You wish to test the following claim $\left(H_{a}\right)$ at a significance level of $\alpha=0.01$. \[ \begin{array}{l} H_{0}: \mu=80.8 \\ H_{a}: \mu>80.8 \end{array} \] You believe the population is normally distributed and you know the standard deviation is $\sigma=6.6$. You obtain a sample mean of $M=83.9$ for a sample of size $n=38$. What is the test statistic for this sample? (Report answer accurate to three decimal places.) \[ \text { test statistic }= \] What is the p-value for this sample? (Report answer accurate to four decimal places.) \[ \mathrm{p} \text {-value }= \] The p-value is... less than (or equal to) $\alpha$ greater than $\alpha$ This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the population mean is greater than 80.8 . There is not sufficient evidence to warrant rejection of the claim that the population mean is greater than 80.8 . The sample data support the claim that the population mean is greater than 80.8 . There is not sufficient sample evidence to support the claim that the population mean is greater than 80.8 . Question Help: Post to forum 1) $64^{\circ} \mathrm{F}$ sunny
Solution
Step 1 ：Given the sample mean (M) is 83.9, the population mean under the null hypothesis (μ) is 80.8, the population standard deviation (σ) is 6.6, and the sample size (n) is 38.
Step 2 ：We can calculate the test statistic (Z) using the formula: \(Z = \frac{M - μ}{σ / \sqrt{n}}\)
Step 3 ：Substituting the given values into the formula, we get: \(Z = \frac{83.9 - 80.8}{6.6 / \sqrt{38}} \approx 2.816\)
Step 4 ：The p-value is the probability of observing a test statistic as extreme as, or more extreme than, the observed test statistic, under the null hypothesis. Since this is a one-tailed test, we find the p-value by looking up the test statistic in a standard normal (Z) table or using a Z-distribution calculator.
Step 5 ：The p-value is \(P(Z > 2.816)\). Using a standard normal table or a Z-distribution calculator, we find that the p-value is approximately 0.0024.
Step 6 ：Since the p-value (0.0024) is less than the significance level (0.01), we reject the null hypothesis.
Step 7 ：\(\boxed{\text{Therefore, there is sufficient evidence to warrant rejection of the claim that the population mean is 80.8. The sample data support the claim that the population mean is greater than 80.8.}}\)
