Question

Test the indicated claim about the means of two populations. Assume that the two samples are independent an have been randomly selected. 3) Two types of flares are tested for their burning times (in minutes) and sample results are 3). given below. Brand X Brand Y n=35 n = 40 x = 19.4 x = 15.1 s = 1.4 s 0.8 Refer to the sample data to test the claim that the two populations have equal means. Use a 0.05 significance level. 1) claim Mo = M2 - CASO 1.or2 2) Ho: M. = M₂ 3) Hi Miz M₂ ~ Tuo tail 4) t =-.0022 Dt: 34 9002% 0.025 -2.032 2.032 Fail to reglect Ho : Case 2 7) Not surficient evident to reject the claim M, = M₂ 4) 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 mean H1 is smaller than the control population mean H2. Test the claim using a significance level of 0.01. Treatment Group ni = 85 Control Group n2 = 75 X2 = 203.7 52 = 39.2 *1 = 189.1 SI = 38.7

Answer 1

4)

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 12 Ha: Mi <ul2 This corresponds to a left-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 = 158. 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 left-tailed test is te = -2.35, for a = 0.01 and df = 158. The rejection region for this left-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: t=- X1 - X2 (0 -1) at t-ng-1) 3 (1 + 5) 189.1 - 203.7 (85-1)38.72+(75-1)39.2 (1 85+75-2 == -2.367 +) (4) Decision about the null hypothesis Since it is observed that t = -2.367 <te = -2.35, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0096, and since p=0.0096 < 0.01, it is concluded that the null hypothesis is rejected.