It's proposed that the average life expectancy these days is 75.2 , but recent data suggests it might be less than that. A sample size of 84 countries found a sample mean life expectancy of 74.4 years with a standard deviation of 12.7 years. You want to formulate and test a hypothesis that can be used to challenge the validity of the claim at a significance level of 0.02 . With hypotheses $H_{0}: \mu \geq 75.2$ vs $H_{1}: \mu<75.2$, what is/are the critical value(s)? (Report answer accurate to 2 decimal places.) What is the test statistic for this sample? (Report answer accurate to 3 decimal places.) What is the p-value for this sample? (Report answer accurate to 4 decimal places.)

Step 1 ：We are given that the average life expectancy these days is 75.2 years, but recent data suggests it might be less than that. A sample size of 84 countries found a sample mean life expectancy of 74.4 years with a standard deviation of 12.7 years. We want to formulate and test a hypothesis that can be used to challenge the validity of the claim at a significance level of 0.02. The hypotheses are \(H_{0}: \mu \geq 75.2\) vs \(H_{1}: \mu<75.2\).

Step 2 ：First, we need to calculate the critical value. The critical value is the z-score that corresponds to the 0.02 significance level in the standard normal distribution. The critical value is \(-2.05\).

Step 3 ：Next, we calculate the test statistic using the formula \((\text{sample mean} - \text{population mean}) / (\text{standard deviation} / \sqrt{\text{sample size}})\). The test statistic is \(-0.577\).

Step 4 ：Finally, we calculate the p-value. The p-value is the probability of observing a value as extreme as, or more extreme than, the test statistic under the null hypothesis. The p-value is \(0.7181\).

Step 5 ：Final Answer: The critical value is \(\boxed{-2.05}\), the test statistic is \(\boxed{-0.577}\), and the p-value is \(\boxed{0.7181}\).