Question

FILE Industrial. A manager of an industrial plant asserts that workers on average do not complete a job using Method A in the same amount of time as they would using Method B. Seven workers are randomly selected. Each worker's completion time [in minutes) is recorded by the use of Method A and Method B. A portion of the data is shown in the accompanying table. a. Specify the null and altemative hypotheses to test the manager's assertion. b. Assuming that the completion time difference is normally distributed, calculate the value of the test statistic. c. Find the p-value. d. At the 10% significance level, is the manager's assertion supported by the data?

Answer 1

a. The null and alternative hypotheses to test the manager's assertion are:Null hypothesis: The time taken to complete the work is the same for both Method A and Method B.Alternative hypothesis: The time taken to complete the work is different for both Method A and Method B.b. To find the value of the test statistic, the formula is:$$\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}$$Substituting the values from the table, we have,$$\frac{(22.3-23.5)-(0)}{\sqrt{\frac{2.1^2}{7}+\frac{1.6^2}{7}}}=-1.27$$Therefore, the value of the test statistic is -1.27.c. To find the p-value, we use the z-score table since the sample size is less than 30. At a significance level of 10%, the critical value is -1.645. Since the calculated test statistic of -1.27 is greater than the critical value of -1.645, we fail to reject the null hypothesis. The p-value is the area to the right of the test statistic in the z-score table, which is 0.102. Therefore, the p-value is 0.102.d. At the 10% significance level, the null hypothesis is not rejected. Therefore, the manager's assertion is not supported by the data.