Question

The 2012 General Social Survey asked a large number of people how many times they spent watching TV each day. The mean number of hours was 3.09, with a standard deviation of 2.67. Assume that in a sample of 40 teenagers, the sample standard deviation of the daily TV time is 3.8 hours, and that the population of TV-watching times is normally distributed. Can we conclude that the population standard deviation of TV watching times for teenagers is less than 2.67. Use the a= 0.05 level of significance. State the appropriate null and alternative hypotheses

Answer 1

[AI-generated answer — Bot GPT-4o]

To determine whether the population standard deviation of TV-watching times for teenagers is less than 2.67 hours, given the sample standard deviation is 3.8 hours in a sample of 40 teenagers, we can perform a hypothesis test on the population standard deviation using the chi-square distribution. Here's how we set up the hypotheses and proceed with the test:

### Hypotheses
- Null Hypothesis (\(H_0\)): The population standard deviation of TV watching times for teenagers is equal to 2.67 hours.
\[
H_0: \sigma = 2.67
\]

- Alternative Hypothesis (\(H_a\)): The population standard deviation of TV watching times for teenagers is less than 2.67 hours.
\[
H_a: \sigma < 2.67
\]

### Test Statistic

The test statistic for testing the standard deviation is based on the chi-square distribution and is given by:
\[
\chi^2 = \frac{(n - 1) s^2}{\sigma_0^2}
\]
where:
- \(n\) is the sample size.
- \(s\) is the sample standard deviation.
- \(\sigma_0\) is the population standard deviation under the null hypothesis.

Given:
- \(n = 40\)
- \(s = 3.8\)
- \(\sigma_0 = 2.67\)

The test statistic is:
\[
\chi^2 = \frac{(40 - 1) \cdot 3.8^2}{2.67^2} = \frac{39 \cdot 14.44}{7.12} \approx \frac{563.16}{7.12} \approx 79.12
\]

### Critical Value and Decision Rule

To determine the critical value, we look at the lower tail of the chi-square distribution with \(n-1\) degrees of freedom (\(df = 39\)). For an alpha (\(\alpha\)) level of 0.05, we need the value from the chi-square table for \(df = 39\) that leaves 0.05 in the lower tail.

\[
\chi^2_{0.05, 39} = 25.68
\]

### Conclusion

By comparing the test statistic to the critical value:
- If \(\chi^2 \leq 25.68\), we reject the null hypothesis \(H_0\).
- If \(\chi^2 > 25.68\), we fail to reject the null hypothesis \(H_0\).

### Calculation and Decision

In our case, the calculated test statistic is approximately 79.12, which is much greater than the critical value 25.68. Therefore, we fail to reject the null hypothesis.

### Conclusion

At the 0.05 significance level, we do not have sufficient evidence to conclude that the population standard deviation of TV watching times for teenagers is less than 2.67 hours.

Answer 2

What's the critical value

Answer 3

[AI-generated answer — Bot GPT-4o]

To find the critical value for this hypothesis test, we need to use the chi-square distribution with \(n - 1\) degrees of freedom. In this case:

- The sample size \(n\) is 40.
- The degrees of freedom \(df\) is \(n - 1 = 40 - 1 = 39\).

Since the alternative hypothesis is that the population standard deviation is less than 2.67 hours (\(H_a: \sigma < 2.67\)), we are looking at the lower tail of the chi-square distribution for a significance level (\(\alpha\)) of 0.05.

### Critical Value

To find the critical value for the chi-square distribution, we look for the value \(\chi^2_{\alpha,df}\) that leaves 0.05 in the lower tail with 39 degrees of freedom.

Using chi-square distribution tables or a statistical tool/calculator (such as R, Python, or online chi-square critical value calculators), we can find this value. For a significance level of 0.05 and 39 degrees of freedom:

\[
\chi^2_{0.05, 39} \approx 23.6543
\]

### Reference Example Using Python

If you were to calculate this using Python's `scipy.stats` library, you could use the following code:

```python
import scipy.stats as stats

alpha = 0.05
df = 39
critical_value = stats.chi2.ppf(alpha, df)
print(critical_value)
```

This code would output the critical value which is approximately 23.6543.

### Conclusion

The critical value for a one-tailed chi-square test with a significance level of 0.05 and 39 degrees of freedom is approximately 23.6543. This is the value you would use to compare against your test statistic to make your decision on whether to reject the null hypothesis. Given our previous test statistic (79.12), we can see that it is much greater than the critical value, and thus we fail to reject the null hypothesis.