To calculate the confidence interval for the population standard deviation \( \sigma \) using the sample standard deviation \( s \), we can use the chi-square distribution. The formula for the confidence interval for the population standard deviation is given by:
\[ \left( \sqrt{\frac{(n-1)s^2}{\chi^2_{1 - \alpha/2}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2}}} \right) \]
Where:
- \( n \) = sample size
- \( s \) = sample standard deviation
- \( \alpha \) = level of significance (1 - confidence level)
- \( \chi^2 \) = chi-square critical values
In this case:
- \( n = 19 \)
- \( s = 0.16 \)
- Confidence level = 95% → \( \alpha = 0.05 \)
We need to find the critical chi-square values for \( n-1 = 18 \) degrees of freedom.
Using a chi-square table or calculator:
- \( \chi^2_{0.025, 18} \) (for \( \alpha/2 = 0.025 \)) approximately equals \( 34.171 \)
- \( \chi^2_{0.975, 18} \) (for \( 1 - \alpha/2 = 0.975 \)) approximately equals \( 8.551 \)
Now we can plug these values into the formula:
-
Lower Limit: \[ \sqrt{\frac{(n-1)s^2}{\chi^2_{0.975}}} = \sqrt{\frac{(19-1)(0.16)^2}{8.551}} = \sqrt{\frac{18 \times 0.0256}{8.551}} = \sqrt{\frac{0.4608}{8.551}} = \sqrt{0.0539} \approx 0.232 \quad (\text{rounded to 3 decimal places}) \]
-
Upper Limit: \[ \sqrt{\frac{(n-1)s^2}{\chi^2_{0.025}}} = \sqrt{\frac{(19-1)(0.16)^2}{34.171}} = \sqrt{\frac{18 \times 0.0256}{34.171}} = \sqrt{\frac{0.4608}{34.171}} = \sqrt{0.0135} \approx 0.116 \quad (\text{rounded to 3 decimal places}) \]
Thus, the confidence interval for the population standard deviation \( \sigma \) is approximately:
\[ \left( 0.116, 0.232 \right) \]
Now rounding the limits to two decimal places, we find:
The final confidence interval is:
\[ \left( 0.12, 0.24 \right) \]
Therefore, the correct answer is:
C) \( 0.12 < \sigma < 0.24 \).