Statistics: Confidence Intervals

Question

6. Independent random samples were drawn from two normal populations assumed to have the approximately
the same standard deviation, resulting in the following summary statistics:
n1 = 25, s1 = 4.8, n2 = 16, s2 = 5.1
(a) Calculate a pooled estimate of the common standard deviation of the two populations.
(b) What is the margin of error for a 99% CI for the difference of the means?

Answer 1

Hello Vincent! I will first start by saying that most statisticians wouldn't pool their standard deviations, as it typically wouldn't provide a more precise estimation of the sampling standard deviation. For the purposes of this question, there are multiple ways to pool the standard deviation for two populations. I calculated the pooled standard deviation using two different methods and they were only off by .04. So, I'll explain the easier of the two formulas.

First, square both sample standard deviations to combine them at the variance level per combination of random variables rules. Add these two solutions together for your numerator and then divide by 2. Take this solution and square it, which would provide you with an average of both sample variances and the square root converts it to the sample standard deviation, which is our goal. Specifically, [(4.8)²+(5.1)²]/2 = 24.525, which is your pooled variance. Take the square root of 24.525, which is 4.952 or the sample pooled standard deviation. That is the answer to part A of your question.

Part B of your question is the margin of error, which by definition is the critical value times the sample standard pooled standard deviation that we calculated in part A. Use InvT (.005, 15) with .005 representing a tail of your t distribution and 15 representing the smaller of your two sample sizes. The critical value or t star is 2.947, which then would be multiplied by your answer in A (4.952), to get a margin of error of 14. 594. I hope this helps you along your statistical journey. Take care now.