Question

suppose you want to test the claim that $\mu_1 < \mu_2$. two samples are randomly selected from normal pop sample statistics are given below. assume that $\sigma_1^2 = \sigma_2^2$. at a level of significance of $\alpha = 0.10$, when s $h_0$?
$n_1 = 15, n_2 = 15, \bar{x}_1 = 22.28, \bar{x}_2 = 24.83, s_1 = 2.9, s_2 = 2.8$

a. reject $h_0$ if the standardized test statistic is less than -1.313.
b. reject $h_0$ if the standardized test statistic is less than -2.467.
c. reject $h_0$ if the standardized test statistic is less than -0.683.
d. reject $h_0$ if the standardized test statistic is less than -1.701.

Explanation:

Step1: Identify Test Type and Degrees of Freedom

This is a two - sample t - test (since \(\sigma_1^2=\sigma_2^2\) and we are testing means from normal populations). The degrees of freedom \(df=n_1 + n_2-2=15 + 15-2 = 28\). The alternative hypothesis is \(H_a:\mu_1<\mu_2\), so it's a left - tailed test. The significance level \(\alpha = 0.10\).

Step2: Find Critical t - Value

For a left - tailed t - test with \(df = 28\) and \(\alpha=0.10\), we look up the critical value in the t - distribution table. The critical value \(t_{\alpha,df}=t_{0.10,28}\). From the t - table, \(t_{0.10,28}\approx - 1.313\) (we take the negative because it's a left - tailed test). So we reject \(H_0\) if the standardized test statistic is less than - 1.313.

Answer:

A. Reject \(H_0\) if the standardized test statistic is less than - 1.313.