Questions: The General Social Survey polled a sample of 1048 adults in the year 2010, asking them how many hours per week they spent on the Internet. The sample mean was 9.79 with a population standard deviation of 13.41. A second sample of 1399 adults was taken in the year 2014. For this sample, the mean was 11.62 with a population standard deviation of 15.02. We are interested in testing whether the mean number of hours per week spent on the Internet increased between 2010 and 2014. The p-value for this test is: .025 .0016 .05 .0008

Solution Steps

We want to test whether the mean number of hours spent on the Internet has increased from 2010 to 2014. Thus, we define our hypotheses as follows:

• Null Hypothesis \( H_0: \mu_1 \leq \mu_2 \) (The mean hours in 2010 is less than or equal to the mean hours in 2014)
• Alternative Hypothesis \( H_a: \mu_1 > \mu_2 \) (The mean hours in 2010 is greater than the mean hours in 2014)

The standard error \( SE \) is calculated using the formula: \[ SE = \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{\frac{13.41^2}{1048} + \frac{15.02^2}{1399}} = \sqrt{\frac{179.8281}{1048} + \frac{225.6004}{1399}} = 0.5769 \]

The test statistic \( z \) is calculated using the formula: \[ z = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{9.79 - 11.62}{0.5769} = -3.172 \]

The p-value is calculated using the formula for a one-tailed test: \[ P = 2 \times (1 - Z(|z|)) = 0.0015 \]

Based on the calculated p-value of \( 0.0015 \), we compare it to the significance level \( \alpha = 0.05 \). Since \( P < \alpha \), we reject the null hypothesis \( H_0 \). This indicates that there is sufficient evidence to conclude that the mean number of hours spent on the Internet has increased from 2010 to 2014.

Final Answer