A very large company is interested in its employees' productivity. The company reports from its historical data that its employees spend a meanof 164 minutes per employee (on a typical day) dealing with email. To test this claim, an independent consultant chooses 26 employees atrandom and finds that those employees spend a sample mean of 172 minutes dealing with email, with a sample standard deviation of 21minutes. Assume that the population 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 claimthat u, the mean number of minutes employees spend dealing with email, is equal to 164.(a) State the null hypothesis H, and the alternative hypothesis H1 that you would use for the test.Họ: 0H: 0OSOO=0?(b) Perform a t test and find the p-value.Here is some information to help you with your i test.The value of the test statistic is given by t =The p-value is two times the area under the curve to the right of the value of the test statistic. Student's t DistributionStep 1: Enter the number of degreesaf freedom.Step 2: Select ae-lailed or twa-tailed.O Oe-taledO Tmo tailedStep 3: Enter the test statistic.(Round to 3 decimal places.)Step 4: Shade the area represanted bythe p-value.Step 5: Enter the p-value.(Round to 3 decimal places.)(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 is equal to164.O since the p-value is less than (or equal to) the level of significance, the null hypothesis is nat rejected. So, there isnot enough evidence to reject the claim that the mean number of minutes employees spend dealing with email isequal 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 to 164.O since the p-value is greater than the level of significance, the null hypothesis is not rejected. So, there is not enoughevidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to 164.

Given, Sample size = n = 26 Sample mean = x¯ = 172 Sample standard deviation = s = 21 Claim : The mean number of minutes employees spend dealing with email is equal to 164 Hypothesis : H0 : μ = 164 H1 : μ ≠ 164 Two tailed test. Test statistic : t =x¯-μsn t =172-1642126 t =1.9425P-value : Using Excel, "=TDIST(t, df, tails)" we got , t =1.9425 df = n - 1 = 26 - 1 = 25 tails = 2 -----> Since two tailed test. So, p-value = TDIST(1.9425, 25, 2) P-value = 0.0634Decision about null hypothesis : Decision rule : 1) Reject null hypothesis if p-value less than significance level α 2) Otherwise fail to reject null hypothesis. Given, significance level = α = 0.05 It is observed that p-value ( 0.0634 ) is greater than α = 0.05 So, the null hypothesis is not rejected. Conclusion : There is not enough evidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to 164