# Find the 99% confidence interval (Interval and test for proportion)

I have the following question and I am getting different results from a friend (I think he forgot to halve $\alpha$). Problem is:

A random sample of 100 students from a large school was taken. It was found 38 went on a trip last month, 62 had not. Obtain a 99% confidence interval for the proportion of students who went on a trip last month.

Solution:

$$N=100, X=38$$ $$\hat{p} = {X\over{N}}$$ $$\hat{p} \pm z_{\alpha/2} \sqrt{{\hat{p}(1-\hat{p})}\over{N}}$$

I then take $z_{\alpha/2}$ from the t-table corresponding to $\alpha = 0.005$ and $n=\infty$ which is 2.576. Is this correct?

Therefore, I get the interval of (0.255, 0.505).

• What is the result of your friend? – callculus Aug 5 '18 at 18:47
• After I´ve made my own calculation I got the same result like you. Therefore it is correct :) The bounds are $0.38\pm 2.576\cdot \sqrt{\frac{0.62\cdot 0.38}{100}}$ – callculus Aug 5 '18 at 18:53
• The results are here – callculus Aug 5 '18 at 18:59
• @callculus he didn't halve $\alpha$ so he used $z=1.96$ – s5s Aug 5 '18 at 19:00
• That´s indeed not right, since it is a two sided interval. But you are right. – callculus Aug 5 '18 at 19:02

An improved "Agresti-Coull" 99% confidence interval uses $\check p = (X+.5c^2)/(n+c^2),$ where $X$ is the number of successes in $n$ trials, and $c = 2.576.$ Then the CI is $$\check p \pm c\sqrt{\frac{\check p(1- \check p)}{n+c^2}}.$$ Then, for your data, $\check p = 0.3875$ and the 99% CI is $( 0.266, .509).$
For very large $n$ (say $n \ge 1000$) the difference between the two types of CIs disappears. For more on binomial confidence intervals see this Q & A and its References.