Question

20) A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure by following a particular diet. Use the sample data below to test the claim that the treatment population , is larger than the control population wz. Test the claim using a significance level of a = 0.01. The population variances are equal. Control Group Treatment Group ng = 79 n = 85 X = 189.1 X = 178.7 S = 37.2 s, = 38.7 B) Reject H. IEETR the claim A) Reject H, IEETS the claim C) FTR H. NEETS the claim NEETR the claim D) FTR H. E) None of the Above

Answer 1

20)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: ui = 2 Ha: ui > 2 This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples with unknown population standard deviations will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.01, and the degrees of freedom are df = 162. In fact, the degrees of freedom are computed as follows, assuming that the population variances are equal: Hence, it is found that the critical value for this right-tailed test is te = 2.35, for a = 0.01 and df = 162. The rejection region for this right-tailed test is R={t:t > 2.35}. (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X2 t= (n-1)33+(12-1)3 (+ 112 189.1 - 178.7 - 1.752 (85–1138.72 = (19–1)37.22 (+) (4) Decision about the null hypothesis Since it is observed that t = 1.752 <te = 2.35, it is then concluded that the null hypothesis is not rejected Using the P-value approach: The p-value is p=0.0408, and since p = 0.0408 > 0.01, it is concluded that the null hypothesis is not rejected. (5) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean y is greater than M2, at the 0.01 significance level.

We fail to reject the $H_0$

21) $H_0: \mu_1=\mu_2$

211111: "H

This is a two tailed test

The P value for the two tailed test = 0.0768

The significance level is $\alpha=0.10$

Since the P value is less than 0.10, we shall reject the null hypothesis.

Hence, we have evidence to say that Yes we support the claim