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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.Ho:μ≥10Ha:μ<10Test statistic: z=9-102.8/50=-2.525Critical value: zcr=-1.645Reject the null hypothesis. We have sufficient evidence to conclude that those joining Weight Reducers on average will lose less than 10 pounds.p-value = 0.005832. 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 qualitycontrol 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.Ho:μ≤16Ha:μ>16Test statistic: z=16.05-160.03/50=11.785Critical value: zcr=1.645Reject the null hypothesis. We have sufficient evidence to conclude that the mean weight is greater than 16 ounces.p-value = 0.000038. 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 5.6At 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.Ho:μ≥6Ha:μ<6x=5.638s=0.635Test statistic: t=5.638-60.635/8=-1.616degree of freedom = 8 – 1 = 7Critical value: tcr=-2.998We cannot reject the null hypothesis. We don’t have sufficient evidence to conclude that the 30-year mortgage rate for small banks is less than 6 percent.p-value = 0.075127. 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 Men WomenSample mean 24.51 22.69Population standard deviation 4.48 3.86Sample size 35 40At 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?Ho:μ1-μ2=0Ha:μ1-μ2≠0Test statistic: z=24.51-22.694.48235+3.86240=1.871Critical value: zcr=2.326We cannot reject the null hypothesis. We don’t have sufficient evidence to conclude that there is a difference in the mean number of times men and women order take-out dinners in a month.p-value = 0.03146. 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 Waiting TimeLittle River 31.73 28.77 29.53 22.08 29.47 18.60 32.94 25.18 29.82 26.49Murrells Inlet 22.93 23.92 26.92 27.20 26.44 25.62 30.61 29.44 23.09 23.10 26.69 22.31Assume the population standard deviations are not the same. At the .05 significance level, is there a difference in the mean waiting time?Ho:μ1-μ2=0Ha:μ1-μ2≠0x1=27.461s1=4.440n1=10x2=25.689s2=2.685n2=12Test statistic: t=27.461-25.6894.440210+2.685212=1.105degree of freedom = 14Critical value: tcr=?2.145We cannot reject the null hypothesis. We don’t have sufficient evidence to conclude that there is a difference in the mean waiting times.52. The president of the American Insurance Institute wants to compare the yearly costs ofauto 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 eachfamily a stipend to contact the two companies and ask for a price quote. To make thedata 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 GEICOFamilyCar Insurance Mutual InsuranceBecker $2,090 $1,610Berry 1,683 1,247Cobb 1,402 2,327Debuck 1,830 1,367DuBrul 930 1,461Eckroate 697 1,789German 1,741 1,621Glasson 1,129 1,914King 1,018 1,956Kucic 1,881 1,772Meredith 1,571 1,375Obeid 874 1,527Price 1,579 1,767Phillips1,577 1,636Tresize 860 1,188Ho:μD=0Ha:μD≠0xD=246.33sD=546.96Test statistic: t=246.33-0546.96/15=1.744Degree of freedom = 15 – 1 = 14Critical value: tcr=1.761We cannot reject the null hypothesis. We don’t have sufficient evidence to conclude 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?Ho:σ12≤σ22Ha:σ12>σ22Test statistic: F=45,600221,3302=4.57degree of freedom numerator = 21 – 1 = 20degree of freedom denominator 18 – 1 = 17Critical value: Fcr=3.607We reject the null hypothesis. We have sufficient evidence to conclude that there is more variation in the selling prices of the oceanfront homes.28. The following is a partial ANOVA table.Sum of MeanSource Squares df Square FTreatment 32021608Error 280920 Total 500 11Complete the table and answer the following questions. Use the .05 significance level.a. How many treatments are there?b. What is the total sample size?c. What is the critical value of F?d. Write out the null and alternate hypotheses.e. What is your conclusion regarding the null hypothesis?a. 3b. 12c. 4.256d. Ho: μ1 = μ2 = μ3Ha: at least one of the treatment means is differente. 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?Ho: There is no difference Ha: There is differenceExpected number of viewers =1503=50Test statistic: X2=53-50250+64-50250+33-50250=9.88Degree of freedom = 3 – 1 = 2Critical value: Xcr2=5.991We reject the null hypothesis. We have sufficient evidence to conclude that there is a difference in the proportion of viewers watching the three channels.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 FrequencyMain Street 140Broad Street 120Cherry Street 90Walnut Street 50Total 400Ho: There is no difference Ha: There is differenceExpected number of viewers =4004=100Test statistic: X2=140-1002100+120-1002100+90-1002100+50-1002100=46Degree of freedom = 4 – 1 = 3Critical value: Xcr2=11.34We reject the null hypothesis. We have sufficient evidence to conclude that there is a difference in the use of the four entrances. ................
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