Question

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.
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

Answer:

Step-by-step explanation:

According to the given question ,  two samples are selected from populations with equal means with the following information:

a) The estimated difference between two mean is \large 4 which is calculated as:

(standard error of the sample mean differences)

$\large S_{\bar{x}_{1}-\bar{x}_{2}}=\sqrt{\left ( \frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}} \right )}$

Now when $\large n_{1}=n_{2}=5$ and first sample variance $\large s^{2}_{1}=38$ and second sample variance \large $s^{2}_{2}=42$ ,

Then the estimated difference between two mean is calculated as:

$\large S_{\bar{x}_{1}-\bar{x}_{2}}=\sqrt{\left ( \frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}} \right )} \\\\\large =\sqrt{\frac{38}{5}+\frac{42}{5}} \\\\\large =\sqrt{16} \\\\\large =4$

b) The estimated difference between two mean is \large 2 which is calculated as:

(standard error of the sample mean differences)

$\large S_{\bar{x}_{1}-\bar{x}_{2}}=\sqrt{\left ( \frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}} \right )}$

Now when $\large n_{1}=n_{2}=20$ and first sample variance $\large s^{2}_{1}=38$ and second sample variance $\large s^{2}_{2}=42$ ,

Then the estimated difference between two mean is calculated as:

$\large S_{\bar{x}_{1}-\bar{x}_{2}}=\sqrt{\left ( \frac{s^{2}_{1}}{n_{1}}+\frac{s^{2}_{2}}{n_{2}} \right )} \\\\\large =\sqrt{\frac{38}{20}+\frac{42}{20}} \\\\\large =\sqrt{4} \\\\\large =2$

c) In order compare the sample size from a) to b) that is from $\large n=5\, to\, \large n=20$ , that is four times increases leads to decreases estimated standard error from

$\large S_{\bar{x}_{1}-\bar{x}_{2}}\mid _{n=5}=4>S_{\bar{x}_{1}-\bar{x}_{2}}\mid _{n=20}=2$

Therefore when variances are unchanged, then increase in sample size, inversely (decreases) affect the size of the standard error for the sample mean difference.

Sample size increases= SE decreases