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31. A new weight-watching company, Weight Reducers International, advertises that those who join will lose, on the average, 10 pounds the first two weeks with a standard deviation of 2.8 pounds. A random sample of 50 people who joined the new weight reduction program revealed the mean loss to be 9 pounds. At the .05 level of significance, can we conclude that those joining Weight Reducers on average will lose less than 10 pounds? Determine the p-value. Hypotheses: H0: µ ≥ 10 Ha: µ < 10 (Claim) Critical value: For a left-sided, one-tailed test with = 0.05, z(crit) = -1.645 Test value: z(test) = (9 – 10) / (2.8 / √50) = -2.5254 p-value = 0.0058 Decision: z(test) < z(crit) Reject the null hypothesis Summary: There is sufficient evidence at the 0.05 level of significance to support the claim that those joining Weight Reducers will lose less than 10 pounds on average. 32. Dole Pineapple, Inc., is concerned that the 16-ounce can of sliced pineapple is being overfilled. Assume the standard deviation of the process is .03 ounces. The quality control department took a random sample of 50 cans and found that the arithmetic mean weight was 16.05 ounces. At the 5 percent level of significance, can we conclude that the mean weight is greater than 16 ounces? Determine the p-value. Hypotheses: H0: µ ≤ 16 Ha: µ > 16 (Claim) Critical value: For a right-sided, one-tailed test with = 0.05, z(crit) = 1.645 Test value: z(test) = (16.05 – 16) / (0.03 / √50) = 11.7851 p-value = 0.00 (when rounded off) Decision: z(test) > z(crit) Reject the null hypothesis Summary: There is sufficient evidence at the 0.05 level of significance to support the claim that the mean weight is greater than 16 ounces. 38. A recent article in The Wall Street Journal reported that the 30-year mortgage rate is now less than 6 percent. A sample of eight small banks in the Midwest revealed the following 30-year rates (in percent): 4.8 5.3 6.5 4.8 6.1 5.8 6.2 5.6 At the .01 significance level, can we conclude that the 30-year mortgage rate for small banks is less than 6 percent? Estimate the p-value. Mean: x-bar = (4.8 + 5.3 + 6.5 + 4.8 + 6.1 + 5.8 + 6.2 + 5.6) / 8 = 5.6375 Standard Deviation: s = 0.6346 Hypotheses: H0: µ ≥ 6 Ha: µ < 6 (Claim) Critical value: For a left-sided, one-tailed test with t(crit) = -2.998 = 0.01, and 7 degrees of freedom, Test value: z(test) = (5.6375 – 6) / (0.6346 / √8) = -1.6157 p-value = 0.0751 Decision: z(test) > z(crit) Do not reject the null hypothesis Summary: There is not sufficient evidence at the 0.01 level of significance to support the claim that the mean 30-year mortgage rate for small banks is less than 6%. 27. A recent study focused on the number of times men and women who live alone buy take-out dinner in a month. The information is summarized below. Statistic Sample mean Population standard deviation Sample size Men 24.51 3.86 35 4.48 Women 22.69 40 At the .01 significance level, is there a difference in the mean number of times men and women order take-out dinners in a month? What is the p-value? Hypotheses: H0: µ1 = µ2 Ha: µ1 ≠ µ2 (Claim) Critical value: For a two-tailed test with = 0.01, z(crit) = ±2.576 Test value: ztest x1 x 2 2 1 2 s s 2 n1 n 2 24.51 22.69 4.48 2 3.86 2 35 40 1.8713 p-value = 0.0613 Decision: z(test) < z(crit) Do not reject the null hypothesis Summary: There is not sufficient evidence at the 0.01 level of significance to support the claim that there is a difference in the mean number of times men and women order take-out dinners in a month. 46. Grand Strand Family Medical Center is specifically set up to treat minor medical emergencies for visitors to the Myrtle Beach area. There are two facilities, one in the Little River Area and the other in Murrells Inlet. The Quality Assurance Department wishes to compare the mean waiting time for patients at the two locations. Samples of the waiting times, reported in minutes, follow: Location Little River Murrells Inlet Waiting Time 28.77 29.53 22.08 29.47 31.73 22.93 23.92 26.92 27.20 26.44 25.62 18.60 30.61 32.94 29.44 25.18 23.09 29.82 23.10 26.49 26.69 22.31 Assume the population standard deviations are not the same. At the .05 significance level, is there a difference in the mean waiting time? Little River : x1 27.5689 s1 4.6959 n1 9 Murrels Inlet : x2 25.9270 s2 2.6788 n 2 12 Hypotheses: H0: µ1 = µ2 Ha: µ1 ≠ µ2 (Claim) Critical value: For a two-tailed test with = 0.05 and 8 degrees of freedom, t(crit) = ± 2.306 Test value: t test x1 x 2 2 1 2 s s 2 n1 n 2 27.5689 25.9270 4.6959 2 2.6788 2 9 12 0.9404 p-value = 0.3745 Decision: t(test) < t(crit) Do not reject the null hypothesis Summary: There is not sufficient evidence at the 0.05 level of significance to support the claim that there is a difference in the mean waiting time. 52. The president of the American Insurance Institute wants to compare the yearly costs of auto insurance offered by two leading companies. He selects a sample of 15 families, some with only a single insured driver, others with several teenage drivers, and pays each family a stipend to contact the two companies and ask for a price quote. To make the data comparable, certain features, such as the deductible amount and limits of liability, are standardized. The sample information is reported below. At the .10 significance level, can we conclude that there is a difference in the amounts quoted? Progressive Car Insurance $2,090 1,683 1,402 1,830 930 697 1,741 1,129 1,018 1,881 1,571 874 1,579 1,577 860 Family Becker Berry Cobb Debuck DuBrul Eckroate German Glasson King Kucic Meredith Obeid Price Phillips Tresize Progressive : x1 1391.00 s1 437.7562 n1 15 GEICO Mutual Insurance $1,610 1,247 2,327 1,367 1,461 1,789 1,621 1,914 1,956 1,772 1,375 1,527 1,767 1,636 1,188 Geico : x2 1637 s2 299.0611 n 2 15 Hypotheses: H0: µ1 = µ2 Ha: µ1 ≠ µ2 (Claim) Critical value: Rather than testing the variances to determine if they are equal or unequal, which is controversial according to some statisticians, simply calculate both: Assuming that the variances are unequal: For a two-tailed test with = 0.10 and 14 degrees of freedom, t(crit) = ±1.761 Assuming that the variances are equal: For a two-tailed test with = 0.10 and 28 degrees of freedom, t(crit) = ±1.701 Test value: Variances unequal: t test x1 x 2 s12 s2 2 n1 n 2 1391 1637 437.7562 2 299.06112 15 15 1.7971 p-value = 0.0939 Variances equal: t test x1 x 2 n1 1s12 n2 1s22 n1 n 2 2 t(test) 1 1 n1 n 2 1391 1637 14437.75622 14 299.06112 1 15 15 2 1.7971 1 15 15 p-value = 0.0831 Decision: Variances unequal: t(test) < t(crit) Reject the null hypothesis Variances equal: t(test) < t(crit) Reject the null hypothesis Summary: There is sufficient evidence at the 0.10 level of significance to support the claim that there is a difference in the amounts quoted. 23. A real estate agent in the coastal area of Georgia wants to compare the variation in the selling price of homes on the oceanfront with those one to three blocks from the ocean. A sample of 21 oceanfront homes sold within the last year revealed the standard deviation of the selling prices was $45,600. A sample of 18 homes, also sold within the last year, that were one to three blocks from the ocean revealed that the standard deviation was $21,330. At the .01 significance level, can we conclude that there is more variation in the selling prices of the oceanfront homes? Use the F-test to compare variances. The sample with the greater variance is designated as “sample 1” and this variance goes in the numerator of the calculated F-statistic. In this scenario, the oceanfront homes are designated as “sample 1”. Hypotheses: H0: s12 ≤ s22 Ha: s12 > s22 (Claim) Critical value: For a right-sided one-tailed test with and n2 = 18 – 1 = 17 = 0.01, n1 = 21 – 1 = 20, F(crit) = 3.16 Test value: F test s12 45,600 4.5703 s2 2 21,3302 2 Decision: F(test) > F(crit) Reject the null hypothesis Summary: There is sufficient evidence at the 0.01 level of significance to support the claim that there is more variation in the selling prices of the oceanfront homes. 28. The following is a partial ANOVA table. Source Treatment Error Total Sum of Squares df 2 Mean Square F 20 500 11 Complete the table and answer the following questions. Use the .05 significance level. The completed chart: Source Treatment Error Total Sum of Squares 320 180 500 df 2 9 11 Mean Square 160 20 F 8 a. How many treatments are there? There are 3 treatments b. What is the total sample size? The total sample size is 12 c. What is the critical value of F? The critical value of F is 4.26 d. Write out the null and alternate hypotheses. H0: µ1 = µ2 = µ3 Ha: At least one of the means is different from the others. (Claim) e. What is your conclusion regarding the null hypothesis? Since F(test) is greater than the critical value, reject the null hypothesis. 19. In a particular market there are three commercial television stations, each with its own evening news program from 6:00 to 6:30 P.M. According to a report in this morning’s local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on WNAE (channel 5), 64 watched on WRRN (channel 11), and 33 on WSPD (channel 13). At the .05 significance level, is there a difference in the proportion of viewers watching the three channels? If the proportions were all equal between the three tv channels, then they would each be equal to 1/3. Hypotheses: H0: p1 = p2 = p3 = 1/3 Ha: At least one proportion is different from 1/3 (Claim) Critical value: Using a Chi-Squared Test, the critical of freedom, at a = 0.05 is: 2 2 value for k – 1 = 3 – 1 = 2 degrees (crit) = 5.991 Test value: The expected number of viewer of each channel, if the proportions are equal is: np = (150)(1/3) = 50 The 2 test statistic is: 2 (test) = (53 – 50)2 / 50 + (64 – 50)2 / 50 + (33 – 50)2 / 50 = 9.88 Decision: 2 (test) > 2 (crit) Reject the null hypothesis Summary: There is sufficient evidence at the 0.05 level of significance to support the claim that there is a difference in the proportion of viewers watching each channel. 20. There are four entrances to the Government Center Building in downtown Philadelphia. The building maintenance supervisor would like to know if the entrances are equally utilized. To investigate, 400 people were observed entering the building. The number using each entrance is reported below. At the .01 significance level, is there a difference in the use of the four entrances? Entrance Main Street Broad Street Cherry Street Walnut Street Total Frequency 140 120 90 50 400 As in the previous problem, if the use of all four entrances was the same, then the proportion using each door would be the same. The expected value for each door should then be 400 /4 = 100 Hypotheses: H0: p1 = p2 = p3 = p4 Ha: At least one proportion is different from 1/4 (Claim) Critical value: Using a Chi-Squared Test, the critical freedom, at 2 2 value for k – 1 = 4 – 1 = 3 degrees of = 0.01 is: (crit) = 11.345 Test value: The 2 test statistic is: 2 (test) = (140 – 100)2 / 100 + (120 – 100)2 / 100 + (90 – 100)2 / 100 + (50 – 100)2 / 100 = 46 2 (test) > Decision: 2 (crit) Reject the null hypothesis Summary: There is sufficient evidence at the 0.05 level of significance to support the claim that there is a difference in the use of the four entrances.