Question

Consider the hypothesis test HO : Mi - M2 O against H: Mi - H2 + 0 samples below: I 36 38 31 32 33 30 31 29 38 38 31 37 37 30 39 32 36 38 II 32 30 33 33 31 2930 36 31 33 30 30 30 33 33 35 Variances: 0] = 3.8, 02 = 1.7. Use a = 0.05. (a) Test the hypothesis and find the P-value. Find the test statistic. Round your answers to four decimal places (e.g. 98.7654). 2.4309 ZO P - value 0.0151 (b) Explain how this test could be conducted with a confidence interval. Round your answers to three decimal places (e.g. 98.765). We are 95% confident that value 1 exceeds value 2 by between 0.467 SH - H2 <T4.353 (c) What is the power of the test for the true difference in means of 2? Round your answer to three decimal places (e.g. 98.765). 0.477 The power of the test is 1-B =

Answer 1

a)

X1 = = 34.2778 X2 = 31.9375 Also, the provided population standard deviations are: 01 = 3.9 02 = 2.9 and the sample sizes are ni = 18 and n2 = 16. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 42 Ha: My = M2 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: 121 > 1.96}

(3) Test Statistics The z-statistic is computed as follows: 2= X - X) Voni +ož/n2 34.2778 – 31.9375 13.92 /18 + 2.92/16 1.999 (4) Decision about the null hypothesis 1.96, it is then concluded that the null hypothesis is Since it is observed that 21 = 1.999 > Ze = rejected. Using the P-value approach: The p-value is p= 0.0456, and since p = 0.0456 < 0.05, it is concluded that the null hypothesis is rejected.

b)

Confidence Interval

The 95% confidence interval for 0.046<μ1​−μ2​<4.635.

The critical value for a = 0.05 is ze = 21-a/2 = 1.96. The corresponding confidence interval is computed as shown below: 1-70 X1 - X2 СІ o 2c X1 - X2 + zc1 + ni n2 ni n2 3.92 2.92 3.92 2.92 34.278 – 31.938 1.96 + 34.278 – 31.938 +1.961 -1.96V + 18 16 18 16 = (0.046, 4.635)

c)

Power of test 1-β = 0.4008

Power = P(Reject Ho | μ1-μ2=2)

Power= P(X1bar - X2bar < -2.2946)+P(X1bar - X2bar > 2.2946)

=P(Z<-3.6)+P(Z>0.2516)

=0.4008