Question

section 10.3

Provided below are summary statistics for independent simple random samples from two populations. Use the nonpooled t-test and the nonpooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval. *4 = 10,5 - 2, 0, -20, 72 - 11,5 - 5, n2 - 20 a. Two-tailed test, a = 0.01 b. 99% confidence interval a. What are the hypotheses for the t-test? O A HOHH2 H₂H12 Oc. Ho: H = 12 Hai H1 H2 B. HH,2 HH12 OD. Holi 24h H:45h Find the test statistic. 1 - - 0.830 (Round to three decimal places as needed.) Find the P-value - (Round to four decimal places as needed.)

Answer 1

b)

Solution: The provided sample means are shown below: X1 = 10 X2 = 11 Also, the provided sample standard deviations are: 81 2 82 = 5 and the sample sizes are ni = 20 and n2 = 20. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: Mi = 42 Ha: 1+12 This corresponds to a two-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 = 38. 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 two-tailed test is te=2.712, for a=0.01 and df = 38. The rejection region for this two-tailed test is R = {t: t > 2.712} (3) Test Statistics Since it is assumed that the population variances are equal, the t-statistic is computed as follows: X1 - X2 (11-1): +(12-1) T +12-2 1 + 10 - 11 (20-1)22 +(20-1)52 20-20-2 -0.83 (20+20) (4) Decision about the null hypothesis Since it is observed that t = 0.83 <te=2.712, it is then concluded that the null hypothesis is not rejected. Using the P-value approach: The p-value is p=0.4115, and since p=0.4115 > 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 mi is different than 12, at the 0.01 significance level.

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Solution: We need to construct the 99% confidence interval for the difference between the population means ui - u2, for the case that the population standard deviations are not known. The following information has been provided about each of the samples: Sample Mean 1 (X1) = 10 Sample Standard Deviation 1 (si) = 2 Sample Size 1 (N) = 20 Sample Mean 2 (X2) 11 Sample Standard Deviation 2 (82) = 5 Sample Size 2 (N2) = 20 Based on the information provided, we assume that the population variances are equal, so then the number of degrees of freedom are df = n1 + n2 - 2 = 20+ 20 - 2 = 38. The critical value for a = 0.01 and df = 38 degrees of freedom is te = t1-a/2;n-1 = 2.712. The corresponding confidence interval is computed as shown below: Since the population variances are assumed to be equal, we need to compute the pooled standard deviation, as follows: Sp (ni - 1)si + (n2 - 1)s ni + n2 - 2 (20 - 1) 22 + (20 - 1) x 52 20 + 20 2 3.808 Since we assume that the population variances are equal, the standard error is computed as follows: 1 1 se = 8pm + n2 1 3.808 x 1 20 20 1.204 Now, we finally compute the confidence interval: CI = (X1 - 2 - tex se, X1 - X2 + te x se) (10 - 11 - 2.712 x 1.204, 10 - 11 +2.712 x 1.204) = (-4.265, 2.265) Therefore, based on the data provided, the 99% confidence interval for the difference between the population means Hi - H2 is -4.265 <H1 - H22.265, which indicates that we are 99% confident that the true difference between population means is contained by the interval (-4.265, 2.265).