Question

Consider the hypothesis test Ho : H1 = H2 against H1 : HI # Hz with known variances oj = 1 1 and oz = 4. Suppose that sample sizes ni = 11 and n2 = 16 and that X = 4,7 and X2 = 7.9. Use a = 0.05.

Question 1 of 1 < > -/1 View Policies Current Attempt in Progress Consider the hypothesis test Ho: M1 = H2 against H: MM with known variances o = 11 and 62 = 4. Suppose that sample sizes n = 11 and 12 = 16 and that X1 = 4.7 and I = 7.9. Use a = 0.05. (a) Test the hypothesis and find the P-value. (b) What is the power of the test in part (a) for a true difference in means of 3? (c) Assuming equal sample sizes, what sample size should be used to obtain p = 0.05 if the true difference in means is 3? Assume that a = 0.05. (a) The null hypothesis rejected. The P-value is Round your answer to three decimal places les 98.765). (b) The power is Round your answer to two decimal places (eg. 98.76). (c) n = n2 = Round your answer up to the nearest Integer Statistical Tables and Charts e Textbook and Media Save for Later Attempts:0 of 3 used Submit Answer

Answer 1

The provided sample means are shown below: X = 4.7 X2 = = 7.9 Also, the provided population standard deviations are: 01 = 11 02 = 4 11 and n2 = and the sample sizes are ni 16. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M1 = M2 Ha: uy = 42 This corresponds to a two-tailed test, for which a z-test for two population means, with known population standard deviations will be used.

(2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a two-tailed test is ze = 1.96. The rejection region for this two-tailed test is R = {z : [2] > 1.96} (3) Test Statistics The z-statistic is computed as follows: z = X1 - 2 Voi/ni +ož/n2 = 4.7 – 7.9 V112/11 +42/16 = -0.924

(4) Decision about the null hypothesis Since it is observed that [2] = 0.924 < ze = 1.96, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p = 0.3556, and since p 0.3556 > 0.05, it is concluded that the null hypothesis is not rejected. Confidence Interval The 95% confidence interval for Mi - M2 is -9.99 Mi - M2 < 3.59. Graphically p-value = Test 0.3556 0.40 0.35 0.30 0.25 0.20 0.15 0.10 0.05 0.00 -4.0 -3.5 -3.0-2.5 -2.0 -1.5 -1.0 -0.5 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0

Thus it may conclude that two population have equal means.