researcher believed that the proportion of women not wearing seat belts is less than the proportion of men not wearing seat belts. The data collected represents a random sample of U.S. adults and is summarized in . Is the proportion of women not wearing seat belts less than the proportion of men not wearing seat belts? Test at a 1% level of significance. Men Women Does not wear seat belts 183 156 Total number surveyed 2231 2169 This is a test of two population proportions. Let M and F be the subscripts for men and women. Then p M and p F are the desired population proportions. Random Variable: p′ F − p′ M = difference in the proportions of men and women who do not wear seat belts. H 0 : p F = p M H 0 : p F – p M = 0 H a : p F < p M H a : p F – p M < 0 The words "less than" tell you the test is left-tailed. Distribution for the test: Since this is a test of two population proportions, the distribution is normal: p c = x F + x M n F + n M = 156 + 183 2169 + 2231 = 0 .077 1 − p c = 0.923 Therefore, p ′ F – p ′ M ∼ N ( 0 , ( 0.077 ) ( 0.923 ) ( 1 2169 + 1 2231 ) ) p′ F – p′ M follows an approximate normal distribution. Calculate the p -value using the normal distribution: p -value = 0.1045 Estimated proportion for women: 0.0719 Estimated proportion for men: 0.082 Graph: Decision: Since α < p -value, Do not reject H 0 Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of women not wearing seat belts is less than the proportion of men not wearing seat belts. Try It A survey was conducted about the favorable beverage as tea. The data collected is summarized in the table. Is the proportion of men favoring tea more than women favoring tea? Test at a 1% level of significance. Men Women Favor tea 16 18 Total surveyed 230 218 Let p M and p F be desired population proportions. H 0 : p F = p M H 0 : p F - p M = 0 H 0 : p F < p M H 0 : p F - p M < 0 The test is for “less than,” so it is left-tailed. Distribution for the test: p C = x F + x M n F + n M = 18 + 16 218 + 230 = 0 . 076 1 - p C = 0 . 924 p ' F - p ' M ~ N 0 , ( 0 . 076 ) ( 0 . 924 ) 1 218 + 1 230 This follows an approximate normal distribution. Calculate the p-value using the normal distribution: p -value = 0.1023 Decision: Since α < p - v a l u e , do not reject H 0 . Conclusion: At 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of females favoring tea is less than the proportion of males favoring tea. Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER . Arrow down to Calculate and press ENTER. The p -value is P = 0.1045 and the test statistic is z = -1.256. A marketing firm claims that the proportion of younger adults who own electric vehicles is greater than the proportion of older adults who own electric vehicles. A random sample of U.S. adults was taken, and the results of the survey indicate the following: Out of a sample of 232 older adults (aged 35 or older), 5% own electric vehicles. Out of a sample of 1,343 young adults (aged 34 or younger), 10% own electric vehicles. Test at the 5% level of significance. Is the proportion of younger adults greater than the proportion of older adults with respect to owning electric vehicles? This is a test of two population proportions. Let Y and O be the subscripts for younger adults and older adults, respectively. Then pY and pO are the desired population proportions. Random Variable: p’Y and p’O = difference in the proportions of younger and older adults who own electric vehicles. H 0 : p Y = p O H 0 : p Y − p O = 0 H a : p Y > p O H a : p Y - p O > 0 The words "greater than" indicate that the test is right-tailed. Distribution for the test: The distribution is approximately normal: p c = x y + x o n y + n o = 134 + 12 1343 + 232 = 0 . 0927 1 - p c = 0 . 9073 Therefore, p ' y - p ' O ~ N 0 , 0 . 0927 ) ( 0 . 9073 ) ( 1 1343 + 1 232 ) p ' y - p ' O follows an approximate normal distribution. Calculate the p -value using the normal distribution: p -value = 0.0077 Estimated proportion for group Y: 0.10 Estimated proportion for group O: 0.05 Graph: Decision: Since  > p -value, reject the H 0 . Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that a larger proportion of younger adults own electric vehicles as compared to older adults. TI-83+ and TI-84: Press STAT . Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 135 for x1, 1343 for n1, 12 for x2, and 232 for n2. Arrow down to p1: and arrow to greater than p2. Press ENTER . Arrow down to Calculate and press ENTER . The P-value is P = 0.0092 and the test statistic is Z = 2.33. Try It A government researcher is investigating whether there is a difference in the use of helmets by motorcyclists in different geographic regions for those states where use of helmets is required by law. The research shows that for motorcyclists in the northeast U.S., 7622 out of 113,231 motorcyclists did not wear helmets. In the southeast U.S., 7439 out of 104,873 motorcyclists did not wear helmets. Test at a 5% significance level. Answer the following questions: a. Is this a test of two means or two proportions? two proportions b. Which distribution do you use to perform the test? normal for two proportions c. What is the random variable? Subscripts: 1 = Northeast, 2 = Southeast P ′ 2 - P ′ 2 d. What are the null and alternative hypothesis? Write the null and alternative hypothesis in symbols. Subscripts: 1 = Northeast, 2 = Southeast H 0 : p 1 = p 2 H 0 : p 1 − p 2 = 0 H a : p 1 ≠ p 2 H a : p 1 − p 2 ≠ 0 e. Is this test right-, left-, or two-tailed? two-tailed f. What is the p-value? p -value = 0.00086 g. Do you reject or not reject the null hypothesis? Reject the H 0 . h. At the ___ level of significance, from the sample data, there ______ (is/is not) sufficient evidence to conclude that ____________. At the 5% significance level, from the sample data, there is sufficient evidence to conclude that there is a difference between the proportion of motorcyclists who did not wear helmets for northeast and southeast regions of the U.S. References Data from Educational Resources , December catalog. Data from Hilton Hotels. Available online at http://www.hilton.com (accessed June 17, 2013). Data from Hyatt Hotels. Available online at http://hyatt.com (accessed June 17, 2013). Data from Statistics, United States Department of Health and Human Services. Data from Whitney Exhibit on loan to San Jose Museum of Art. Data from the American Cancer Society. Available online at http://www.cancer.org/index (accessed June 17, 2013). Data from the Chancellor’s Office, California Community Colleges, November 1994. “State of the States.” Gallup, 2013. Available online at http://www.gallup.com/poll/125066/State-States.aspx?ref=interactive (accessed June 17, 2013). “West Nile Virus.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/ncidod/dvbid/westnile/index.htm (accessed June 17, 2013). Chapter Review Test of two population proportions from independent samples. Random variable: p ^ A – p ^ B = difference between the two estimated proportions Distribution: normal distribution Formula Review Pooled Proportion: p c = x F + x M n F + n M Distribution for the differences: p ′ A − p ′ B ∼ N [ 0 , p c ( 1 − p c ) ( 1 n A + 1 n B ) ] where the null hypothesis is H 0 : p A = p B or H 0 : p A – p B = 0. Test Statistic ( z -score): z = ( p ′ A − p ′ B ) p c ( 1 − p c ) ( 1 n A + 1 n B ) where the null hypothesis is H 0 : p A = p B or H 0 : p A − p B = 0. where p′ A and p′ B are the sample proportions, p A and p B are the population proportions, P c is the pooled proportion, and n A and n B are the sample sizes. Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS 1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS 2 had system failures within the first eight hours of operation. OS 2 is believed to be more stable (have fewer crashes) than OS 1 . Is this a test of means or proportions? What is the random variable? P ′ OS1 – P ′ OS2 = difference in the proportions of phones that had system failures within the first eight hours of operation with OS 1 and OS 2 . State the null and alternative hypotheses. What is the p -value? 0.1018 What can you conclude about the two operating systems? Use the following information to answer the next twelve exercises. In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota. Is this a test of means or proportions? proportions State the null and alternative hypotheses. H 0 : _________ H a : _________ Is this a right-tailed, left-tailed, or two-tailed test? How do you know? right-tailed What is the random variable of interest for this test? In words, define the random variable for this test. The random variable is the difference in proportions (percents) of the populations that are of two or more races in Nevada and North Dakota. Which distribution (normal or Student's t ) would you use for this hypothesis test? Explain why you chose the distribution you did for the Exercise 10.56 . Our sample sizes are much greater than five each, so we use the normal for two proportions distribution for this hypothesis test. Calculate the test statistic. Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p -value. Answers may vary. Find the p -value. At a pre-conceived α = 0.05, what is your: Decision: Reason for the decision: Conclusion (write out in a complete sentence): Reject the null hypothesis. p -value < alpha At the 5% significance level, there is sufficient evidence to conclude that the proportion (percent) of the population that is of two or more races in Nevada is statistically higher than that in North Dakota. Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not? Homework DIRECTIONS: For each of the word problems, use a solution sheet to do the hypothesis test. The solution sheet is found in . Please feel free to make copies of the solution sheets. For the online version of the book, it is suggested that you copy the .doc or the .pdf files. NOTE If you are using a Student's t -distribution for one of the following homework problems, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, however.) A recent drug survey showed an increase in the use of drugs and alcohol among local high school seniors as compared to the national percent. Suppose that a survey of 100 local seniors and 100 national seniors is conducted to see if the proportion of drug and alcohol use is higher locally than nationally. Locally, 65 seniors reported using drugs or alcohol within the past month, while 60 national seniors reported using them. We are interested in whether the proportions of conferred physical science degrees for people aged 21 to 24 are the same for White and Black people in the United States. The number of conferred degrees for White people in a given year is 4,930. Five hundred eighty were aged 21 to 24. The estimate for Black people is 330. Forty were aged 21 to 24. We will let people with conferred physical science degrees be our population. H 0 : P W = P B H a : P W ≠ P B The random variable is the difference in the proportions of White and Black people with conferred degrees in a given year, aged 21 to 24. normal for two proportions test statistic: –0.1944 p -value: 0.8458 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportions of White and Black people with conferred physical science degrees, aged 21 to 24, are different. Elizabeth Mjelde, an art history professor, was interested in whether the value from the Golden Ratio formula, ( larger + smaller dimension larger dimension ) was the same in the Whitney Exhibit for works from 1900 to 1919 as for works from 1920 to 1942. Thirty-seven early works were sampled, averaging 1.74 with a standard deviation of 0.11. Sixty-five of the later works were sampled, averaging 1.746 with a standard deviation of 0.1064. Do you think that there is a significant difference in the Golden Ratio calculation? A recent year was randomly picked from 1985 to the present. In that year, there were 2,051 Hispanic/Latino students at Cabrillo College out of a total of 12,328 students. At Lake Tahoe College, there were 321 Hispanic/Latino students out of a total of 2,441 students. In general, do you think that the percent of Hispanic/Latino students at the two colleges is basically the same or different? Subscripts: 1 = Cabrillo College, 2 = Lake Tahoe College H 0 : p 1 = p 2 H a : p 1 ≠ p 2 The random variable is the difference between the proportions of Hispanic/Latino students at Cabrillo College and Lake Tahoe College. normal for two proportions test statistic: 4.29 p -value: 0.00002 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for decision: p -value < alpha Conclusion: There is sufficient evidence to conclude that the proportions of Hispanic/Latino students at Cabrillo College and Lake Tahoe College are different. Use the following information to answer the next three exercises. Some individuals who were exposed to the COVID-19 virus experienced a loss of taste. In a sample of COVID patients during 2022, there were 629 reported cases of loss of taste out of a total of 1,021 reported cases. In 2021, there were 486 reported cases of loss of taste out of a sample of 712 cases. Is the 2022 proportion of loss of taste more than the 2021 proportion? Using a 1% level of significance, conduct an appropriate hypothesis test. “2022” subscript: 2022 group. “2021” subscript: 2021 group This is: a test of two proportions a test of two independent means a test of a single mean a test of matched pairs. An appropriate null hypothesis is: p 2022 ≤ p 2021 p 2022 ≥ p 2021 μ 2022 ≤ μ 2021 p 2022 > p 2021 a The p -value is 0.0022. At a 1% level of significance, the appropriate conclusion is There is sufficient evidence to conclude that the proportion of people in the United States in 2022 who experienced loss of taste is less than the proportion of people in the United States in 2021 who experienced loss of taste. There is insufficient evidence to conclude that the proportion of people in the United States in 2022 who experienced loss of taste is more than the proportion of people in the United States in 2021 who experienced loss of taste. There is insufficient evidence to conclude that the proportion of people in the United States in 2022 who experienced loss of taste is less than the proportion of people in the United States in 2021 who experienced loss of taste. There is sufficient evidence to conclude that the proportion of people in the United States in 2022 who experienced loss of taste is more than the proportion of people in the United States in 2021 who experienced loss of taste. Researchers conducted a study to find out if there is a difference in the use of eReaders by different age groups. Randomly selected participants were divided into two age groups. In the 16- to 29-year-old group, 7% of the 628 surveyed use eReaders, while 11% of the 2,309 participants 30 years old and older use eReaders. Test: two independent sample proportions. Random variable: p ′ 1 - p ′ 2 Distribution: H 0 : p 1 = p 2 H a : p 1 ≠ p 2 The proportion of eReader users is different for the 16- to 29-year-old users from that of the 30 and older users. Graph: two-tailed p -value : 0.0033 Decision: Reject the null hypothesis. Conclusion: At the 5% level of significance, from the sample data, there is sufficient evidence to conclude that the proportion of eReader users 16 to 29 years old is different from the proportion of eReader users 30 and older. Adults aged 18 years old and older were randomly selected for a survey on working from home. The researchers wanted to determine if the proportion of women who work from home is less than the proportion of men who work from home. The results are shown in . Test at the 1% level of significance. Number who work from home Sample size Men 42,769 155,525 Women 67,169 248,775 Two computer users were discussing tablet computers. A higher proportion of people ages 16 to 29 use tablets than the proportion of people age 30 and older. details the number of tablet owners for each age group. Test at the 1% level of significance. 16–29 year olds 30 years old and older Own a Tablet 69 231 Sample Size 628 2,309 Test: two independent sample proportions Random variable: p′ 1 − p′ 2 Distribution: H 0 : p 1 = p 2 H a : p 1 > p 2 A higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. Graph: right-tailed p -value: 0.2354 Decision: Do not reject the H 0 . Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that a higher proportion of tablet owners are aged 16 to 29 years old than are 30 years old and older. A group of friends debated whether more people aged 21-30 use wearable fitness devices than people aged 31-40. They consulted a research study of wearable fitness devices use among adults. The results of the survey indicate that of the 973 randomly sampled people in their twenties, 379 use wearable fitness devices. For people in their thirties, 404 of the 1,304 who were randomly sampled use wearable fitness devices. Test at the 5% level of significance. While her husband spent 2½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 67 men, 24 said they enjoyed the activity. Eight of the 24 women surveyed claimed to enjoy the activity. Interpret the results of the survey. Subscripts: 1: men; 2: women H 0 : p 1 ≤ p 2 H a : p 1 > p 2 P ′ 1 − P ′ 2 is the difference between the proportions of men and women who enjoy shopping for electronic equipment. normal for two proportions test statistic: 0.22 p -value: 0.4133 Answers may vary. Alpha: 0.05 Decision: Do not reject the null hypothesis. Reason for Decision: p -value > alpha Conclusion: At the 5% significance level, there is insufficient evidence to conclude that the proportion of men who enjoy shopping for electronic equipment is more than the proportion of women. We are interested in whether children’s educational computer software costs less, on average, than children’s entertainment software. Thirty-six educational software titles were randomly picked from a catalog. The mean cost was $31.14 with a standard deviation of $4.69. Thirty-five entertainment software titles were randomly picked from the same catalog. The mean cost was $33.86 with a standard deviation of $10.87. Decide whether children’s educational software costs less, on average, than children’s entertainment software. Joan Nguyen recently claimed that the proportion of college-age men with at least one pierced ear is as high as the proportion of college-age women. She conducted a survey in her classes. Out of 107 men, 20 had at least one pierced ear. Out of 92 women, 47 had at least one pierced ear. Do you believe that the proportion of men has reached the proportion of women? H 0 : p 1 = p 2 H a : p 1 ≠ p 2 P ′ 1 − P ′ 2 is the difference between the proportions of men and women that have at least one pierced ear. normal for two proportions test statistic: –4.82 p -value: zero Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% significance level, there is sufficient evidence to conclude that the proportions of men and women with at least one pierced ear is different. Use the data sets found in to answer this exercise. Is the proportion of race laps Terri completes slower than 130 seconds less than the proportion of practice laps she completes slower than 135 seconds? "To Breakfast or Not to Breakfast?" by Richard Ayore In the American society, birthdays are one of those days that everyone looks forward to. People of different ages and peer groups gather to mark the 18th, 20th, …, birthdays. During this time, one looks back to see what they have achieved for the past year and also focuses ahead for more to come. If, by any chance, I am invited to one of these parties, my experience is always different. Instead of dancing around with my friends while the music is booming, I get carried away by memories of my family back home in Kenya. I remember the good times I had with my brothers and sister while we did our daily routine. Every morning, I remember we went to the shamba (garden) to weed our crops. I remember one day arguing with my brother as to why he always remained behind just to join us an hour later. In his defense, he said that he preferred waiting for breakfast before he came to weed. He said, “This is why I always work more hours than you guys!” And so, to prove him wrong or right, we decided to give it a try. One day we went to work as usual without breakfast, and recorded the time we could work before getting tired and stopping. On the next day, we all ate breakfast before going to work. We recorded how long we worked again before getting tired and stopping. Of interest was our mean increase in work time. Though not sure, my brother insisted that it was more than two hours. Using the data in , solve our problem. Work hours with breakfast Work hours without breakfast 8 6 7 5 9 5 5 4 9 7 8 7 10 7 7 5 6 6 9 5 H 0 : µ d = 0 H a : µ d > 0 The random variable X d is the mean difference in work times on days when eating breakfast and on days when not eating breakfast. t 9 test statistic: 4.8963 p -value: 0.0004 Answers may vary. Alpha: 0.05 Decision: Reject the null hypothesis. Reason for Decision: p -value < alpha Conclusion: At the 5% level of significance, there is sufficient evidence to conclude that the mean difference in work times on days when eating breakfast and on days when not eating breakfast has increased. Pooled Proportion estimate of the common value of p 1 and p 2 .

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