. For each of the following, assume that the two samples are selected from populations with equal means and calculate how much difference should be expected, on average, between the two sample means.

Question

a. Each sample has n _ 5 scores with s2 _ 38 for the first sample and s2 _ 42 for the second. (Note: Because the two samples are the same size, the pooled variance is equal to the average of the two sample variances.)
b. Each sample has n _ 20 scores with s2 _ 38 for the first sample and s2 _ 42 for the second.
c. In part b, the two samples are bigger than in part a, but the variances are unchanged. How does sample size affect the size of the standard error for the sample mean difference?

Answer 1

a. To calculate the standard error for the mean difference (SEM_diff), we start with the formula:

SEM_diff = sqrt((pooled variance / n1) + (pooled variance / n2))

First, we need to calculate the pooled variance for samples of the same size:

pooled variance = (s1^2 + s2^2) / 2 = (38 + 42) / 2 = 40

Now we can calculate the SEM_diff:

SEM_diff = sqrt((40 / 5) + (40 / 5)) = sqrt(16) = 4

So, on average, there should be a difference of 4 between the two sample means.

b. Using the same formula with n = 20 for both samples:

SEM_diff = sqrt((40 / 20) + (40 / 20)) = sqrt(4) = 2

So, on average, we should expect a difference of 2 between the two sample means when the sample size is 20.

c. The standard error for the mean difference decreases as the sample size increases. In part a, the SEM_diff was 4 with a sample size of 5. In part b, the SEM_diff was 2 with a sample size of 20. As the sample size increases, the standard error decreases, meaning that the sample means become more precise and close to the true population means. This is because larger samples provide more information about the population and reduce the impact of random sampling error.