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
b. with 95% confidence, you can say that the
population variance is greater than
population variance is less than
c. with 5% confidence, you can say that the
d. with 95% confidence, you can say that the
population variance is between
population variance is between
8232853 and 27573365
what is the confidence interval for the population standard deviation \\( \sigma \\)?
( ) (round to the nearest integer as needed.)

Explanation:

Step1: Recall the formula for confidence interval of standard deviation

If the confidence interval for variance \(\sigma^{2}\) is \((L, U)\), then the confidence interval for standard deviation \(\sigma\) is \((\sqrt{L},\sqrt{U})\)

Step2: Calculate the lower and upper bounds

Given \(L = 8232853\) and \(U=27573365\)
The lower bound for \(\sigma\) is \(\sqrt{8232853}\approx2869\)
The upper bound for \(\sigma\) is \(\sqrt{27573365}\approx5251\)

Answer:

\((2869,5251)\)