The General Social Survey reported that in a sample of 68 men aged $18-25$, the mean number of hours of television watched per day was 2.76 with a standard deviation of 2.21 . In a sample of 72 women aged $18-25$, the mean number of hours of television watched per day was 2.88 with a standard deviation of 2.43 . Can you conclude that the mean number of hours of television watched differs between men and women? Use a = 0.05 level of significance.
1) State the Null and Alternative Hypotheses.
A. $\mu_{1}=\mu_{2}$
B. $\mu_{1} \neq \mu_{2}$
C. $\mu_{1}> \mu_{2}$
D. $\mu_{1}< \mu_{2}$

Type A, B, C, or D in the answer box
$\mathrm{Ho:}$

H1:
2) Compute the test statistic. Round the answer in 4 decimal places.

Test statistic $=$
3) Compute the P-value. Round the answer in 2 decimal places.

Step 1 ：State the Null and Alternative Hypotheses. The null hypothesis is that the means are equal, and the alternative hypothesis is that the means are not equal. This corresponds to option B. \( \mu_{1} \neq \mu_{2} \)

Step 2 ：Compute the test statistic. Given values are \( n_{1} = 68, x_{1} = 2.76, s_{1} = 2.21, n_{2} = 72, x_{2} = 2.88, s_{2} = 2.43 \). The test statistic is calculated as \( t = \frac{x_{1} - x_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}} + \frac{s_{2}^{2}}{n_{2}}}} \). The computed test statistic is approximately -0.3060

Step 3 ：Compute the P-value. The degrees of freedom is calculated as \( df = n_{1} + n_{2} - 2 \). The p-value is then calculated using the test statistic and the degrees of freedom. The computed p-value is approximately 0.76

Step 4 ：\(\boxed{\text{Final Answer:}}\) 1) The correct option is B. \( \mu_{1} \neq \mu_{2} \) 2) The test statistic is approximately -0.3060 3) The p-value is approximately 0.76