A very large company is interested in its employees productivity. The company reports from its historical data that its employees spend a mean of 136 minutes per employee (on a typical day) dealing with email. To test this claim, an independent consultant chooses 23 employees at random and finds that those employees spend a sample mean of 165 minutes dealing with email, with a sample standard deviation of 24 minutes. 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 claim that J, the mean number of minutes employees spend dealing with email, is equal to 156. D (a) State the null hypothesis Ho and the alternative hypothesis H, that you would use for the test. H: H X H : I 020 0=0 0#0 X ? (b) Perform a t test and find the p-value. Here is some information to help you with your t test. x - ul . The value of the test statistic is given by t = S Jno The pvalue is two times the area under the curve to the right of the value of the test statistic. Student's t Distribution Step 1: Enter the number of degrees of freedom. :1 Step 2: Select onetailed or twotailed. O Onetailed O Twortailed Step 3: Enter the test statistic. (Rnund t0 3 decimal places.) H Step 4: Shade the area represented by the pValue. 9 Step 5: Enter the prvalue. (RDund to 3 decimal places.) H 0.4 0,3 0.2 0,1 .__ (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. Since the p-value is less than (or equal to) the level of significance, the null hypothesis is rejected. So, there is enough evidence to reject the claim that the mean number of minutes employees spend dealing with email is equal X 5 to 156. O Since the p-value is less than (or equal to) the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to 156. O Since the p-value is greater than the level of significance, the null hypothesis is rejected. So, there is enough evidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to 156. O Since the p-value is greater than the level of significance, the null hypothesis is not rejected. So, there is not enough evidence to reject the claim that the mean number of minutes employees spend dealing with email is equal to 156