Test the claim that the mean GPA of night students is smaller than 3.5 at the 0.025 significance level. The null and alternative hypothesis would be: \[ \begin{array}{l} H_{0}: \mu \geq 3.5 H_{0}: \mu=3.5 H_{0}: p \leq 0.875 H_{0}: p \geq 0.875 \quad H_{0}: \mu \leq 3.5 \quad H_{0}: p=0.875 \\ H_{1}: \mu<3.5 H_{1}: \mu \neq 3.5 H_{1}: p>0.875 \quad H_{1}: p<0.875 \quad H_{1}: \mu>3.5 \quad H_{1}: p \neq 0.875 \\ \end{array} \] The test is: right-tailed two-tailed left-tailed Based on a sample of 55 people, the sample mean GPA was 3.47 with a standard deviation of 0.07 The test statistic is: (to 2 decimals) The p-ralue is: (to 2 decimals) Based on this we: Fail to reject the null hypothesis Reject the null hypothesis Calculator

Step 1 ：Define the null and alternative hypothesis: \(H_{0}: \mu \geq 3.5\) and \(H_{1}: \mu<3.5\). This is a left-tailed test.

Step 2 ：Given a sample of 55 students, a sample mean of 3.47, and a standard deviation of 0.07.

Step 3 ：Calculate the test statistic using the formula: \((sample \: mean - population \: mean) / (standard \: deviation / \sqrt{sample \: size})\). The test statistic is approximately -3.18.

Step 4 ：Calculate the p-value using a z-table or a statistical software. The p-value is approximately 0.00074.

Step 5 ：Compare the p-value with the significance level. Since the p-value (0.00074) is less than the significance level (0.025), we reject the null hypothesis.

Step 6 ：Final Answer: We reject the null hypothesis. The mean GPA of night students is less than 3.5. \(\boxed{Reject \: the \: null \: hypothesis}\).