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.