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?

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)

Now when and first sample variance and second sample variance \large ,

Then the estimated difference between two mean is calculated as:

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

(standard error of the sample mean differences)

Now when and first sample variance and second sample variance ,

Then the estimated difference between two mean is calculated as:

c) In order compare the sample size from a) to b) that is from , that is four times increases leads to decreases estimated standard error from

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

your answer is a

step-by-step explanation:

4*4*4*4=256

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}