Question

A consultant for a large university studied the number of hours per week freshmen watch TV versus the number of hours seniors do. The results of this study follow. Is there enough evidence to show the mean number of hours per week freshmen watch TV is different from the mean number of hours seniors do at a=0.01? Seniors Freshmen 10 xbar 18.5 7.8740 11.9 3.9749 For the hypothesis stated above (in terms of Seniors - Freshmen): Question 1 What is/are the critical value(s)? Question 2 What is the decision? Question 3 What is the p-value? Fill in only one of the following statements. If the Z table is appropriate, p-value If thet table is appropriate, p-value

Answer 1

Solution: Seniors: n = 4 X = 11.9 Su = 3.9749 Freshmen: n2 = 10 x2 = 18.5 S2 = 7.8740 Ho: M1 = uz Ha: M1 M2 Degrees of Freedom: df = = df = (4 (3.97492 4.87402) + 10 ) 1_13.97492 1 (4.874022 4-1 4 ) + 10-11 10 ) df = 10.9 ~ 10 (Rounding down) T (3.9749 Critical Values: Level of significance = a = 0.01 The test is a two-tailed test. Hence, a/2 = 0.01/2 = 0.005

From t-distribution table; the critical value of 't, having area under the curve to the right of it equal to 0.005 and for (df = 10), is given by: 3.169 Since the test is a two-tailed test, critical values are = -3.169, 3.169 Decision Rule: Reject Ho ift <-3.169or t > 3.169 t = Test Statistic: _[(x1 - x2) - (U1 )] sĩ, sở ni n2 [11.9 - 18.5 - 0] t= 3.97492 4.87402 v 4 T 10 =t= -2.072 Decision: Since the value of test statistic is not less than the lower tail critical value of -3.169, we fail to reject the null hypothesis. P-value: From t-distribution table; the area in the lower tail to the left of (t = -2.072) and for (df = 10), is given by: 0.0326 The test is a two-tailed test. Hence, p-value = 0.0326 x 2 = 0.0652 Therefore, 0.05 < p-value < 0.10