# Confidence intervals for ratios of variances and other confusing stats

## Homework Statement:

Well, it's statistics and probabilities, but I couldn't figure out a better sub to post it. It's not exactly homework, it's an old exam exercise. I don't have a lot of time to learn the subject so I'm trying to figure out how to solve exercises from older tests.

The exercise gives you two populations of steel rods, population A containing 4 rods and population B containing 5 rods. The quantity measured for each bar is the weight they can take until they bend, measured in tons. For A, the mean is 18.6 and the standard deviation is 1.8, for B the mean is 17.8 and the standard deviation 2.1.

The first question requires you to find a 95% confidence interval for the ratio of the variances of populations A and B.
The second question asks you to check the hypothesis that the mean for bars of type A is higher than that of type B for 99% confidence (I am less than 99% sure I am confident in my understanding of what this means).
The third question gives you that A follows a distribution with a mean of 17.5 and a standard deviation of 2. It also gives you that a population of bars is considered acceptable if the mean is above 17, and asks you what the probability of a population of bars of type A to be "acceptable" is.

## Relevant Equations:

I don't even know...
Now I don't really know much about the subject, I'm primarily just peaking into my textbook to see how to solve this or that exercise. I believe I can figure out how to solve the third question. However I couldn't find how to solve the first two. I know how to find a 95% confidence interval for, say, the mean or the variance or whatever of a population. But how do I determine the confidence interval for the ratio between the variances for two different populations with different sizes? I'm also a bit confused about what the second question wants you to do conceptually. Any help would be appreciated, and will probably help me pass my exams a lot.

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Mark44
Mentor
Homework Statement:: Well, it's statistics and probabilities, but I couldn't figure out a better sub to post it. It's not exactly homework, it's an old exam exercise. I don't have a lot of time to learn the subject so I'm trying to figure out how to solve exercises from older tests.

The exercise gives you two populations of steel rods, population A containing 4 rods and population B containing 5 rods. The quantity measured for each bar is the weight they can take until they bend, measured in tons. For A, the mean is 18.6 and the standard deviation is 1.8, for B the mean is 17.8 and the standard deviation 2.1.

The first question requires you to find a 95% confidence interval for the ratio of the variances of populations A and B.
The second question asks you to check the hypothesis that the mean for bars of type A is higher than that of type B for 99% confidence (I am less than 99% sure I am confident in my understanding of what this means).
The third question gives you that A follows a distribution with a mean of 17.5 and a standard deviation of 2. It also gives you that a population of bars is considered acceptable if the mean is above 17, and asks you what the probability of a population of bars of type A to be "acceptable" is.
Relevant Equations:: I don't even know...

Now I don't really know much about the subject, I'm primarily just peaking into my textbook to see how to solve this or that exercise. I believe I can figure out how to solve the third question. However I couldn't find how to solve the first two. I know how to find a 95% confidence interval for, say, the mean or the variance or whatever of a population. But how do I determine the confidence interval for the ratio between the variances for two different populations with different sizes? I'm also a bit confused about what the second question wants you to do conceptually. Any help would be appreciated, and will probably help me pass my exams a lot.
Regarding the first question, what you need to look at is the F test, a test that is used to compare the variances of two samples. Here's a link to some information: https://www.itl.nist.gov/div898/handbook/eda/section3/eda359.htm

For the second question, what you need is a two-sample t-test. Here's a link from the same source as above: https://www.itl.nist.gov/div898/handbook/eda/section3/eda353.htm

• AndreasC
Regarding the first question, what you need to look at is the F test, a test that is used to compare the variances of two samples. Here's a link to some information: https://www.itl.nist.gov/div898/handbook/eda/section3/eda359.htm

For the second question, what you need is a two-sample t-test. Here's a link from the same source as above: https://www.itl.nist.gov/div898/handbook/eda/section3/eda353.htm
Hmmm, the interval looks larger than I anticipated... Is there some kind of calculator I can use or something to confirm my result?

To be clear, I believe the second question expects you to conclude that the two variances are equal, in which case you can use the simpler formula to calculate the degrees of freedom for the t distribution (4+5-2=7, which is roughly consistent with what the most complex formula would give you). However the interval I get is pretty big so it doesn't seem like I can draw that conclusion.

Mark44
Mentor
To be clear, I believe the second question expects you to conclude that the two variances are equal
Do you mean "expects you to assume that the two variances are equal"?
From post #1, you said this:
The second question asks you to check the hypothesis that the mean for bars of type A is higher than that of type B for 99% confidence [/quote]
You would conclude that the null hypothesis is true (means are equal) or false (means not equal); i.e., you reject the null hypothesis.
Depending on how you set up your alternate hypothesis can make the difference on your confidence interval.
means unequal -- Two tailed test, with .005 probability in each tail
mean A > mean B - One tailed test, with .01 probability in one end of the distribution.
mean A < mean B - One tailed test, with .01 probability at the other end of the distribution.
AndreasC said:
, in which case you can use the simpler formula to calculate the degrees of freedom for the t distribution (4+5-2=7, which is roughly consistent with what the most complex formula would give you). However the interval I get is pretty big so it doesn't seem like I can draw that conclusion.

Last edited:
Do you mean "expects you to assume that the two variances are equal"?
From post #1, you said this:
The second question asks you to check the hypothesis that the mean for bars of type A is higher than that of type B for 99% confidence
You would conclude that the null hypothesis is true (means are equal) or false (means not equal); i.e., you reject the null hypothesis.
Depending on how you set up your alternate hypothesis can make the difference on your confidence interval.
means unequal -- Two tailed test, with .005 probability in each tail
mean A > mean B - One tailed test, with .01 probability in one end of the distribution.
mean A < mean B - One tailed test, with .01 probability at the other end of the distribution.
No, I mean I believe the answer to the first question should be that the variances are roughly equal. The reason I'm saying that is because if that is not the case, then I'm gonna have to use the more complex formula to calculate the degrees of freedom for the t test, and then the result is something like 6.93 or so. I believe the result should be 7, which is indeed the result if you assume the variances are equal and use the simpler formula n+m-2.