Question

a sample mean, sample size, and sample standard deviation are provided below. use the one - mean t - test to perform the required hypothesis test at the 5% significance level.
$overline{x}=25$, $s = 7$, $n = 32$, $h_{0}:mu=28$, $h_{a}:mu<28$
click here to view a partial table of values of $t_{\alpha}$.

Explanation:

Step1: Calculate the t - statistic

The formula for the one - sample t - statistic is \(t=\frac{\bar{x}-\mu_0}{s/\sqrt{n}}\).
Given \(\bar{x} = 25\), \(\mu_0=28\), \(s = 7\), \(n = 32\).
Substitute the values into the formula:
\(t=\frac{25 - 28}{7/\sqrt{32}}=\frac{- 3}{7/\sqrt{32}}\approx\frac{-3}{1.237}\approx - 2.425\)

Step2: Determine the critical value

The significance level is \(\alpha=0.05\), and the degrees of freedom is \(df=n - 1=32-1 = 31\).
For a left - tailed test (\(H_a:\mu<28\)), the critical value \(t_{\alpha,df}\) with \(\alpha = 0.05\) and \(df = 31\) (using a t - table or statistical software) is approximately \(t_{0.05,31}=-1.696\)

Step3: Make a decision

If \(tSince \(t=-2.425<-1.696\)

Answer:

We reject the null hypothesis \(H_0\) at the \(5\%\) significance level.