Difference in Means

Statistics problems often involve comparisons between two independent sample means. This lesson explains how to compute probabilities associated with differences between means.

Difference Between Means: Theory

Suppose we have two populations with means equal to μ₁ and μ₂. Suppose further that we take all possible samples of size n₁ and n₂. And finally, suppose that the following assumptions are valid.

• The size of each population is large relative to the sample drawn from the population. That is, N₁ is large relative to n₁, and N₂ is large relative to n₂. (In this context, populations are considered to be large if they are at least 20 times bigger than their sample.)
• The samples are independent; that is, observations in population 1 are not affected by observations in population 2, and vice versa.
• The set of differences between sample means is normally distributed. This will be true if each population is normal or if the sample sizes are large. (Based on the central limit theorem, sample sizes of 40 would probably be large enough).

Given these assumptions, we know the following.

• The expected value of the difference between all possible sample means is equal to the difference between population means. Thus,

  E(x₁ - x₂) = μ_d = μ₁ - μ₂.

• The standard deviation of the difference between sample means (σ_d) is approximately equal to:

  σ_d = sqrt( σ₁² / n₁ + σ₂² / n₂ )

It is straightforward to derive the last bullet point, based on material covered in previous lessons. The derivation starts with a recognition that the variance of the difference between independent random variables is equal to the sum of the individual variances. Thus,

σ²_d = σ² _{(x1 - x2)} = σ² _x1 + σ² _x2

If the populations N₁ and N₂ are both large relative to n₁ and n₂, respectively, then

σ² _x1 = σ²₁ / n₁

σ² _x2 = σ²₂ / n₂

σ_d² = σ₁² / n₁ + σ₂² / n₂

σ_d = sqrt( σ₁² / n₁ + σ₂² / n₂ )

Difference Between Means: Sample Problem

In this section, we work through a sample problem to show how to apply the theory presented above. In this example, we will use Stat Trek's Normal Distribution Calculator to compute probabilities.

Problem 1

For boys, the average number of absences in the first grade is 15 with a standard deviation of 7; for girls, the average number of absences is 10 with a standard deviation of 6.

In a nationwide survey, suppose 100 boys and 50 girls are sampled. What is the probability that the male sample will have at most three more days of absences than the female sample?

(A) 0.025
(B) 0.035
(C) 0.045
(D) 0.055
(E) None of the above

Solution

The correct answer is B. The solution involves three or four steps, depending on whether you work directly with raw scores or z-scores. The "raw score" solution appears below:

• Find the mean difference (male absences minus female absences) in the population.

  μ_d = μ₁ - μ₂ = 15 - 10 = 5

• Find the standard deviation of the difference.

  σ_d = sqrt( σ₁² / n₁ + σ₂² / n₂ )

  σ_d = sqrt(7²/100 + 6²/50)

  σ_d = sqrt(49/100 + 36/50)

  σ_d = sqrt(0.49 + .72) = sqrt(1.21) = 1.1

• Find the probability. This problem requires us to find the probability that the average number of absences in the boy sample minus the average number of absences in the girl sample is less than 3. To find this probability, we use Stat Trek's Normal Distribution Calculator. Specifically, we enter the following inputs: 3, for the normal random variable; 5, for the mean; and 1.1, for the standard deviation.

We find that the probability of the mean difference (male absences minus female absences) being 3 or less is about 0.035.

Alternatively, we could have worked with z-scores (which have a mean of 0 and a standard deviation of 1). Here's the z-score solution:

• Find the mean difference (male absences minus female absences) in the population.

  μ_d = μ₁ - μ₂ = 15 - 10 = 5

• Find the standard deviation of the difference.

  σ_d = sqrt( σ₁² / n₁ + σ₂² / n₂ )

  σ_d = sqrt(7²/100 + 6²/50) = sqrt(49/100 + 36/50)

  σ_d = sqrt(0.49 + .72) = sqrt(1.21) = 1.1

• Find the z-score that is produced when boys have three more days of absences than girls. When boys have three more days of absences, the number of male absences minus female absences is three. And the associated z-score is

  z = (x - μ)/σ = (3 - 5)/1.1 = -2/1.1 = -1.818

• Find the probability. To find this probability, we use Stat Trek's Normal Distribution Calculator. Specifically, we enter the following inputs: -1.818, for the normal random variable; 0, for the mean; and 1, for the standard deviation.

We find that the probability of probability of a z-score being -1.818 or less is about 0.035. Of course, the result is the same, whether you work with raw scores or with z-scores.

Note: Some analysts might have used the t-distribution to compute probabilities for this problem. We chose the normal distribution because the population variance was known and the sample size was large; but either choice would have been acceptable. In a previous lesson, we offered some guidelines for choosing between the normal and the t-distribution.