Question

the prices of a random sample of 23 new motorcycles have a sample standard deviation of $3710. assume the sample is from a normally distributed population. construct a confidence interval for the population variance \\(\sigma^{2}\\) and the population standard deviation \\(\sigma\\). use a 95% level of confidence. interpret the results.
what is the confidence interval for the population variance \\(\sigma^{2}\\)?
(\\(\square\\),\\(\square\\)) (round to the nearest integer as needed.)

Explanation:

Step1: Determine the degrees of freedom

The degrees of freedom \(df=n - 1\), where \(n = 23\). So \(df=23-1 = 22\).

Step2: Find the critical values

For a \(95\%\) confidence level (\(\alpha=1 - 0.95=0.05\)), \(\frac{\alpha}{2}=0.025\) and \(1-\frac{\alpha}{2}=0.975\).
Using the \(\chi^{2}\) - distribution table or a calculator, \(\chi_{L}^{2}=\chi_{0.025,22}^{2}=36.781\) and \(\chi_{R}^{2}=\chi_{0.975,22}^{2}=10.982\).

Step3: Calculate the confidence interval for the population variance \(\sigma^{2}\)

The formula for the confidence interval for the population variance is \(\frac{(n - 1)s^{2}}{\chi_{U}^{2}}<\sigma^{2}<\frac{(n - 1)s^{2}}{\chi_{L}^{2}}\), where \(s = 3710\) and \(n = 23\).
First, \((n - 1)s^{2}=(23 - 1)\times3710^{2}=22\times13764100 = 302810200\).
Then \(\frac{(n - 1)s^{2}}{\chi_{U}^{2}}=\frac{302810200}{36.781}\approx8233333\) and \(\frac{(n - 1)s^{2}}{\chi_{L}^{2}}=\frac{302810200}{10.982}\approx27573333\).

Answer:

\((8233333,27573333)\)