Test the claim that the mean GPA of night students is larger than 3 at the 0.01 significance level.
The null and alternative hypotheses would be:
\[
\begin{array}{llllll}
H_{0}: \mu=3 & H_{0}: p=0.75 & H_{0}: p=0.75 & H_{0}: p=0.75 & H_{0}: \mu=3 & H_{0}: \mu=3 \\
H_{1}: \mu< 3 & H_{1}: p \neq 0.75 & H_{1}: p> 0.75 & H_{1}: p< 0.75 & H_{1}: \mu \neq 3 & H_{1}: \mu> 3
\end{array}
\]
The test is:
two-tailed right-tailed left-tailed
Based on a sample of 40 people, the sample mean GPA was 3.03 with a standard deviation of 0.14
The $p$-value is:
(to 3 decimals)
The significance level is:
(to 2 decimals)
Based on this we:
Reject the null hypothesis

Step 1 ：Define the null hypothesis as \(H_{0}: \mu=3\) and the alternative hypothesis as \(H_{1}: \mu>3\).

Step 2 ：The given sample size is 40, the sample mean is 3.03, and the sample standard deviation is 0.14.

Step 3 ：Calculate the test statistic using the formula \(Z = \frac{\bar{X} - \mu}{\frac{\sigma}{\sqrt{n}}}\), where \(\bar{X}\) is the sample mean, \(\mu\) is the population mean, \(\sigma\) is the standard deviation, and \(n\) is the sample size. The calculated test statistic is approximately 1.36.

Step 4 ：Calculate the p-value, which is the probability of observing a result as extreme as the test statistic, assuming the null hypothesis is true. The calculated p-value is approximately 0.088.

Step 5 ：Compare the p-value with the significance level (0.01). Since the p-value is greater than the significance level, we fail to reject the null hypothesis.

Step 6 ：\(\boxed{\text{Final Answer: We fail to reject the null hypothesis. The data does not provide strong evidence to support the claim that the mean GPA of night students is larger than 3.}}\)