Qnt561 One and Two Samples Sets Week 5 free essay sample

One- and Two-Sample Tests of Hypothesis, Variance, and Chi-squared Analysis Problem Sets University of Phoenix Applied Business Research and Statistics QNT 561 August 5, 2011 Chapter 10 Exercise Question 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 To calculate the test statistics: Z=(9-10)/(2. 8/sqrt(50))= -2. 525 From the z-table, we find P9z-2. 525)=0. 0058 So we reject the null hypothesis. There is strong evidence to suggest that the average weight loss at Weight Reducers is less than 10 pounds. Exercise Question 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. Ho: ? ? 16| | | | | Ha: ? 16| | | | | ? = 0. 05| | | | | critical value z = 1. 645| | | | test statistic = (16. 05-16)/(0. 03/sqrt(50)) = 11. 785| p-value=0. 0000 We reject the null hypothesis since the test statistic is greater than the critical value. We have sufficient evidence to conclude that the mean weight is greater than 16 ounces. Exercise Question 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 rate (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. The p-value is 7. 5% which is greater than significance level of 1%. We fail to reject the null hypothesis but we can’t conclude that the rates are less than 6%. Chapter 11 Exercise Question 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: StatisticMen Women Sample Mean24. 5122. 69 Population standard 4. 483. 86 Deviation Sample Size3540 At the . 01 significance level, is there a difference in the mean number of time mean and women order take-out dinners in a month? What is the p-value? The test statistic: (24. 51-22. 69)/sqrt(4. 48^2/35 + 3. 86^2/40) = 1. 871 The p-value is about 0. 06 This is well above the 0. 01 significance, so we do not reject the null hypothesis. There is no statistic difference. Exercise Question 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| Waiting Time| Little River| 31. 73 28. 77 29. 53 22. 08 29. 47 18. 60 32. 94 25. 18 29. 82 26. 49| Murrells Inlet| 22. 93 23. 92 26. 92 27. 20 26. 44 25. 62 30. 61 29. 44 23. 09 23. 10 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? P(T=t) one tail 0. 143947512 P(T=t) two-tail 0. 287895024558682 Since the p value is much higher than 0. 05, we do not reject the null hypothesis. The means are statistically equal. Exercise Question 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? Geico Mutual Insurance Family| Progressive Car Insurance| GEICO Mutual Insurance| Becker| 2,090| 1,610| Berry| 1,683| 1,247| Cobb| 1,402| 2,327| Debuck| 1,830| 1,367| DuBrul| 930| 1,461| Eckroate| 697| 1,789| German| 1,741| 1,621| Glasson| 1,129| 1,914| King| 1,018| 1,956| Kucic| 1,881| 1,772| Meredith| 1,571| 1,375| Obeid| 874| 1,527| Price| 1,579| 1,767| Phillips| 1,577| 1,636| Tresize| 860| 1,188| Chapter 12 Exercise Question 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 process 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. As the . 1 significance level, can we conclude that there is more variation in the selling prices of the oceanfront homes? The critical F value, with 0. 01 significance is: 3. 1615 The test statistic: sigma1^2 / sigma2^2 = 45600^2/21330^2 = 4. 57033 The test statistic is greater than the critical value, so we reject the null hypothesis and conclude that the variances are different. Exercise Question 28: The following is a pa rtial ANOVA table. Complete the table and answer the following questions. Use the . 05 significance level. a. How many treatments are there? 2+1=3 b. What is the total sample size? 1+1=12 c. What is the critical value of F? 4. 2565 d. Write out the null and alternate hypotheses. H0: means are equal; Ha: means are not equal e. What is your conclusion regarding the null hypothesis? Since our F statistic (8) is greater than the critical value, we reject the null. The means are different. Source| Sum of Squares| Df| Mean Square| F| Treatment| 320| 2| 160| 8| Error| 180| 9| 20| | Total| 500| 11| | | Chapter 17 Exercise Question 19: In a particular market there are three commercial television stations, each with its own evening news program from 6:00 to 6:30PM. 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 portion of views watching the three channels? The statistic is much higher than the critical value, so we reject the null hypothesis. There is a difference between the 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| Frequency| Main Street| 140| Broad Street| 120| Cherry Street| 90| Walnut Street| 50| Total| 400| Since the calculated Chi square – statistics is equal to 46. 00, which are greater than 11. 3449, the we reject the null hypothesis because there is sufficient evidence at a 1% level of significance that there is a difference in the use of the four entrances.