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 . Then give me the critical value

Answer 1

To determine whether we can conclude that the population standard deviation of TV-watching times for teenagers is less than 2.67 hours, we will conduct a hypothesis test for the population standard deviation.

### Step 1: State the Hypotheses
**Null Hypothesis (H₀):** The population standard deviation of TV-watching times for teenagers is not less than 2.67 hours (i.e., it is 2.67 hours or more).
\[ H_0: \sigma \geq 2.67 \]

**Alternative Hypothesis (H₁):** The population standard deviation of TV-watching times for teenagers is less than 2.67 hours.
\[ H_1: \sigma < 2.67 \]

### Step 2: Determine the Test Statistic
We'll use the Chi-Square test for the population standard deviation:
\[ \chi^2 = \frac{(n-1)s^2}{\sigma_0^2} \]
where:
- \( n \) is the sample size,
- \( s \) is the sample standard deviation,
- \( \sigma_0 \) is the hypothesized population standard deviation.

From the problem:
- \( n = 40 \),
- \( s = 3.8 \),
- \( \sigma_0 = 2.67 \).

So,
\[ \chi^2 = \frac{(40-1) \cdot (3.8)^2}{(2.67)^2} \]
\[ \chi^2 = \frac{39 \cdot 14.44}{7.1289} \]
\[ \chi^2 \approx \frac{563.16}{7.1289} \]
\[ \chi^2 \approx 79.01 \]

### Step 3: Determine the Critical Value
We need the critical value for a Chi-Square distribution with \( n - 1 = 39 \) degrees of freedom at the 0.05 significance level (one-tailed test).

Using a Chi-Square distribution table or calculator, you can find the critical value (χ²) at the 0.05 significance level for 39 degrees of freedom. Because this is a left-tailed test:

\[ \chi^2_{\text{critical}} = \inf{39, (1 - 0.05)} \]

For a 0.95 confidence level (since it's the lower tail):
\[ \chi^2_{0.95,39} \approx 24.433 \]

### Step 4: Compare the Test Statistic to the Critical Value
- **If** \( \chi^2 \) calculated (79.01) is greater than the critical value (24.433), we **fail to reject** the null hypothesis.
- **If** \( \chi^2 \) calculated is less than the critical value, we **reject** the null hypothesis.

Since \( \chi^2 \approx 79.01 \) is much greater than 24.433, we **fail to reject** the null hypothesis.

### Conclusion
Based on our calculations, we do not have enough evidence at the 0.05 significance level to conclude that the population standard deviation of TV watching times for teenagers is less than 2.67 hours.