Question

The General Social Survey asked a sample of 1314 people how much time they spent watching TV each day. The mean number of hours was 2.85 with a standard deviation of 2.62. A sociologist claims that people watch a mean of 3 hours of TV per day. Do the data provide sufficient evidence to conclude that the mean hours of TV watched per day is less than the claim? Use the a=0.05 level of significance and the p-value method with the TI-84 Plus calculator.

Compute the value of the test statistic.

Answer 1

To test the sociologist's claim that people watch a mean of 3 hours of TV per day, we will perform a hypothesis test using the following null and alternative hypotheses:

- Null hypothesis \((H_0)\): \(\mu = 3\) (The mean number of hours spent watching TV per day is equal to 3 hours)
- Alternative hypothesis \((H_a)\): \(\mu < 3\) (The mean number of hours spent watching TV per day is less than 3 hours)

Given:
- Sample size \(n = 1314\)
- Sample mean \(\bar{x} = 2.85\)
- Sample standard deviation \(s = 2.62\)
- Population mean under the null hypothesis \(\mu_0 = 3\)

We will use a one-sample t-test, and the test statistic can be calculated using the formula:

\[
t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}
\]

Substituting the values:

\[
t = \frac{2.85 - 3}{2.62 / \sqrt{1314}}
\]

Calculating the standard error:

\[
s / \sqrt{n} = \frac{2.62}{\sqrt{1314}} \approx \frac{2.62}{36.244} \approx 0.0723
\]

Now substitute back into the test statistic formula:

\[
t = \frac{2.85 - 3}{0.0723} \approx \frac{-0.15}{0.0723} \approx -2.07
\]

Thus, the value of the test statistic is approximately:

\[
t \approx -2.07
\]