p 2: Select one-tailed or two-tailed.One-tailedTwo-tailedStep 3: Enter the test statistic.(Round to 3 decimal places.)Step 4: Shade the area represented bythe p-value.Step 5: Enter the p-value.(Round to 3 decimal places.)0.3+0.20.1X(c) Based on your answer to part (b), choose what can be concluded, at the 0.05 level of significance, about the claim made by the company.o Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there isenough evidence to reject the claim that the mean number of minutes employees spend dealing with email isequal to 164.Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, thereis not enough evidence to reject the claim that the mean number of minutes employees spend dealing withemail is equal to 164.o Since the p-value is greater than the level of significance, the null hypothesis is rejected. So, there is enoughevidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to164.o Since the p-value is greater than the level of significance, the null hypothesis is not rejected. So, there is notenough evidence to reject the claim that the mean number of minutes employees spend dealing with email isequal to 164.xŚ A very large company is interested in its employees' productivity. The company reports from its historical data that its employees spend a mean of 164minutes per employee (on a typical day) dealing with email. To test this claim, an independent consultant chooses 26 employees at random and findsthat those employees spend a sample mean of 172 minutes dealing with email, with a sample standard deviation of 21 minutes. Assume that thepopulation of amounts of time employees spend dealing with email is approximately normally distributed.Complete the parts below to perform a hypothesis test to see if there is enough evidence, at the 0.05 level of significance, to reject the claim that μ, themean number of minutes employees spend dealing with email, is equal to 164.(a) State the null hypothesis H. and the alternative hypothesis H, that you would use for the test.Ho DH₁: 0HOO(b) Perform at test and find the p-value.Here is some information to help you with your / test.ローロ• The value of the test statistic is given by t= -x-μS√nThe p-value is two times the area under the curve to the right of the value of the test statistic.0*0Not equal to0.4+0.3+E

The sample size is 26, sample mean is 172, sample standard deviation is 21.(a) The claim is that μ the mean number of minutes employees spend dealing with email is equal to 164. The hypothesis is, Null hypothesis: H0:μ=164 Alternative hypothesis: H1:μ≠164The test statistic is, t=x¯-μsn=172-1642126=1.9425 Thus, the test statistic is 1.9425. The degrees of freedom is, df=n-1=26-1=25 Since the test is two-tailed, by using the EXCEL the p-value is obtained below: Thus, the p-value is 0.0634. In the given table, enter degrees of freedom as 25, select two-tailed, enter test statistic as 1.942, choose shade area above the graph, and enter the p-value as 0.063.(c) Since the p-value is greater than the level of significance, the null hypothesis is not rejected indicating that there is not enough evidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to 164. Correct Answer: Option 4.