Question

a one sample t - test is conducted on ho: \\( \mu = 81.6 \\) and ha > 81.6 the sample has a sample mean = 84.1, s = 3.1, n = 25, and \\( \alpha =.01 \\).

Explanation:

Step1: Calculate the t - statistic

The formula for the one - sample t - statistic is \(t=\frac{\bar{x}-\mu}{s/\sqrt{n}}\).
Given \(\bar{x} = 84.1\), \(\mu=81.6\), \(s = 3.1\), and \(n = 25\).
Substitute the values into the formula: \(t=\frac{84.1 - 81.6}{3.1/\sqrt{25}}=\frac{2.5}{3.1/5}=\frac{2.5}{0.62}\approx4.03\).

Step2: Determine the degrees of freedom and critical value

The degrees of freedom \(df=n - 1=25-1 = 24\).
For a one - tailed test with \(\alpha = 0.01\) and \(df = 24\), using the t - distribution table or a statistical calculator, the critical value \(t_{\alpha,df}=t_{0.01,24}= 2.492\).

Step3: Make a decision

Since the calculated t - statistic (\(t\approx4.03\)) is greater than the critical value (\(t_{0.01,24}=2.492\)), we reject the null hypothesis \(H_{0}\).

Answer:

Reject \(H_{0}\)