Determine the P-value: Use a t-test to test the claim about the population mean $\mu$ at the given level of significance $\alpha$ using the given sample statistics. Assume the population is normally distributed. Claim: $\mu \geq 8100 ; \alpha=0.05$ Sample statistics: $\bar{x}=7900, s=480, n=23$ What are the null and alternative hypotheses? A. \[ \begin{array}{l} H_{0}: \mu=8100 \\ H_{a}: \mu \neq 8100 \end{array} \] c. \[ \begin{array}{l} H_{0}: \mu \leq 8100 \\ H_{a}: \mu>8100 \end{array} \] B. \[ \begin{array}{l} H_{0}: \mu \geq 8100 \\ H_{a}: \mu<8100 \end{array} \] D. \[ \begin{array}{l} H_{0}: \mu \neq 8100 \\ H_{a}: \mu=8100 \end{array} \] What is the value of the standardized test statistic? The standardized test statistic is -2.0 . (Round to two decimal places as needed.) What is the P-value? $\mathrm{P}=\square$ (Round to three decimal places as needed.)
Solution
Step 1 ：The null and alternative hypotheses are: \(H_{0}: \mu \geq 8100\) and \(H_{a}: \mu<8100\)
Step 2 ：The standardized test statistic is given as -2.0
Step 3 ：The degrees of freedom for the t-distribution is the sample size minus 1, which is 23 - 1 = 22
Step 4 ：To calculate the P-value, we can use the cumulative distribution function (CDF) of the t-distribution. The CDF gives the probability that a random variable drawn from the t-distribution is less than or equal to a given value. Since the test statistic is negative, we need to find the probability that a random variable drawn from the t-distribution is less than or equal to -2.0
Step 5 ：Using the test statistic and degrees of freedom, the P-value is calculated to be approximately 0.029
Step 6 ：Final Answer: The P-value is \(\boxed{0.029}\)
