1. ## Hello everyone, I need your help...

Hello, studying for a test and I have a review question that I just can't seem to get.. if anyone can help me it would be greatly appreciated...

A container of oil is supposed to hold 1000 ml of oil. We need to be sure that the standard deviation of the oil containers is less than 20 ml. We randomly select 10 cans of oil with a mean of 997 ml and a standard deviation of 32 ml. Use these samples to construct a 95% confidence interval for the true value of the standard deviation.

2. Originally Posted by iza0iza
Hello, studying for a test and I have a review question that I just can't seem to get.. if anyone can help me it would be greatly appreciated...

A container of oil is supposed to hold 1000 ml of oil. We need to be sure that the standard deviation of the oil containers is less than 20 ml. We randomly select 10 cans of oil with a mean of 997 ml and a standard deviation of 32 ml. Use these samples to construct a 95% confidence interval for the true value of the standard deviation.
Read this: Estimating a population standard devation or variance

3. Originally Posted by iza0iza
Hello, studying for a test and I have a review question that I just can't seem to get.. if anyone can help me it would be greatly appreciated...

A container of oil is supposed to hold 1000 ml of oil. We need to be sure that the standard deviation of the oil containers is less than 20 ml. We randomly select 10 cans of oil with a mean of 997 ml and a standard deviation of 32 ml. Use these samples to construct a 95% confidence interval for the true value of the standard deviation.
n is small (n < 30), so use the following confidence interval:

$\displaystyle \bar{x} \pm t_{\alpha/2}(n-1)\frac{s}{\sqrt{n}}$

Where $\displaystyle \bar{x} = 997, s = 32, \alpha = 0.05, n = 10$

4. Originally Posted by colby2152
n is small (n < 30), so use the following confidence interval:

$\displaystyle \bar{x} \pm t_{\alpha/2}(n-1)\frac{s}{\sqrt{n}}$

Where $\displaystyle \bar{x} = 997, s = 32, \alpha = 0.05, n = 10$
Hi colby. Unfortunately, the question asks for a confidence interval for the population standard deviation, not for the population mean ...... (I nearly made the same mistake).

Since n is small (n < 30), the required confidence interval is of the form:

$\displaystyle \frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2}, ~ df=n-1}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1 - \frac{\alpha}{2}, ~ df=n-1}}$.

Note that there are two critical values that must be read from the appropriate tables (due to the asymmetry of the chi-square distribution .....)