1. ## calculating confidence

I have this problem in my homework and I can't seem to arrive at the correct answer

In a random sample of 500 college students at WVU, 72% indicated they preferred to study while listening to music or some kind of background noise while others indicated they liked to study in a quiet setting. What is the 99% confidence interval for music/background noise studying?

a. 72 ± 3.44

b. 72 ± 14.14

c. 72 ± .72

d. 72 ± 9.83

e. None of the above

Can anyone teach me how to solve these kinds of questions?

2. ## Re: calculating confidence

Originally Posted by DocNougat
I have this problem in my homework and I can't seem to arrive at the correct answer
In a random sample of 500 college students at WVU, 72% indicated they preferred to study while listening to music or some kind of background noise while others indicated they liked to study in a quiet setting. What is the 99% confidence interval for music/background noise studying?
a. 72 ± 3.44
b. 72 ± 14.14
c. 72 ± .72
d. 72 ± 9.83
e. None of the above
Approx normal dist to binomial dist:

p=.72
q=1-p

approx SD = $\displaystyle \sigma \text{:=}\sqrt{p q} = 0.449$

99% confidence interval of mean = $\displaystyle 0.72\pm 2.58\sqrt{p q} = 0.72\pm 1.15842$

3. ## Re: calculating confidence

So it would be none of the above?

4. ## Re: calculating confidence

(b) would be good too!

5. ## Re: calculating confidence

Also how do you get the 2.58?

7. ## Re: calculating confidence

Thanks, this makes more sense now. Any advice on the other question I posted?

8. ## Re: calculating confidence

Originally Posted by MaxJasper
Approx normal dist to binomial dist:

p=.72
q=1-p

approx SD = $\displaystyle \sigma \text{:=}\sqrt{p q} = 0.449$

99% confidence interval of mean = $\displaystyle 0.72\pm 2.58\sqrt{p q} = 0.72\pm 1.15842$
Doesn't make sense, values are too large and go over 100%! Did you perhaps forget the 1/N in the square root? $\displaystyle 0.72\pm 2.58\sqrt{p q/N}$ where N=500?