How to calculate the confidence interval?

Question

We are measuring the IQ of students at a college. We choose a sample of 30 students, and find for the same a mean IQ of 118.3, with a standard deviation of 11.4. Find the 95% confidence interval for the population mean. I have the answer [114.221 to 122.379] but I want to know the step by step formula to use. I don't know the statistical symbols so if someone can write it out for me I would greatly appreciate it.

Answer 1

Hi,
Sounds as if you want a blow-by-blow account, so I'll try to give it to you. First of all, technically, to use the normal distribution, we need n>30 and the POPULATION standard deviation. So, even though it going to make very little difference, we'll use the Student t distribution.

Secondly, because of the limitations of this word processor, I need to define a couple terms:
a) Let a (should be alpha) be 1-degree of confidence, 1-.95 = .05.
b) Let t(a/2) be the t-value for a two-tailed distribution.
c) x-bar is the sample mean, an x with a bar over it.

Now, let's do the work.
1) The maximum error is:
E = t(a/2) * s/sqrt(n)
We know that s = 11.4 and n = 30, but we need t(a/2).
2) To find t(a/2), we look in a Student t distribution table with .05 in two tails and with 29 (that's 30-1) degrees of freedom. We get 2.045.
3) Now back to our formula in step 2.
E = 2.045 *11.4/Sqrt(30)
=4.256
4) Finally, the interval is:
x-bar +- E
= 118.3 +4.256
and
= 118.3 - 4.256
I'll leave that arithmetic to you.
As you can see, using Student t made very little difference.

Hope this helps.
FE

Answer 2

The general formula for a confidence interval around a population mean (µ) is:
Xbar ± Zα/2[s/√n] Where Xbar is the mean of your sample; s is the sample standard deviation. Z(sub)α/2 is the Z-value in in the standard normal table that "cuts-off" the α/2 probability of 0.025. In this case Zα/2 = 1.96.
So your confidence interval is:
118.3 ± 1.96[11.4/5.477] = 118.3 ± 4.08 and the interval is 114.2 < µ < 122.38.