# How to solve this Confidence Level Question?

A company has conducted a test on $$400$$ people. Out of $$400$$ people, $$224$$ people like chocolate. Find the percentage of people who like chocolate in the $$99\%$$ confidence interval.

The problem is the question doesn't give the standard deviation for the calculation. I believe the mean is \begin{align*} \frac{224}{400} &= 0.56\quad\text{(the mean)}\\[5pt] N &= 400\\[5pt] Z &= \frac{2.576}{(99\% CI)} \\[5pt] 4^n &= 262,144 \\[5pt] CI &= \text{mean} \mp \frac{s \cdot Z}{\sqrt{N}} \\[5pt] \end{align*}

How do I find the $$s$$ value?

your distribution is a bernulli $$Bern(0.56)$$

Its standard deviation is $$\sigma=\sqrt{p(1-p)}$$

A good estimation of $$\sigma$$ is

$$\hat{\sigma}=\sqrt{\overline{X}(1-\overline{X})}$$

This is your

$$s=\sqrt{0.56\cdot\left(1-0.56\right)}\approx0.4964$$

The standard deviation of a sample proportion is $$\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}.$$

The confidence interval should then be

$$\left(\hat{p} - Z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \, \hat{p} + Z^*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\right).$$