Statistical Techniques in Business and Economics - Essay Example

Chapter 10 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. Ans) Hypothesis Formulation: The null hypothesis is that the population mean is 10 pounds. The alternate hypothesis is that the population mean is less than 10 pounds. We can express the null and alternate hypotheses as follows: Significance Level: The significance level is 5% or Decision rule: We will reject the null hypothesis if the z-test statistic is less than -1.645 (z < 1.645) since the test is one-tailed (Left tailed) and we want to determine whether there has been a reduction in the weight of individuals who joined Weight Reducers International. Test Statistic: Conclusion: Since the z-test statistic is less than -1.645 therefore at the 5% significance level, we have sufficient evidence to conclude that those who will join Weight Reducers International on average will lose less than 10 pounds. P-Value: The minimum significance level (p-value) at which the null hypothesis can be rejected is (0.5-0.4943) = 0.57% 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. Ans) Hypothesis Formulation: The null hypothesis is that the population mean is 16 ounces. The alternate hypothesis is that the population mean is less than 16 ounces. We can express the null and alternate hypotheses as follows: Significance Level: The significance level is 5% or Decision rule: We will reject the null hypothesis if the z-test statistic is greater than 1.645 (z > 1.645) since the test is one-tailed (Right tailed) and we want to determine whether 16-ounce can of sliced pineapple is being overfilled. Test Statistic: Conclusion: Since the z-test statistic is greater than 1.645 therefore at the 5% significance level, we have sufficient evidence to conclude that 16-ounce can of sliced pineapple is being overfilled. P-Value: The minimum significance level (p-value) at which the null hypothesis can be rejected is (0.5-0.5) = 0.00% 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. Ans) Hypothesis Formulation: The null hypothesis is that the population mean is 6 percent. The alternate hypothesis is that the population mean is less than 6 percent. We can express the null and alternate hypotheses as follows: Significance Level: The significance level is 1% or Decision rule: The t-test statistic at 1 % significance level with df = (n – 1) = 8 – 1 = 7 We will reject the null hypothesis if t-test statistic is less than -2.998 since the test is one-tailed (left tailed) and we want to determine whether the 30-year mortgage rate is now less than 6 percent Test Statistic: Conclusion: Since the t-test statistic is greater than -2.998 therefore we will not reject the null hypothesis. At 1% significance level we have sufficient evidence to conclude that the 30-year mortgage rate is not less than 6 percent. P-Value: The minimum significance level (p-value) at which the null hypothesis can be rejected is between 5% and 10% since the value -1.61 lies in between -1.415 and -1.895 for a t-distribution with degrees of freedom = 7 Chapter 11 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 Men Women Sample mean 24.51 22.69 Population standard deviation 4.48 3.86 Sample Size 35 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? Ans) Hypothesis Formulation: The null hypothesis is that there is no difference in the population mean number of times men and women buy take-out dinner in a month. The alternate hypothesis is that there is difference in the population mean number of times men and women buy take-out dinner in a month. We can express the null and alternate hypotheses as follows: Significance Level: The significance level is 1% or Decision rule: We will reject the null hypothesis if the z-test statistic is lesser than -2.576 or greater than 2.576 since it is a two tailed test. Test Statistic: Conclusion: Since the z-test statistic 1.87 does not fall in the rejection region therefore we will not reject the null hypothesis. At the 1% significance level, we have sufficient evidence to conclude there is no difference in the population mean number of times men and women buy take-out dinner in a month. P-Value: The minimum significance level (p-value) at which the null hypothesis could have been rejected is (0.5-0.4693) x 2 = 6.14% 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? Ans) Hypothesis Formulation: The null hypothesis is that there is no difference in the population mean waiting times in Little River & Murrells Inlet. The alternate hypothesis is that there is difference in the population mean waiting times in Little River & Murrells Inlet. We can express the null and alternate hypotheses as follows: Significance Level: The significance level is 5% or Decision rule: The t-test statistic with df = 14 is We will reject the null hypothesis if it is is lesser than -2.145 or greater than 2.145 since it is a two tailed test. Test Statistic: Conclusion: Since the t-test statistic 1.11 does not fall in the rejection region therefore we will not reject the null hypothesis. At 5% significance level, we have sufficient evidence to conclude there is no difference in the population mean waiting times of Little River & Murrells Inlet P-Value: The minimum significance level (p-value) at which the null hypothesis could have been rejected is less than 10%. 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? 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 Ans) Hypothesis Formulation: The null hypothesis is that there is no difference in the population mean yearly costs of auto insurance offered by two leading companies. The alternate hypothesis is that there is difference in the population mean yearly costs of auto insurance offered by two leading companies. We can express the null and alternate hypotheses as follows: Significance Level: The significance level is 10% or Decision rule: The t-test statistic with df = n – 1 = 15 – 1 = 14 is We will reject the null hypothesis if t-test statistic is lesser than -1.761 or greater than 1.761 since it is a two tailed test. Test Statistic: Family Difference Becker $480 Berry $436 Cobb ($925) Debuck $463 DuBrul ($531) Eckroate ($1,092) German $120 Glasson ($785) King ($938) Kucic $109 Meredith $196 Obeid ($653) Price ($188) Phillips ($59) Tresize ($328) Conclusion: Since the t-test statistic 1.744 does not fall in the rejection region therefore we will not reject the null hypothesis. At 10% significance level, we have sufficient evidence to conclude that yearly costs of auto insurance offered by two leading companies are not different. P-Value: The minimum significance level (p-value) at which the null hypothesis could have been rejected is less than 10%. Chapter 12 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? Ans) Hypothesis Formulation: The null hypothesis is that there is no variation in the selling price of homes on the oceanfront with those one to three blocks from the ocean. The alternate hypothesis is that there is variation in the selling price of homes on the oceanfront with those one to three blocks from the ocean. We can express the null and alternate hypotheses as follows: Significance Level: The significance level is 1% or Decision rule: The F-test statistic with degrees of freedom in the numerator = = 21 – 1 = 20 and degrees of freedom in the denominators are = 18 – 1 = 17 We will reject the null hypothesis if the f-test statistic is greater than 3.16 Test Statistic: Conclusion: Since the F-test statistic 4.57 is greater than 3.16 therefore we will reject the null hypothesis. At the 1% significance level, we have sufficient evidence to conclude that there is more variation in the selling price of homes on the oceanfront with those one to three blocks from the ocean. 28. The following is a partial ANOVA table. Source Sum of Squares df Mean Square F Treatment 2 Error 20 Total 500 11 Complete 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? Ans) a) k – 1 = 2 k = 3 There are three treatments in the hypothesis. b) n – 1 = 11 n = 12 The total sample size is 12. c) The critical values for F-test statistic at 5% significance level is with degrees of freedom in numerator = k – 1 = 3 – 1 = 2 and degrees of freedom in the denominator = n – k = 12 – 3 = 9 F-test statistic = 4.26 Reject the null hypothesis if the F-test statistic is greater than 4.26. d) e) SSE / n – k = MSE MSE = 20 SSE = 20 x (12 – 3) SSE = 180 SST = 500 – 180 = 320 MST = SST / k – 1 MST = 320 / 2 MST = 160 F-test statistic = MST / MSE = 160 / 20 = 8 Since the F-test statistic is greater than 4.26 therefore we will reject the null hypothesis. At 5% significance level we have sufficient evidence to conclude that the mean scores are not equal. Source Sum of Squares df Mean Square F Treatment 320 2 160 8 Error 180 9 20 Total 500 11 Chapter 17 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? Ans) Hypothesis Formulation: Significance Level: The significance level is 5% or Decision rule: The Chi square test statistic with df = 3 – 1 = 2 is We will reject the null hypothesis if Chi square test statistic is greater than 5.991 Test Statistic: Channel Number of Viewers (fo) Expected Frequency (fe) WNAE 53 50 WRRN 64 50 WSPD 33 50 Total 150 150 Conclusion: Since the Chi- square test statistic is greater than 5.991 therefore we will reject the null hypothesis. At 5% significance level 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 Frequency Main Street 140 Broad Street 120 Cherry Street 90 Walnut Street 50 Total 400 Ans) Hypothesis Formulation: Significance Level: The significance level is 1% or Decision rule: The Chi square test statistic with df = 4 – 1 = 3 is We will reject the null hypothesis if Chi square test statistic is greater than 11.345 Test Statistic: Entrance Frequency (fo) Expected Frequency(fe) Main Street 140 100 Broad Street 120 100 Cherry Street 90 100 Walnut Street 50 100 Total 400 400 Conclusion: Since the Chi- square test statistic is greater than 11.345 therefore we will reject the null hypothesis. At 1% significance level we have sufficient evidence to conclude that there is a difference in the utilization of each entrance. References: Lind, D. A., Marchal, W. G. & Wathen, S. A. (2008). Statistical Techniques in Business & Economics, NY: McGraw Hill. Read More