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the waiting times (in minutes) of a random sample of 20 people at a ban…

Question

the waiting times (in minutes) of a random sample of 20 people at a bank have a sample standard deviation of 3.2 minutes. construct a confidence interval for the population variance \\( \sigma^{2} \\) and the population standard deviation \\( \sigma \\). use a 95% level of confidence. assume the sample is from a normally distributed population.
what is the confidence interval for the population standard deviation \\( \sigma \\)?
(2.4, 4.7) (round to one decimal place as needed.)
interpret the results. select the correct choice below and fill in the answer box(es) to complete your choice.
(round to one decimal place as needed.)
a. with 95% confidence, you can say that the population standard deviation is between and minutes.
b. with 5% confidence, you can say that the population standard deviation is less than minutes.
c. with 95% confidence, you can say that the population standard deviation is greater than minutes.
d. with 5% confidence, you can say that the population standard deviation is between minutes and minutes.

Explanation:

Step1: Recall the confidence interval formula for population standard deviation

For a normally - distributed population, the confidence interval for the population standard deviation \(\sigma\) is \(\sqrt{\frac{(n - 1)s^{2}}{\chi_{\alpha/2}^{2}}}<\sigma<\sqrt{\frac{(n - 1)s^{2}}{\chi_{1-\alpha/2}^{2}}}\), where \(n = 20\) (sample size), \(s = 3.2\) (sample standard deviation), and for a \(95\%\) confidence interval, \(\alpha=1 - 0.95=0.05\), \(\alpha/2 = 0.025\), \(1-\alpha/2=0.975\). The degrees of freedom \(df=n - 1=19\).
From the \(\chi^{2}\) - distribution table, \(\chi_{0.025,19}^{2}=32.852\) and \(\chi_{0.975,19}^{2}=8.907\).

Step2: Calculate the lower and upper bounds

First, calculate \((n - 1)s^{2}=(20 - 1)\times3.2^{2}=19\times10.24 = 194.56\).
The lower bound for \(\sigma\): \(\sqrt{\frac{194.56}{32.852}}\approx\sqrt{5.922}\approx2.4\).
The upper bound for \(\sigma\): \(\sqrt{\frac{194.56}{8.907}}\approx\sqrt{21.844}\approx4.7\).

Answer:

A. With \(95\%\) confidence, you can say that the population standard deviation is between \(2.4\) and \(4.7\) minutes.