Question

find the critical value, $t_0$, to test the claim that $mu_1 < mu_2$. two samples are randomly selected and that are normal. the sample statistics are given below. assume that $sigma_1^2 = sigma_2^2$. use $alpha = 0.05$.$n_1 = 15, n_2 = 15, \bar{x}_1 = 25, \bar{x}_2 = 27.55, s_1 = 2.9, s_2 = 2.8$\\(\bigcirc\\) a. $-1.701$\\(\bigcirc\\) b. $2.467$\\(\bigcirc\\) c. $-1.313$\\(\bigcirc\\) d. $0.683$

Explanation:

Step1: Determine Degrees of Freedom

The formula for degrees of freedom (\(df\)) when \(\sigma_1^2 = \sigma_2^2\) is \(df = n_1 + n_2 - 2\). Given \(n_1 = 15\) and \(n_2 = 15\), we calculate:
\(df = 15 + 15 - 2 = 28\).

Step2: Identify Test Type and Significance Level

We are testing \(\mu_1 < \mu_2\), so it is a left - tailed \(t\) - test. The significance level \(\alpha = 0.05\).

Step3: Find Critical Value from t - Distribution Table

Using a \(t\) - distribution table (or calculator) for a left - tailed test with \(df = 28\) and \(\alpha = 0.05\), we look up the critical value. From the \(t\) - table, the critical value \(t_0\) for \(df = 28\) and \(\alpha = 0.05\) (left - tailed) is approximately \(- 1.701\).

Answer:

A. - 1.701