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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?

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2 Answers 2

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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

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

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

The confidence interval should then be

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

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