Calculating variance / std deviation

MacDanger

Hi

Working on this question:

In a sample of 200 registered voters, 108 plan to vote for candidate A. Construct a 95% confidence interval for the proportion of voters that plan to vote for Candidate A.

Two questions on this:
1) Do I use (p)(q) = (108/200)(92/200) to calculate variance & std dev? Or should I use (n)(p)(q)??
2) Is this result considered "S" or "Sigma" squared? i.e. do I subsequently calculate the Z-statistic or the t-statistic?

Thanks

Yakuza

http://en.wikipedia.org/wiki/Binomial_distribution#Normal_approximation

B(n,p)~ N(np,npq) (Variance is npq), standard dev is [latex]\sqrt{npq}[/latex]

Your interval is 1.96 standard devs either side of the mean.

MacDanger

Good stuff, thanks Yakuza

MacDanger

Hi

Are you sure about that? When I calculate it out, the result doesn't make any sense. Or am I making a mistake somewhere??

95% CI for the proportion of voters voting for candidate A:

Z97.5 = 1.96; n = 200; p = 0.54; q = 0.46
μ = p = 0.54;
σ2 = (n)(p)(q) = (200)(0.54)(0.46) = 49.68; σ = 7.048

CI = μ ± (Z97.5)(σ/√n)
= 0.54 ± (1.96)(7.048√200) = 0.54 ± 0.977 = [-0.437, 1.517]

A negative value and a value > 1 for proportion of voters makes no sense?

Calculating using σ2 = (p)(q) = (0.54)(0.46) = 0.2484; σ = 0.4984
Gives:
CI = μ ± (Z97.5)(σ/√n)
= 0.54 ± (1.96)(0.4984/√200) = 0.54 ± 0.069 = [0.471, 0.609]

Which appears to be a reasonable CI although perhaps tighter than I would have expected.

Any help appreciated :-)

Yakuza

μ isn't 0.54, it's np, or 108.
I'd have given 108 ± 1.96 * 7.048 as the CI for the number of voters, or roughly [94.185,121.815].
The question actually asks for the proportion so you need to divide the above by 200, yielding [0.471,0.609].

We both got there in the end

MacDanger

Good man, thanks for clarifying that for me :-)