To test the claim that the mean GPA of night students is smaller than 3.2 at the 0.05 significance level, let's outline the hypotheses and the test type.
Hypotheses
We want to determine if the mean GPA (μ) is less than 3.2.
- Null Hypothesis (H₀): \( \mu \geq 3.2 \) (the mean GPA is greater than or equal to 3.2)
- Alternative Hypothesis (H₁): \( \mu < 3.2 \) (the mean GPA is less than 3.2)
Thus, the correct choice for the hypotheses would be:
E. H₀ : μ ≥ 3.2
H₁ : μ < 3.2
Test Type
Since we are testing for whether the mean is less than 3.2, this is a left-tailed test. Therefore, the correct choice is: A. left-tailed
Test Statistic
To compute the test statistic, we will use the formula for the t-statistic since the sample size is small (n < 30) and we assume the population from which the sample is drawn is approximately normally distributed:
\[ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} \]
Where:
- \(\bar{x} = 3.19\) (sample mean)
- \(\mu_0 = 3.2\) (hypothesized population mean)
- \(s = 0.08\) (sample standard deviation)
- \(n = 40\) (sample size)
Substituting the values into the formula:
\[ t = \frac{3.19 - 3.2}{0.08 / \sqrt{40}} \]
Calculating the denominator first:
\[ \sqrt{40} \approx 6.32 \] \[ 0.08 / 6.32 \approx 0.01265 \]
Now calculate \( t \):
\[ t = \frac{-0.01}{0.01265} \approx -0.791 \]
Thus, the test statistic is: Test statistic is: -0.79 (to 2 decimals)
P-value
Now we will calculate the p-value for a left-tailed t-test with the test statistic \( t \approx -0.79 \) and \( n - 1 = 39 \) degrees of freedom.
Using a t-table or calculator for the t-distribution:
- For \( t = -0.79 \) and \( 39 \) degrees of freedom, we find the corresponding p-value.
Using a calculator or t-table, we find the one-tailed p-value is approximately \( 0.215 \).
Thus: P-value is: 0.22 (to 2 decimals)
Conclusion
- Since the p-value \( (0.22) \) is greater than the significance level \( (0.05) \), we fail to reject the null hypothesis.
The final answer for the conclusion would be: B. Fail to reject the null hypothesis.