Question

test the claim that the mean gpa of night students is significantly different than 2.4 at the 0.05 significance level.
the null and alternative hypothesis would be:
\\( \

$$\begin{array}{l} h_{o}: \\mu = 2.4 \\\\ h_{a}: \\mu > 2.4 \\end{array}$$

\\)
\\( \

$$\begin{array}{l} h_{o}: \\mu = 2.4 \\\\ h_{a}: \\mu < 2.4 \\end{array}$$

\\)
\\( \

$$\begin{array}{l} h_{o}: \\mu = 2.4 \\\\ h_{a}: \\mu \ eq 2.4 \\end{array}$$

\\)

the test is:
right - tailed two - tailed left - tailed

based on a sample of 19 people, the sample mean gpa was 2.35 with a standard deviation of 0.08
the test statistic is: \\( t = \\) \\( \square \\) (to 2 decimals)

the p - value is = \\( \square \\) (to 4 decimals)

the p - value is...
\\( \circ \\) less than alpha
\\( \circ \\) greater than alpha

this test statistic leads to a decision to...
\\( \circ \\) fail to reject the null
\\( \circ \\) reject the null

Explanation:

Step1: State hypotheses

We test for a difference, so:
$H_0: \mu = 2.4$
$H_a: \mu
eq 2.4$

Step2: Identify test type

Since $H_a$ uses $
eq$, this is a two-tailed test.

Step3: Calculate t-statistic

Use $t = \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$, where $\bar{x}=2.35$, $\mu=2.4$, $s=0.08$, $n=19$

$$ t = \frac{2.35 - 2.4}{\frac{0.08}{\sqrt{19}}} = \frac{-0.05}{\frac{0.08}{4.3589}} = \frac{-0.05}{0.01835} \approx -2.73 $$

Step4: Find p-value

For $t=-2.73$, $df=19-1=18$, two-tailed test. Using t-table or calculator, p-value $\approx 0.0148$

Step5: Compare p-value to $\alpha$

$\alpha=0.05$, $0.0148 < 0.05$, so p-value < alpha.

Step6: Make decision

Reject the null hypothesis since p-value < alpha.

Answer:

The null and alternative hypothesis would be:
$H_0: \mu = 2.4$
$H_a: \mu
eq 2.4$

The test is:
two-tailed

The test statistic is: $t = -2.73$

The p-value is: $0.0148$

The p-value is...
less than alpha

This test statistic leads to a decision to...
reject the null