Question

For each of the following, assume that the two samples are obtained from populations with the same mean, and calculate how much difference should be expected, on average, between the two sample means. Each sample has n =4 scores with s^2 = 68 for the first sample and s^2 = 76 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).
a) 4.24.
b) 0.24.
c) 8.48.
d) 6.00.
Next, each sample has n=16 scores with s^2 = 68 for the first sample and s^2 = 76 for the second.
a) 0.12.
b) 2.12.
c) 4.24.
d) 3.00.

Answer 1

Answer:

d)6.00

d)3.00

Step-by-step explanation:

We are given that

n=4 scores

$S^2_1=68$

$S^2_2=76$

We have to find the  difference should be expected, on average, between the two sample means.

$S_{M_1-M_2}=\sqrt{\frac{S^2_1}{n_1}+\frac{S^2_2}{n_2}}$

$n_1=n_2=4$

Using the formula

$S_{M_1-M_2}=\sqrt{\frac{68}{4}+\frac{76}{4}}$

$S_{M_1-M_2}=\sqrt{\frac{68+76}{4}}$

$S_{M_1-M_2}=\sqrt{36}=6$

Option d is correct.

Now, replace n by 16

$n_1=n_2=16$

$S_{M_1-M_2}=\sqrt{\frac{68}{16}+\frac{76}{16}}$

$S_{M_1-M_2}=\sqrt{\frac{68+76}{16}}$

$S_{M_1-M_2}=\sqrt{9}=3$

Option d is correct.