College tuition: The mean annual tuition and fees for a sample of 14 private colleges in California was $\$ 33,500$ with a standard deviation of $\$ 7350$. A dotplot shows that it is reasonable to assume that the population is approximately normal. Can you conclude that the mean tuition and fees for private institutions in California is less than $\$ 35,000$ ? Use the $\alpha=0.10$ level of significance and the $P$-value method with the TI-84 Plus calculator. Part 1 of 5 (a) State the appropriate null and alternate hypotheses. \[ \begin{array}{l} H_{0}: \mu=35,000 \\ H_{1}: \mu<35,000 \end{array} \] This hypothesis test is a left-tailed $\mathbf{v}$ test. Part 2 of 5 (b) Compute the value of the test statistic. Round the answer to two decimal places. \[ t=-0.76 \] Part 3 of 5 (c) Compute the $P$-value. Round the $P$-value to at least four decimal places. \[ P \text {-value }=0.2294 \] Part: $3 / 5$ Part 4 of 5 (d) Determine whether to reject $H_{0}$. (Choose one) $\mathbf{v}$ the null hypothesis $H_{0}$.
Solution
Step 1 ：State the appropriate null and alternate hypotheses. The null hypothesis \(H_{0}: \mu=35,000\) and the alternate hypothesis \(H_{1}: \mu<35,000\). This hypothesis test is a left-tailed test.
Step 2 ：Compute the value of the test statistic. The test statistic is \(t=-0.76\).
Step 3 ：Compute the P-value. The P-value is \(P \text {-value }=0.2294\).
Step 4 ：Determine whether to reject \(H_{0}\). The rule is that if the p-value is less than the significance level, we reject the null hypothesis. In this case, the p-value is 0.2294 and the significance level is 0.10. So, we need to compare these two values to make a decision. The p-value is greater than the significance level (0.2294 > 0.10), so we do not reject the null hypothesis.
Step 5 ：\(\boxed{\text{Final Answer: We do not reject the null hypothesis } H_{0}}\)
