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cldubitsky · Mar 27, 2009, 12:39 PM

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

galactus · Mar 27, 2009, 12:44 PM

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...

cldubitsky · Mar 27, 2009, 03:01 PM

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

galactus · Mar 27, 2009, 03:48 PM

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

cldubitsky · Mar 27, 2009, 05:14 PM

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. :)

galactus · Mar 27, 2009, 07:08 PM

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.

cldubitsky · Mar 27, 2009, 07:47 PM

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.

cldubitsky · Mar 27, 2009, 07:49 PM

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