Verified Answer :
- Identify the degrees of freedom \( u = 15\) and critical values for \(\alpha = 0.05\) using the Chi-squared distribution.
- Calculate the variance interval using the formula \(\left[ \frac{(n-1)s^2}{\chi^2_{\alpha/2, u}}, \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, u}} \right]\).
- Take the square root of the variance interval to obtain the confidence interval for the standard deviation.
- The 95% confidence interval for the standard deviation is \(\boxed{[8.86, 18.57]}\) minutes.
Explanation
- Analyze the problem and identify parameters
To estimate the population standard deviation \(\sigma\) using a sample standard deviation, we assume the population is normally distributed. With a sample of size \(n = 16\), a sample standard deviation \(s = 12\), and a confidence level of \(95\%\), we have:
- Degrees of freedom: \( u = n - 1 = 15\).
- Significance level: \(\alpha = 1 - 0.95 = 0.05\), which means we need the critical values \(\chi^2_{0.025, 15}\) and \(\chi^2_{0.975, 15}\).
Using statistical tables or calculation, we find:
- \(\chi^2_{0.975, 15} \approx 6.262\)
- \(\chi^2_{0.025, 15} \approx 27.488\)
- Calculate the confidence interval for the variance \(\sigma^2\)
The confidence interval for the population variance \(\sigma^2\) is given by:
\[ \left[ \frac{(n-1)s^2}{\chi^2_{0.025, 15}}, \frac{(n-1)s^2}{\chi^2_{0.975, 15}} \right] \]
Substituting the known values \(n=16\), \(s=12\), \(\chi^2_{0.025, 15} \approx 27.488\), and \(\chi^2_{0.975, 15} \approx 6.262\):
\[ \sigma^2 \in \left[ \frac{15 \cdot 12^2}{27.488}, \frac{15 \cdot 12^2}{6.262} \right] \]
\[ \sigma^2 \in \left[ \frac{15 \cdot 144}{27.488}, \frac{15 \cdot 144}{6.262} \right] \approx [78.58, 344.93] \]
- Calculate the confidence interval for the standard deviation \(\sigma\)
To find the confidence interval for the standard deviation \(\sigma\), we take the square root of the interval endpoints calculated for the variance:
\[ \sigma \in \left[ \sqrt{78.58}, \sqrt{344.93} \right] \approx [8.86, 18.57] \]
Rounding to two decimal places, the 95% confidence interval for the standard deviation of the time taken to eat lunch is approximately \([8.86, 18.57]\) minutes.
Examples
In restaurant management, understanding the variability in customer dining time is crucial for optimizing table turnover. If the manager knows the standard deviation of lunch times is between 8.86 and 18.57 minutes with 95% confidence, they can better schedule staff and manage peak hour seating, ensuring efficient service and customer satisfaction. The math uses the Chi-squared distribution formula: \(\sigma \in \left[ \sqrt{\frac{(n-1)s^2}{\chi^2_{\alpha/2, u}}}, \sqrt{\frac{(n-1)s^2}{\chi^2_{1-\alpha/2, u}}} \right]\).