1. Mental health score (GHQ) of a sample of 49 unemployed men are sample mean $=10.54$ (Note: small values indicate better mental health), standard deviation $=5.20$. Test to see if the true mean GHQ score for all unemployed men exceeds 10. Use $a=0.05$. Also, find the $p$-value of the test.

Step 1 ：Identify a sample statistic. Since we are trying to estimate a population mean, we choose the sample mean (10.54) as the sample statistic.

Step 2 ：State the hypotheses. The first hypothesis is \(H_0: \mu = 10\). The second hypothesis is \(H_1: \mu > 10\).

Step 3 ：Compute the standard error (SE). SE is the standard deviation divided by the square root of the sample size. \(SE = \frac{5.20}{\sqrt{49}} = \frac{5.20}{7} = 0.7428571428571429\)

Step 4 ：Compute the test statistic. The formula of the test statistic is (sample mean - population mean) / standard error. \(t = \frac{10.54 - 10}{0.7428571428571429} = 0.7285714285714285\)

Step 5 ：Find the p-value. The p-value is the probability that a t statistic greater than the observed value would occur by chance if the null hypothesis were true. For a t-distribution with 48 degrees of freedom (sample size - 1), a t-value of 0.7285714285714285 corresponds to a one-tailed p-value of approximately 0.235.

Step 6 ：Since the p-value (0.235) is greater than the significance level (0.05), we cannot reject the null hypothesis. Therefore, we do not have enough evidence to say that the true mean GHQ score for all unemployed men exceeds 10.

Step 7 ：\(\boxed{\text{The p-value of the test is 0.235.}}\)