To test the claim that the mean GPA of night students is smaller than 3.2, we can conduct a one-sample t-test.
Step 1: Set up the hypotheses
- Null hypothesis (H0): \(\mu = 3.2\)
- Alternative hypothesis (H1): \(\mu < 3.2\)
Step 2: Collect the sample statistics
- Sample size (\(n\)) = 40
- Sample mean (\(\bar{x}\)) = 3.18
- Population mean (\(\mu_0\)) = 3.2
- Sample standard deviation (\(s\)) = 0.06
Step 3: Calculate the test statistic
The formula for the t-test statistic is:
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
Substituting in the values:
\[ t = \frac{3.18 - 3.2}{0.06 / \sqrt{40}} \]
Calculating the denominator:
\[ s / \sqrt{n} = 0.06 / \sqrt{40} \approx 0.06 / 6.32 \approx 0.0095 \]
Now substituting this back into the formula for the t statistic:
\[ t = \frac{3.18 - 3.2}{0.0095} \approx \frac{-0.02}{0.0095} \approx -2.11 \]
So the test statistic is:
\[ \text{Test statistic (t)} = -2.11 \]
Step 4: Calculate the p-value
To find the p-value for a one-tailed t-test with \(n-1 = 40-1 = 39\) degrees of freedom, we can use a t-distribution table or a calculator.
Using a t-table or calculator, we can find the p-value corresponding to our test statistic of \(t = -2.11\) with 39 degrees of freedom.
Using statistical software or a calculator to find the one-tail p-value:
- The p-value associated with \( -2.11 \) for \( 39 \) degrees of freedom is approximately \( 0.02 \).
Conclusion
- Test statistic: \(-2.11\)
- p-value: \(0.02\)
Thus, the answers are:
- The test statistic is: -2.11
- The p-value is: 0.02