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 5% of 75%.

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. @ganeshie8

Answer 1

I really need help with this problem @ganeshie8

Answer 2

@ganeshie8 Please ganishie

Answer 3

What does "nine times out of ten" tell you?

Answer 4

The margin of error should be fairly obvious.

Answer 5

As for the confidence interval, it would have the form
\[\left(\bar{x}-Z_{\alpha/2}\sqrt{\frac{\sigma^2}{n}},~\bar{x}+Z_{\alpha/2}\sqrt{\frac{\sigma^2}{n}}\right)\]
\(\bar{x}\) is the estimated population mean, \(Z_{\alpha/2}\) is a critical value for a \((1-\alpha)\%\) confidence level, and \(\sqrt{\dfrac{\sigma^2}{n}}\) is the margin of error.

\(\bar{x}\) is given explicitly. The confidence level is just a matter of interpreting that phrase I pointed out. The margin of error is given.

Answer 6

Yes it is but I somewhat need help setting this up.

Answer 7

If you can help me set this up @SithsAndGiggles

Answer 8

You have the formula and all the necessary info. There's not much I can do in the way of helping without just telling the answer.

Answer 9

If some event occurs "nine out of ten times," what does that mean? How many times out of a hundred does the event occur?

Answer 10

Ok so let me check if am actually doing this correctly 1.96√p(1-p)/N

Answer 11

So if I were to plug in values it would be 1.96√9/10(1-(9/10)/?

Answer 12

right @SithsAndGiggles

Answer 13

Not quite, the experiment in the question is trying to estimate the population mean, not a population proportion. The 'nine times out of ten' clause should tell you something about the confidence level.

Answer 14

That it would 9/10

Answer 15

Right, and \(\dfrac{9}{10}=\dfrac{90}{100}=90\%\), so the interval has a 90% confidence level.

Answer 16

Ok then so would I plug that into the equation

Answer 17

What are the critical values for a 90% confidence interval?

Answer 18

1.645

Answer 19

@SithsAndGiggles

Answer 20

Correct, so \(Z_{\alpha/2}=Z_{.05}=1.645\). (Don't worry about the notation if you haven't seen it before.)

The margin of error is given to be 5%, or 0.05, since an average score is expected to fall within 5% of the estimated mean.

The estimated mean itself is?

Answer 21

5%

Answer 22

No, that's the margin of error. What's the estimated average score?

Answer 23

Is it 5%+1.645? @SithsAndGiggles

Answer 24

No.
\[\left(\bar{x}-Z_{\alpha/2}\sqrt{\frac{\sigma^2}{n}},~\bar{x}+Z_{\alpha/2}\sqrt{\frac{\sigma^2}{n}}\right)\]
\(\bar{x}\) is the estimated average score. According to the given info, that would be 75%, or 0.75.

The exam creator says the scores will fall within 5% of a 75% score, so the margin of error is 5%, or 0.05. In the formula, \(\sqrt{\dfrac{\sigma^2}{n}}\) is the margin of error.

The critical value for a 90% confidence interval is 1.645, so \(Z_{\alpha/2}=1.645\).

So to reiterate, the confidence interval would be
\[\left(0.75-1.645\times0.05,~0.75+1.645\times0.05\right)\]
Simplify.

Answer 25

oh ok and that would be my answer to this problem? @SithsAndGiggles

Answer 26

Your answer would be the interval, yes. I suggest you study up on the topic. It doesn't seem like you're really grasping the procedure here.

Answer 27

Confidence Interval: 9 out of 10 = 90%
The exam creator says the scores will fall within 5% of a 75% score, so the margin of error is 5%, or 0.05.
The confidence interval is (.67, .83)
90% confident that the average scores fall in between 67% and 83%.

Answer 28

you're welcome