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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).

s5s
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1 Answers1

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Your computation is correct for the type of confidence interval you are using. However, this kind of confidence interval is known not to provide the promised level of confidence, in your case 99%.

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.

BruceET
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