Hi everyone,
I'm having a little trouble with these two questions...

1. Find the 99% confidence interval for the variance and standard deviation of the weights of 25 one-gallon containers of motor oil if a sample of 14 containers has a variance of 3.2. The weights are given in ounces. Assume the variables is normally distributed.

2. Find the 95% confidence interval for the variance and standard deviation of the ounces of coffee that a machine dispenses in 12-ounce cups. Assume the variable is normally distributed. Use the following data... (12.03,12.10,12.02,11.98,12.00,12.05,11.97,11.99).


I'm not looking for the answer just how to get to the answer. I know how find chi-square distributions. There are no other questions in the book like #1 to compare whether I am doing it right and #2 I attempted but kept on getting stuck on the end when I was dividing (n-1)s2 by the chi-square distribution for the right side. It kept giving me a really low # like .000069943. So I figured I was doing it wrong.


Thank you anyone that can help me! :)

-Cara

For #1, do you have a mean or data from which we can get a mean?

The standard deviation is the square root of the variance. In this case 1.7888...

The question I typed here is the exact question from the book. Is it able to be solved?

It says 'weights are given in ounces'. What weights? That's what makes me think something is missing.

I think what they want me to do is use that information...
If 14oz has a variance of 3.2 than what would 25gallons (3200oz) have a variance of. I just can't figure out how to get there. Since there are no other problems like that I can find I can't check my answer to make sure I am right. Thank goodness I'm doing my homework ahead of time. :)

1. Find the 99% confidence interval for the variance and standard deviation of the weights of 25 one-gallon containers of motor oil if a sample of 14 containers has a variance of 3.2. The weights are given in ounces. Assume the variables is normally distributed.


We don't need the mean. I was suffering from temporary cerebral flatulence.

s=\sqrt{3.2}=1.7888

The area to the right of {\chi}_{R}^{2}=\frac{1-c}{2}=\frac{1-.99}{2}=.005

The area to the left is {\chi}_{L}^{2}=\frac{1+c}{2}=.995

From the chi distribution table, using the values n=14, \;\ d.f.=13, \;\ c=.99, the critical values for {\chi}_{R}^{2}=29.819 \;\ and \;\ {\chi}_{L}^{2}=3.565

Using these values and s=1.7888, the C.I. for {\sigma}^{2} is as follows:

left endpoint:

\frac{(n-1)s^{2}}{{\chi_{R}}^{2}}=\frac{(14-1)(3.2)}{29.819}=1.395

right endpoint:

\frac{(14-1)(3.2)}{3.565}=11.669

1.395<{\sigma}^{2}<11.669

The confidence interval for the variance is:

1.181<{\sigma}<3.416

So, with 99% confidence, we can say the population variance is between 1.395 and 11.669.

The ppopulation standard deviation is between 1.181 and 3.416.

So the part where it says "25 gallons" is point less to have in the question? That's the part that confused me the most.

Thank you very much for the taking the time out to help me with this too!