Test the claim that the mean GPA of night students is significantly different than 3 at the 0.2 significance level. The null and alternative hypothesis would be: \[ \begin{array}{llllccc} H_{0}: \mu=3 & H_{0}: \mu \leq 3 & H_{0}: p \geq 0.75 & H_{0}: p \leq 0.75 & H_{0}: \mu \geq 3 & H_{0}: p=0.75 \\ H_{1}: \mu \neq 3 & H_{1}: \mu>3 & H_{1}: p<0.75 & H_{1}: p>0.75 & H_{1}: \mu<3 & H_{1}: p \neq 0.75 \\ \bigcirc & \bigcirc & \ddots & & & & \end{array} \] The test is: left-tailed right-tailed two-tailed Based on a sample of 20 people, the sample mean GPA was 2.98 with a standard deviation of 0.08 The $p$-value is: (to 2 decimals) Based on this we: Reject the null hypothesis Fail to reject the null hypothesis

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

Step 2 ：This is a two-tailed test because we are testing for a difference and not a specific direction (greater or less than).

Step 3 ：Given that the sample mean (\(\bar{x}\)) is 2.98, the standard deviation (s) is 0.08, and the sample size (n) is 20.

Step 4 ：Calculate the test statistic (t) using the formula: \(t = (\bar{x} - \mu) / (s / \sqrt{n})\)

Step 5 ：Substitute the given values into the formula to get \(t = (2.98 - 3) / (0.08 / \sqrt{20})\), which gives \(t = -1.12\)

Step 6 ：Find the p-value corresponding to the calculated t-value. The p-value is approximately 0.28.

Step 7 ：Since the p-value (0.28) is greater than the significance level (0.2), we fail to reject the null hypothesis.

Step 8 ：This means that we do not have enough evidence to support the claim that the mean GPA of night students is significantly different than 3.

Step 9 ：Final Answer: We \(\boxed{\text{fail to reject the null hypothesis}}\). The mean GPA of night students is not significantly different than 3 at the 0.2 significance level.