Question

A field test for a new exam was given to randomly selected freshmen. 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.

I think I can calculate the margin of error and the confidence interval but I don't know which values to use.

Answer 1

I also know the z score for 90% is 1.645, 95% is 1.96, 99% is 2.575

Answer 2

@ganeshie8 Can you help me?

Answer 3

the exam creator claims that on the same exam, nine times out of ten, seniors will have an average score within 6% of 80%.

Answer 4

that means Confidence Interval is 90%

Answer 5

Can you explain further please?

Answer 6

9 / 10

means 90 / 100

so, the Confidence Interval is 90% accurate, okay ?

Answer 7

Oh, I see. So how would I go about finding the margin of error?

Answer 8

and,

confidence interval = \(\large 80\% \pm 6\%\)

Answer 9

so "Margin of Error" is \(\large 6 \%\)

Answer 10

in interval notation,

confidence interval = \(\large (74\%, ~86\%)\)

Answer 11

see if that makes more or less sense.. .

Answer 12

Yes that makes a lot of sense

Answer 13

sure :)

Answer 14

everything looks correct, except for ur interpretation of confidence interval in the last line

Answer 15

the "actual average score" of seniors is NOT a floating target

Answer 16

actual average score is a population parameter

Answer 17

the right interpretation is below :-
we are 90% certain that the actual average score of seniors will lie between 74% and 86%

Answer 18

in other words :
if u take multiple samples of freshmen, 90% of the samples will capture the "actual average score" of seniors

Answer 19

Ok I see. Thank you so much @ganeshie8 :)

Answer 20

u wlc :)