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. x overbar 1equals11, s 1equals3, n 1equals25, x overbar 2equals10, s 2equals8, n 2equals20 a. Right-tailed test, alphaequals0.05 b. 90% confidence interval a. What are the hypotheses for the t-test? A. H0: mu1equalsmu2 Ha: mu1not equalsmu2 B. H0: mu1equalsmu2 Ha: mu1greater thanmu2 Your answer is correct.C. H0: mu1equalsmu2 Ha: mu1less thanmu2 D. H0: mu1greater than or equalsmu2 Ha: mu1less thanmu2 Find the test statistic. tequals . 530 (Round to three decimal places as needed.) Find the P-value. Pequals . 3006 (Round to four decimal places as needed.) What is the conclusion of the hypothesis test? A. Reject H0. There is insufficient evidence that mu1 is greater than mu2. B. Do not reject H0. There is sufficient evidence that mu1 is greater than mu2. Your answer is not correct.C. Do not reject H0. There is insufficient evidence that mu1 is greater than mu2. This is the correct answer.D. Reject H0. There is sufficient evidence that mu1 is greater than mu2. b. The 90% confidence interval is from nothing to nothing. (Round to three decimal places as needed.)
Accepted Solution
A:
Using the t-distribution, we have that:a)The null hypothesis is [tex]H_0: x_1 - x_2 \leq 0[/tex].The alternative hypothesis is [tex]H_1: x_1 - x_2 > 0[/tex]i) The test statistic is t = 0.746.ii) The p-value is of 0.2299.b) The 90% confidence interval is from -1.25 to 3.25.Item a:For a right-tailed test, we test if [tex]x_1[/tex] is greater than [tex]x_2[/tex], hence:The null hypothesis is [tex]H_0: x_1 - x_2 \leq 0[/tex].The alternative hypothesis is [tex]H_1: x_1 - x_2 > 0[/tex]The standard errors are:[tex]S_{e1} = \frac{5}{\sqrt{25}} = 1[/tex][tex]S_{e2} = \frac{4}{\sqrt{20}} = 0.8944[/tex]The distribution of the difference has mean and standard error given by:[tex]\overline{x} = x_1 - x_2 = 11 - 10 = 1[/tex][tex]s = \sqrt{S_{e1}^2 + S_{e2}^2} = \sqrt{1^2 + 0.8944^2} = 1.34[/tex]We have the standard deviation for the samples, hence, the t-distribution is used.The test statistic is given by:[tex]t = \frac{\overline{x} - \mu}{s}[/tex]In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.Hence:[tex]t = \frac{\overline{x} - \mu}{s}[/tex][tex]t = \frac{1 - 0}{1.34}[/tex][tex]t = 0.746[/tex]The test statistic is t = 0.746.The p-value is found using a t-distribution calculator, with t = 0.746, 25 + 20 - 2 = 43 df and 0.05 significance level.Using the calculator, it is of 0.2299.Item b:The critical value for a 90% confidence interval with 43 df is [tex]t = 1.6811[/tex].The interval is:[tex]\overline{x} \pm ts[/tex]Hence:[tex]\overline{x} - ts = 1 - 1.6811(1.34) = -1.25[/tex][tex]\overline{x} + ts = 1 + 1.6811(1.34) = 3.25[/tex]The 90% confidence interval is from -1.25 to 3.25.A similar problem is given at