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
Best Answer
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.