Math Snap

STEP 1

1. The sample size is $n = 8$.
2. The sample standard deviation is $s = 6$.
3. The distribution of IQ scores is normal.
4. We are constructing a $95\%$ confidence interval for the population standard deviation $\sigma$.

STEP 2

1. Determine the chi-square critical values.
2. Calculate the confidence interval for the population variance.
3. Calculate the confidence interval for the population standard deviation.

STEP 3

Determine the chi-square critical values for a $95\%$ confidence interval with $n-1 = 7$ degrees of freedom.
For a $95\%$ confidence interval, the significance level $\alpha = 0.05$. The critical values are found using a chi-square distribution table:
$$\chi^2_{\alpha/2, n-1} = \chi^2_{0.025, 7} \quad \text{and} \quad \chi^2_{1-\alpha/2, n-1} = \chi^2_{0.975, 7}$$ Using a chi-square table, we find:
$$\chi^2_{0.025, 7} \approx 2.167 \quad \text{and} \quad \chi^2_{0.975, 7} \approx 16.013$$

STEP 4

Calculate the confidence interval for the population variance $\sigma^2$.
The formula for the confidence interval for the variance is:
$$\left( \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}}, \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}} \right)$$ Substitute the values:
$$\left( \frac{7 \times 6^2}{16.013}, \frac{7 \times 6^2}{2.167} \right)$$ Calculate the values:
$$\left( \frac{252}{16.013}, \frac{252}{2.167} \right) \approx (15.74, 116.29)$$

Calculate the confidence interval for the population standard deviation $\sigma$.
Take the square root of the variance interval:
$$\left( \sqrt{15.74}, \sqrt{116.29} \right) \approx (3.97, 10.79)$$ The $95\%$ confidence interval for the population standard deviation $\sigma$ is:
$$\boxed{3.97 < \sigma < 10.79}$$