Question

A field test for a new exam was given to randomly selected seniors. The exams were graded, and the sample mean and sample standard deviation were calculated. Based on the results, the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 6% of 80%. Is the confidence interval at 90%, 95%, or 99%? What is the margin of error? Calculate the confidence interval and explain what it means in terms of the situation.

Answer 1

nine times out of ten, changed to a percent

Answer 2

that would be 0.9 right?

Answer 3

yes. But we are answering the question
Is the confidence interval at 90%, 95%, or 99%?

Answer 4

its 90 percent i get that part. but i dont get how im supposed to find the margin of error without a sample mean or standard deviation

Answer 5

@phi

Answer 6

@Hero

Answer 7

How do i find the margin of error with no sample mean and no standard deviation?
@phi

Answer 8

average score within 6% of 80%.

they tell you it's 6%

Answer 9

I still dont get it like how would i setup the margin of error equation with that information?

Answer 10

@phi

Answer 11

@Hero

Answer 12

@Saifoo.Khan

Answer 13

the problem tells you the expected value (on the test) is 80
with a margin of error of 6, meaning 74 to 86 is the confidence interval
nine times out of ten, i.e. with 90% confidence

Answer 14

oh ok thanks a lot @phi

Answer 15

The margin of error is 6% of 80% which is:
\[\large \frac{80\times6}{100}=4.8\ percent\]

Answer 16

ok so how would i calculate the confidence interval? @kropot72

Answer 17

there is an unfortunate ambiguity here. Do they mean 80 ± 6
or 80 ± 6% (of 80)

Answer 18

im not sure and like i said it says the sample mean and sample standard deviation are the same but theres no value given

Answer 19

though kropot gives the more natural interpretation

Answer 20

so how should i go about finding the std?

Answer 21

They give the sample mean (80% on the test)
they did not give the sample std.

rather they gave (pre-computed) margin of error= Z(for 90%) * std/ sqr(N)

Answer 22

we can't find the sample standard deviation. we need Z for 90% = 1.645
we have the margin of error 6% of 80= 4.8
but we don't have N, the number of people taking the test.

Answer 23

So does that mean I cant do the last question?

Answer 24

But they don't ask for the std. they ask for the confidence interval, which we find using the mean (of 80 ) and the margin of error (4.8)

Answer 25

The margin of error is half the width of the confidence interval. So the confidence interval is:
\[\large (80-4.8, 80+4.8)\]

Answer 26

ohhhh ok see I wouldve never saw that! Thanks alot you guys @phi @kropot72

Answer 27

You're welcome :)