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 4% of 70%.

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.

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 4% of 70%.

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.

anonymous · Answer

@rational @robtobey @freckles

anonymous · Answer

@fallengoddess

anonymous · Answer

@tkhunny

tkhunny · Answer

Stop tagging people and start showing your work.  You can't have NOTHING.

anonymous · Answer

margin of error is 4%, the confidence interval would be 90% since says 9/10

anonymous · Answer

i dont know how to calculate confidence interval

anonymous · Answer

margin of error is 4% or 4% to 70%?

tkhunny · Answer

Margin of Error = $(z-score)\cdot (standard-deviation)$

Look up the z-scores for each of those confidence intervals.  (2-tail)

anonymous · Answer

90 = 1.645 95 = 1.96 99 = 2.575

anonymous · Answer

So 1.645 -
How do you find standard and deviation

tkhunny · Answer

That is awesome.  Good work.

You have to memorize it.
Sample Proportion: Score: 0.70,

Sample Standard Deviation: $\sqrt{\dfrac{0.70\cdot(1 - 0.70)}{n}}$

anonymous · Answer

I'm confused on what I'm supposed to do

anonymous · Answer

That problem is $$\sqrt{\frac{ 0.21 }{ n }}$$

anonymous · Answer

@perl

perl · Answer

ok im catching up here. you agree that the confidence level is 90%, since it says 9 out of 10

anonymous · Answer

yes

perl · Answer

this is a sample mean problem, not a sample proportion

perl · Answer

the margin of error is just 4 I believe

anonymous · Answer

i think so too

perl · Answer

so the confidence interval is ( 70 - 4, 70 + 4)

perl · Answer

We don't have enough information anyway to use that formula . 
the margin of error is already given as 4, since it says the sample mean is within 4 of 70. that means the sample mean ranges between 70 - 4, 70 + 4

perl · Answer

that is your confidence interval , at 90% confidence

anonymous · Answer

Well i understand what you wrote but what do i write for my answer?

anonymous · Answer

confidence interval = 90% margin of error = 4% but what do i write for calculate the confidence interval

anonymous · Answer

oh so just the 70 - 4  to 70 + 4 or 66 to 74

perl · Answer

confidence level 90%
margin of error 4%
The confidence interval is (66%, 74%)
Interpretation:
We are 90% confident that the true mean of the population is between 66% and 74%
Another interpretation:
If we were to take 100 additional samples, in 90 times the mean test score would fall between 66% and 74%.

anonymous · Answer

Ok I understand, thanks for all the help!!

perl · Answer

Your welcome :)

tkhunny · Answer

Whoops.  Reading in my sleep, I guess.