### Section 5.4: Distribution of the Difference Between Two Sample Means

#### Key Concepts

This sampling distribution should be used whenever the situation involves two independent samples from different populations and the sample means are compared. As in all examples presented in this chapter, the central limit theorem allows the use of the normal distribution to find probabilities.

In this situation, there are different formula for the mean and standard error, but the same logic and procedure for solving problems remains the same.

#### Formula

The sampling distribution of is summarized by:

mean( ) = and SE( ) = The shape will be approximately normal for sufficiently large samples. For most practical applications, this will hold if each sample has at least 25 or 30 observations.

#### Example

A method to assess the effectiveness of a drug is to measure its concentration in the urine after a period of time. Suppose that for two brands of aspirin, the mean concentrations are 19.2 and 15.6 (mg%) with standard deviations of 8.6 and 7.8 (mg%) respectively, one hour after ingestion. Twenty people are given the first brand and twenty five are given the second brand. What is the probability that the mean difference in their concentrations ( ) is more than 4.0 (mg%)?

The sampling distribution for the difference in sample means will be approximately normal with a mean of 19.2 - 15.5 = 3.7 and a standard deviation of = 2.48. The z-score is

z = (4.0 - 3.7) / 2.48 = 0.12

Since the area to the right of 0.12 under the standard normal curve is .4522, the probability is about 45% that the mean difference is 4 mg% or higher.

Last modified: Feb 19, 1996

Bret Larget, larget@mathcs.duq.edu