Question

e. Suppose that a random sample of 40 men taking the verbal SAT is selected. What is the 97.5th percentile for the sample mean score of men based on a random sample of 40 men taking the verbal SAT?

Answer 1

4. According to the US College Board, the scores on the verbal Scholastic Aptitude Test (SAT) for women have a mean of 502 and a standard deviation of 123. The scores for the verbal SAT for men have a mean of 506 and a standard deviation of 112. Assume that the data are normally distributed.

Answer 2

n = 40 z = 1.96

Answer 3

By the Central Limit Theorem, for large n, the sampling distribution for the mean of the values in the same will be approximately normal. Thus,

(mean{x} - mean) / std_dev ~ N(0, 1), so...

mean{x} = N(0,1) * std_dev + mean

You want z = 1.96, as you stated, so 1.96 * 123 + 502.

Answer 4

Typo: "same" = "sample"

Answer 5

so 743.08?

Answer 6

no std for men is 112

Answer 7

Seems about right.

Answer 8

so 725

Answer 9

Okay, I guess I need to read the question more carefully. Good catch.

Answer 10

Mean for men is also 506.

Answer 11

Suppose that a random sample of 40 men taking the verbal SAT is selected. What is the 97.5th percentile for the sample mean score of men based on a random sample of 40 men taking the verbal SAT? is also 725.52............

Answer 12

so it makes sense that if we do a smaller sample size

Answer 13

the answer is the same?

Answer 14

Ooops.

Answer 15

I made a mistake. The standard deviation needs to be divided by the square root of n.

Answer 16

i mean sorry wrong copy

Answer 17

b. What is the 97.5th percentile for the scores of men?

Answer 18

Wait, nevermind. I got myself confused. Let me re-read question.

Answer 19

Is the same answer as 725

Answer 20

Ok, so, it's not asking what is the 97.5th percentile score, but the 97.5th percentile for the mean score. So, here's the formula:

Answer 21

mean = 506 + 112/sqrt(40)*1.96

Answer 22

The reason we divide by square root of n has to do with the fact that as n increases, the variance in the mean will decrease. This can be proven if you know how to calculation expectations and variances of, say, the mean function.

Answer 23

And the vairance will decrease by a factor of 1/n, or equivalently, the standard error will decrease by a factor of 1/sqrt(n).

Answer 24

Okay so I got an answer of 49.85. Is this my final answer or do I need to use the Z table?

Answer 25

97.5th percentile of the mean = 506 + 112/sqrt(40)*1.96

Answer 26

So, clearly, it's nowhere close. We know that it must be at least greater than the mean, 506.

Answer 27

You just have to plug in those numbers given.

Answer 28

But here's the formula: sample_mean + 1.96 * (standard_deviation / sqrt(n))

Answer 29

I got 84.56

Answer 30

It is still a lot smaller than the mean 506

Answer 31

What else am I doing wrong?

Answer 32

How are you getting that number?

Answer 33

506 + (some positive number) should be > 506.

Answer 34

Make sure that you are not, for instance, dividing the entire formula by sqrt(n)

Answer 35

OH I plugged it into the wrong formula

Answer 36

is 49.85 mu?

Answer 37

506 + 112/sqrt(40)*1.96 = 540.71

Answer 38

I just plugged it into a calculator. We know, immediately, that anything less than 506 can't be it though.

Answer 39

So, how to interpret this: when given a sample of size 40 from this population, if we take the mean of this sample, 97.5% of the time, it will be less than 540.71.

Answer 40

thank you so much!