Questions: 2. In order to determine how many hours per week freshmen college students watch online news and video, a random sample of 25 students was selected. It was determined that the students in the sample spent an average of 20.3 hours with a sample standard deviation of 3.5 hours watching TV per week. Please answer the following questions: (a) Provide a 95% confidence interval estimate for the average number of hours that all college freshmen spend watching online news and video per week. (b) Assume that a sample of 31 students was selected (with the same mean and the sample standard deviation). Provide a 95% confidence interval estimate for the average number of hours that all college freshmen spend watching online news and video per week.

Solution Steps

For the first sample of 25 students, we calculate the 95% confidence interval for the average number of hours spent watching online news and video per week. The formula used is:

\[ \bar{x} \pm t \frac{s}{\sqrt{n}} \]

Where:

• \(\bar{x} = 20.3\) (sample mean)
• \(t \approx 2.06\) (t-value for 95% confidence and 24 degrees of freedom)
• \(s = 3.5\) (sample standard deviation)
• \(n = 25\) (sample size)

Substituting the values:

\[ 20.3 \pm 2.06 \cdot \frac{3.5}{\sqrt{25}} = 20.3 \pm 2.06 \cdot 0.7 \]

Calculating the margin of error:

\[ 20.3 \pm 1.442 \]

Thus, the confidence interval is:

\[ (20.3 - 1.442, 20.3 + 1.442) = (18.86, 21.74) \]

For the second sample of 31 students, we again calculate the 95% confidence interval using the formula:

\[ \bar{x} \pm z \frac{s}{\sqrt{n}} \]

Where:

• \(\bar{x} = 20.3\) (sample mean)
• \(z \approx 1.96\) (z-value for 95% confidence)
• \(s = 3.5\) (sample standard deviation)
• \(n = 31\) (sample size)

Substituting the values:

\[ 20.3 \pm 1.96 \cdot \frac{3.5}{\sqrt{31}} = 20.3 \pm 1.96 \cdot 0.629 \]

Calculating the margin of error:

\[ 20.3 \pm 1.235 \]

Thus, the confidence interval is:

\[ (20.3 - 1.235, 20.3 + 1.235) = (19.07, 21.53) \]

Final Answer