Use the given degree of confidence and sample data to find a confidence interval for the population standard deviation σ. Assume that the population has a normal distribution. Round the confidence interval limits to the same number of decimal places as the sample standard deviation.

To find the standard deviation of the diameter of wooden dowels, the manufacturer measures 19 randomly selected dowels and finds the standard deviation of the sample to be s = 0.16. Find the 95% confidence interval for the population standard deviation σ.

A) 0.11 < σ < 0.25
B) 0.13 < σ < 0.22
C) 0.12 < σ < 0.24
D) 0.15 < σ < 0.21

1 answer

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:

  1. Lower Limit: \[ \sqrt{\frac{(n-1)s^2}{\chi^2_{0.975}}} = \sqrt{\frac{(19-1)(0.16)^2}{8.551}} = \sqrt{\frac{18 \times 0.0256}{8.551}} = \sqrt{\frac{0.4608}{8.551}} = \sqrt{0.0539} \approx 0.232 \quad (\text{rounded to 3 decimal places}) \]

  2. 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) \( 0.12 < \sigma < 0.24 \).