Confidence interval without std?

Question

I have the following problem:

A survey was done among 138 people. They had to choose between dark chocolate and milk chocolate. 88 of them preferred dark chocolate. Compute the 90% confidence interval for the proportion p of the people who prefer dark chocolate.

I know that the formula of a confidence interval is:

$[X+ Z\cdot\sqrt(\frac{\sigma^2}{n}), X+ Z\cdot\sqrt(\frac{\sigma^2}{n})]$

With X = 88/138 and n = 138. But I am stuck because the standard deviation is not given. I know that the survey is done only once, does that mean that there is no std?

How do I proceed?

Thanks

@PeterForeman: side question: is this an unbiaised estimator of the variance ? — , Aug 13 '20 at 08:45
@YvesDaoust I believe so because any function of an unbiased estimator provides an unbiased estimator. — Peter Foreman, Aug 13 '20 at 08:55
https://en.wikipedia.org/wiki/Binomial_proportion_confidence_interval, https://math.stackexchange.com/q/1448233/321264. — StubbornAtom, Aug 13 '20 at 15:55

score 2 · Accepted Answer · answered Aug 13 '20 at 08:42

2

The confidence interval for the proportion is

$[p-z \sqrt{\frac{p(1-p)}{n}}, p+z \sqrt{\frac{p(1-p)}{n}}]$

Where $p$ is the sample proportion.

answered Aug 13 '20 at 08:42

What is $z$ here ? – Aug 13 '20 at 08:46
1

It is the quantile of the normal distribution and it is given according to confidence level. – Aug 13 '20 at 08:58
thank you for pointing this out! – stats_noob Sep 04 '22 at 07:27

score 2 · Answer 2 · 2020-08-13T09:03:24.090

2

The distribution is a binomial law with estimated probability $p=\frac{88}{138}$. The confidence interval can be obtained as the $5\%$ and $95\%$ quantiles of the binomial distribution $B(n,p)$.

The Gaussian approximation is justified, though the upper limits differs by one unit (low limit in $[78,79]$, high limit in $[96,97]$ or $[97,98]$).

edited Aug 13 '20 at 09:03

answered Aug 13 '20 at 08:48

Confidence interval without std?

2 Answers2

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