Marc wants to know if the mean age of the prison population in his city is less than 26 years.He obtains a random sample of 25 prisoners,and finds a mean age of 24.4 years and a standard deviation of 9.2 years.At a significance level of 0.05,what is his conclusion?

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Marc is conducting a one-sample t-test to determine if the mean age of the prison population is less than 26 years. The null hypothesis (H0) is that the mean age μ = 26 years, and the alternative hypothesis (Ha) is that the mean age is less than 26 years. To find the t-score, we use the formula: $t = \frac{\bar{x} - \mu}{(s/\sqrt{n})}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean under the null hypothesis, $s$ is the sample standard deviation, and $n$ is the sample size. Plugging in the values: $t = \frac{24.4 - 26}{(9.2/\sqrt{25})} = \frac{-1.6}{1.84} \approx -0.8696$. Looking up this t-value in a t-distribution table or using a calculator for a one-tailed test with 24 degrees of freedom (n-1) at a significance level of 0.05, we find that the p-value is greater than 0.05, indicating that we fail to reject the null hypothesis. This means there is not sufficient evidence to conclude that the mean age is less than 26 years. Among the provided options, the one that matches the conclusion of failing to reject the null hypothesis with a p-value that indicates insufficient evidence to support the alternative hypothesis is the correct choice.

Marc Wants to Know If the Mean Age of the Prison