# Thread: Constructing A Confidence Interval

1. ## Constructing A Confidence Interval

Hello,
A random survey of 225 of the 2700 institutions of higher learning in the United States was conducted
in 1976 by the Carnegie Commission. The survey showed a mean enrollment of 3700 students per
institution, and a standard deviation of 6000 students. Construct a 95% confidence interval for the
total student enrollment in all 2700 institutions.

I know the answer is 9990000 +- 2120000 but I'm having trouble coming to this answer.

I know by taking the mean enrollment and multiplying it by the total number of institutions you get the 9990000 but I'm having trouble finding the proper interval.

Sorry about posting in the university math section, if it needs to be moved to the regular statistics thread then please do so.

2. ## Re: Constructing A Confidence Interval

Hey ~berserk.

Hint: The variance of Var[X_bar] = Var[X]/N (N is the number of items in the sample) which implies SE(X_bar) = SE(X)/SQRT(N).

However since you are looking at the total number of institutions you are looking at Var[Sum of X's] = N*Var[X] which gives a standard deviation of SQRT(N)*SE(X).

3. ## Re: Constructing A Confidence Interval

yea I'm still confused. I believe I'm supposed to be using the formula: (x_bar)+- z (stdev/sqrt(n)) the regular confidence interval estimate formula. I keep trying different standard deviations but i can't come up with that interval.

4. ## Re: Constructing A Confidence Interval

For this problem you don't use stdev/sqrt(n): instead you use sqrt(n)*stdev because you are looking at a confidence interval for (X1 + X2 + ... + XN) instead of X_bar = (X1 + X2 + ... + XN)/N.

Remember that you are looking at an interval for the total number of people and not the mean which is why you have to adjust your mean and your standard deviation.

5. ## Re: Constructing A Confidence Interval

I still can't seem to get it. is the x-bar supposed to be 9990000? i know the z is +- 1.96 but is the standard deviation still 6000? I supposed to be using the 2700 as the sample size since they want to know total student enrollment?

6. ## Re: Constructing A Confidence Interval

You are meant to get a distribution of (X1 + X2 + ... + XN) instead of X_bar which is (X1 + X2 + ... + XN)/N.

This means your mean is X_bar*N (which is the big number) and your standard error is now SQRT(N)*SE(X) instead of SE(X)/SQRT(N).

Remember that your distribution is not of the mean, but of the total and this is why your estimate and confidence interval need to be adjusted.