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} \\)? (8,232,853, 27573365) (round to the nearest integer as needed.) interpret the results. select the correct choice below and fill in the answer box(es) to complete your choice. (round to the nearest integer as needed.) a. with 5% confidence, you can say that the population variance is greater than b. with 95% confidence, you can say that the population variance is less than c. with 5% confidence, you can say that the population variance is between and d. with 95% confidence, you can say that the population variance is between and

Explanation:

Step1: Recall the formula for confidence interval of population variance

The formula for the confidence interval of the population variance \(\sigma^{2}\) when the population is normally distributed is \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}<\sigma^{2}<\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}\), where \(n\) is the sample size, \(s\) is the sample standard deviation, \(\chi_{\alpha/2}^{2}\) and \(\chi_{1-\alpha/2}^{2}\) are the critical values of the chi - square distribution.
Given \(n = 23\), so the degrees of freedom \(df=n - 1=22\). For a \(95\%\) confidence interval, \(\alpha=0.05\), \(\alpha/2 = 0.025\), and \(1-\alpha/2=0.975\).
From the chi - square distribution table, \(\chi_{0.025,22}^{2}=36.781\) and \(\chi_{0.975,22}^{2}=10.982\), and \(s = 3710\), \(s^{2}=3710^{2}=13764100\).

Step2: Calculate the lower and upper bounds of the confidence interval for \(\sigma^{2}\)

The lower bound: \(\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}=\frac{22\times13764100}{36.781}\approx8232853\)
The upper bound: \(\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}=\frac{22\times13764100}{10.982}\approx27573365\)

Step3: Interpret the confidence interval

A confidence interval for the population variance (or standard deviation) gives a range of values within which the true population variance (or standard deviation) is likely to fall. A \(95\%\) confidence interval means that if we were to take many samples and construct confidence intervals in the same way, about \(95\%\) of those intervals would contain the true population variance.

Answer:

D. With \(95\%\) confidence, you can say that the population variance is between \(8232853\) and \(27573365\).