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}\\)?
(round to the nearest integer as needed.)

Explanation:

Step1: Determine the degrees of freedom and critical values

The degrees of freedom \(df=n - 1\), where \(n = 23\), so \(df=23-1 = 22\).
For a 95% confidence level, \(\alpha=1 - 0.95=0.05\), and \(\frac{\alpha}{2}=0.025\), \(1-\frac{\alpha}{2}=0.975\).
From the \(\chi^{2}\) - distribution table, \(\chi_{L}^{2}=\chi_{0.025,22}^{2}=36.781\), \(\chi_{R}^{2}=\chi_{0.975,22}^{2}=10.982\).
The sample standard deviation \(s = 3710\), so the sample variance \(s^{2}=3710^{2}=13764100\).

Step2: Calculate the confidence interval for the population variance \(\sigma^{2}\)

The formula for the confidence interval of the population variance is \(\frac{(n - 1)s^{2}}{\chi_{\frac{\alpha}{2},n - 1}^{2}}\leq\sigma^{2}\leq\frac{(n - 1)s^{2}}{\chi_{1-\frac{\alpha}{2},n - 1}^{2}}\).
Substitute the values: \(\frac{(23 - 1)\times13764100}{36.781}\leq\sigma^{2}\leq\frac{(23 - 1)\times13764100}{10.982}\).
First, \(\frac{22\times13764100}{36.781}=\frac{302810200}{36.781}\approx8233339\).
Second, \(\frac{22\times13764100}{10.982}=\frac{302810200}{10.982}\approx27573365\).

Answer:

\((8233339,27573365)\)