1: Solution: a) Here the claim is that, is there evidence of a difference in the mean waiting times between the two offices. Under the claim the null and alternative hypotheses are, So, the correct option is (C). Given level of significance is To test the hypothesis the test statistic is given by, The required calculations are as follows: From the above output the test statistic value is The test is two tailed test, from t-table values at 0.05 level and at 38 degrees of freedom the critical values are -2.0244, 2.0244. Since, the calculated test statistic value does not lies in the critical region we fail to reject the null hypothesis and conclude that there is insufficient evidence that the means are differ. So the correct option is (C). b) From the output in part (a), we have the calculated P-value is 0.7104. The P-value is the probability of getting a test statistic equal to or more extreme than the sample results if there is no difference in the mean problem clearing time of the two offices. So the correct option is (B) c) Since the sample sizes are both less than 30, it must be assumed that both sampled populations are approximately normal. So the correct option is (A). d) Construct the 95% confidence interval for the population mean difference. The confidence interval for the population mean difference is given by, The required calculations are as shown below: From the above output we have the 95% confidence interval for the men difference is From the 95% confidence interval we can conclude that, we are 95% confidence that the population mean time difference will lies between -0.92 and 1.34. So the correct option is (A). Question 2: Solution: a) From the given information we have, Under the given claim that the null and alternative hypotheses are, So the correct option is (A). Given level of significance is To test the hypothesis the test statistic is given by, The required calculations are as shown below: From the above output we have The test is two tailed test, from the normal area table values, at 0.10 level of significance the critical values are -1.64, 1.64. Since the calculated test statistic value is lies in the critical region we reject the null hypothesis. There is sufficient evidence to support the claim that there is a significance difference in the proportion of users under age 50 and users 50 years and older that accessed the news on their cell phones. b) From the output in part (a), we have the calculated P-value is 0.0000. The probability of obtaining a difference in proportions that gives rise to a test statistic that deviates from the difference 0 to 13.1087 is equal to the P-value of 0.000 if there is no difference between the population proportions of users under age 50 and users 50 years and older that accessed the news on their cell phones. c) Construct the 90% confidence interval for difference of the proportions. The confidence interval for the difference of population proportions is given by, The required calculations are as shown below: From the above output we have the 90% confidence interval is (0.2284, 0.2916). The researchers performing this study can be 90% confident that the difference in the population proportion of users under age 50 and users 50 years and older that accessed the news on their cell phones is lies within the interval from 0.2284 to 0.2916. Question 3: Solution: a) Under this test the null and alternative hypotheses are, So the correct option is (D). Given level of significance is To find the required test statistic values we have to perform one-way ANOVA. The required calculations are shown below: From the above output, the test statistic is P-value of the test statistic is. Since the P-value of the test statistic is less than the given level of significance, reject the null hypothesis. There is enough evidence to conclude that there is a difference in the mean waiting time in the four locations. b) The post hoc analysis: From the above output, observe the post hoc analysis main campus, satellite-II and main campus, satellite-I highly differ in the mean weighting time. So the correct option is B and E. c) Here we have to test the difference in variation. Under this test the null and alternative hypotheses are, Given level of significance is To find the required test statistic values we have to perform one-way ANOVA. The required calculations are shown below From the above output we have the test statistic value as and the corresponding P-value is 0.7325. Since the P-value is greater than the level of significance we fail to reject the null hypothesis. There is no sufficient evidence to conclude that there is a difference in the vitiation in waiting time in the four locations. Question 4: Solution: From the given scenario, the null and alternate hypothesis can be stated as follows: So the correct option is (C). Given level of significance is Calculate F test statistic by using the following formula. The required calculations are shown below: From the above output we have the test statistic value as Since the test is two tailed test, at 0.05 level the critical value of F is 9.60. Conclusion: Since the calculated test statistic does not lie in the critical region we fail to reject the null hypothesis. There is no sufficient of difference in the variance of the yield between money market accounts and five-year CDs. Question 6: Solution: a) Under this test the null and alternative hypotheses are, So the correct option is (B). Given level of significance is To find the required test statistic values we have to perform one-way ANOVA. The required calculations are shown below: From the above output we have the test statistic value is Critical value Since the calculated test statistic value is lies in the critical region we reject the null hypothesis . There is sufficient evidence of a difference in the mean amount of food eaten among the various products. b) The post hoc analysis: From the above output we have, No, there is no sufficient evidence that group 1 and group 2 differ in the mean amount. No, there is no sufficient evidence that group 1 and group 3 differ in the mean amount. Yes, there is sufficient evidence that group 1 and group 4 differ in the mean amount. Yes, there is sufficient evidence that group 1 and group 5 differ in the mean amount. Yes, there is sufficient evidence that group 2 and group 4 differ in the mean amount. Yes, there is sufficient evidence that group 2 and group 5 differ in the mean amount. Yes, there is sufficient evidence that group 3 and group 4 differ in the mean amount. Yes, there is sufficient evidence that group 3 and group 5 differ in the mean amount. No, there is no sufficient evidence that group 4 and group 5 differ in the mean amount. c) Here we have to test the difference in variation. Under this test the null and alternative hypotheses are, So the correct option is (B). Given level of significance is To find the required test statistic values we have to perform one-way ANOVA. The required calculations are shown below From the above output we have the test statistic value as and the corresponding P-value is 0.7325. Critical value Since the calculated test statistic value is does not lies in the critical region we fail to reject the null hypothesis . There is no sufficient evidence that there is a difference in the variation in the amount of food eaten among the various products. d) The correct option is (A). Question 7: Solution: a) Under this test the null and alternative hypotheses are, So the correct option is (F). To test the hypotheses the test statistic is given by, The required calculations are shown below From the above output, we have the test statistic value as P-value: 0.793. Since the P-value is greater than the value of , we fail to reject the null hypothesis. There is no sufficient evidence of a difference in the mean measurements. b) The correct option is (B). c) Box plot for the differences The correct option is (B). No, the boxplot indicates skewness. d) Construct the 95% confidence interval: Form the output in part (a) we have the required confidence interval as, The correct option is (C). Question 8: Solution: a) From the given information we have, Under the given claim that the null and alternative hypotheses are, So the correct option is (D). Given level of significance is To test the hypothesis the test statistic is given by, The required calculations are as shown below: From the above output we have Rejection region: Correct option is (A). Conclusion: Correct option is (C). b) Construct the 99% confidence interval for difference of the proportions. The confidence interval for the difference of population proportions is given by, The required calculations are as shown below: From the above output we have the 99% confidence interval is (0.0450, 0.4550). Question 5: Solution: For this problem there is no data set given