Question

retail thefts find the 90% confidence interval for the variance and standard deviation of the number of retail thefts per week at 22 convenience stores if the standard deviation of the sample is 6.6. assume the variable is approximately normally distributed. use the chi - square distribution table to find any chi - square values to three decimal places. round your final answers to two decimal places.

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 22\). So \(df=22-1 = 21\).

Step2: Find the chi - square values

For a \(90\%\) confidence interval, \(\alpha=1 - 0.90=0.10\), and \(\frac{\alpha}{2}=0.05\), \(1-\frac{\alpha}{2}=0.95\).
From the chi - square distribution table, \(\chi^{2}_{L}=\chi^{2}_{0.95,21}=11.591\) and \(\chi^{2}_{R}=\chi^{2}_{0.05,21}=32.671\).

Step3: Calculate the confidence interval for the variance

The formula for the confidence interval for the variance is \(\frac{(n - 1)s^{2}}{\chi^{2}_{R}}<\sigma^{2}<\frac{(n - 1)s^{2}}{\chi^{2}_{L}}\).
Given \(s = 6.6\), then \(s^{2}=43.56\).
Substitute the values: \(\frac{(22 - 1)\times43.56}{32.671}<\sigma^{2}<\frac{(22 - 1)\times43.56}{11.591}\).
\(\frac{21\times43.56}{32.671}<\sigma^{2}<\frac{21\times43.56}{11.591}\).
\(\frac{914.76}{32.671}<\sigma^{2}<\frac{914.76}{11.591}\).
\(28.00<\sigma^{2}<78.92\).

Step4: Calculate the confidence interval for the standard deviation

Take the square root of each part of the variance interval.
\(\sqrt{28.00}<\sigma<\sqrt{78.92}\).
\(5.29<\sigma<8.88\).

Answer:

\(28.00<\sigma^{2}<78.92\)
\(5.29<\sigma<8.88\)