The 

2012

 general Social Survey asked a large number of people how much time 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 

32

 teenagers, the sample standard deviation of daily TV time is 

3.3

 hours, and that the population of TV watching times is normally distributed. Can you conclude that the population standard deviation of TV watching times for teenagers differs from 

2.67

? Use the 

=α0.10

 level of significance.

To determine if the population standard deviation of TV watching times for teenagers differs from 2.67 hours, we can perform a hypothesis test for the population standard deviation using the chi-square distribution. Here are the steps to perform the test:

1. **State the hypotheses:**

   - Null hypothesis (\(H_0\)): \(\sigma = 2.67\)
   - Alternative hypothesis (\(H_a\)): \(\sigma \neq 2.67\)

2. **Calculate the test statistic:**

   The test statistic for the chi-square test for the standard deviation 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, and \( \sigma_0 \) is the population standard deviation under the null hypothesis.

   Given:
   - \( n = 32 \)
   - \( s = 3.3 \)
   - \( \sigma_0 = 2.67 \)

   Plugging in the values:

   \[
   \chi^2 = \frac{(32 - 1) \times (3.3)^2}{(2.67)^2}
   \]

   \[
   \chi^2 = \frac{31 \times 10.89}{7.1289}
   \]

   \[
   \chi^2 = \frac{337.59}{7.1289}
   \]

   \[
   \chi^2 \approx 47.36
   \]

3. **Determine the degrees of freedom:**

   The degrees of freedom (\( df \)) for this test is \( n - 1 \). For our sample:

   \[
   df = 32 - 1 = 31
   \]

4. **Find the critical value and p-value:**

   Since this is a two-tailed test at the \(\alpha = 0.10\) significance level, we need to find the critical values for \( \chi^2 \) at \( df = 31 \) for \(\alpha/2 = 0.05\).

   Using chi-square distribution tables or a calculator:

   - The critical value for the lower tail at 0.05 significance level: \( \chi^2_{0.05, 31} \approx 18.49 \)
   - The critical value for the upper tail at 0.05 significance level: \( \chi^2_{0.95, 31} \approx 45.72 \)

   Our calculated test statistic \( \chi^2 \approx 47.36 \) falls outside this range.

5. **Make the decision:**

   Since our test statistic \( \chi^2 \approx 47.36 \) is greater than the upper critical value of 45.72, we reject the null hypothesis.

**Conclusion:**

At the 0.10 significance level, there is sufficient evidence to conclude that the population standard deviation of TV watching times for teenagers differs from 2.67 hours.

Would you like more details or have any questions about this process?

Here are 8 related questions to expand your understanding:

1. What are the assumptions required for conducting a chi-square test for standard deviation?
2. How would the test change if we used a different significance level, say \(\alpha = 0.05\)?
3. Can you explain the difference between a one-tailed and two-tailed test?
4. How does the sample size affect the chi-square test for standard deviation?
5. What are the implications of rejecting the null hypothesis in this context?
6. How can we calculate the p-value for this test statistic?
7. What is the effect of sample standard deviation being larger than the population standard deviation on the chi-square statistic?
8. Can this method be applied to non-normally distributed populations?

**Tip:** Always check the assumptions of normality when using the chi-square test for standard deviation to ensure the validity of your results.

Learn how to conduct hypothesis testing to determine if the population standard deviation of TV watching times for teenagers differs from 2.67 hours. Detailed step-by-step solution using chi-square test statistic. Suitable for advanced high school or college-level students.

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