College tuition: The mean annual tuition and fees for a sample of 13 private colleges in California was $\$ 38,000$ with a standard deviation of $\$ 7900$. 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 differs from $\$ 35,000$ ? Use the $\alpha=0.01$ level of significance and the $P$-value method with the $\mathrm{TI}-84$ Plus calculator. Part 1 of 5 (a) State the appropriate null and alternate hypotheses. \[ \begin{array}{l} H_{0}: \mu=35,000 \\ H_{1}: \mu \neq 35,000 \end{array} \] This hypothesis test is a two-tailed $\mathbf{\nabla}$ test. Part 2 of 5 (b) Compute the value of the test statistic. Round the answer to two decimal places. \[ t=1.37 \] Part: 2 / 5 Part 3 of 5 (c) Compute the $P$-value. Round the $P$-value to at least four decimal places. \[ P \text {-value }= \]
Solution
Step 1 ：(a) State the appropriate null and alternate hypotheses. The hypotheses are \(H_{0}: \mu=35,000\) and \(H_{1}: \mu \neq 35,000\). This hypothesis test is a two-tailed test.
Step 2 ：(b) Compute the value of the test statistic. The test statistic is \(t=1.37\).
Step 3 ：(c) Compute the P-value. The P-value is the probability that a random variable is equal to or more extreme than the observed value, assuming the null hypothesis is true. In this case, we are conducting a two-tailed test, so we need to find the probability that a t-distributed random variable with 12 degrees of freedom (13 - 1) is equal to or more extreme than 1.37 in both directions.
Step 4 ：The P-value is approximately 0.1958. This is the probability of observing a test statistic as extreme or more extreme than the one we observed, assuming the null hypothesis is true.
Step 5 ：Final Answer: The P-value is approximately \(\boxed{0.1958}\).
