Test the claim that the mean GPA of night students is smaller than 3.2 at the 0.05 significance level.

The null and alternative hypothesis would be:

A. H_0 : p = 0.8
H_1 : p ≠ 0.8

B. H_0 : p ≥ 0.8
H_1 : p < 0.8

C. H_0 : p ≤ 0.8
H_1 : p > 0.8

D. H_0 : μ ≤ 3.2
H_1 : μ > 3.2

E. H_0 : μ ≥ 3.2
H_1 : μ < 3.2

F. H_0 : μ = 3.2
H_1 : μ ≠ 3.2

The test is:

A. left-tailed

B. right-tailed

C. two-tailed

Based on a sample of 40 people, the sample mean GPA was 3.19 with a standard deviation of 0.08

The test statistic is: ____ (to 2 decimals)

The p-value is: ____ (to 2 decimals)

Based on this we:
A. Reject the null hypothesis
B. Fail to reject the null hypothesis

1 answer

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.