# Confidence Intervals, Proportion Estimations

I am stuck on one of my homework questions for my stats class. I was wondering if anyone could give me some insight as to how to find an answer. here is the question:

In a study of perception, 107 men are tested and 24 are found to have red/green color blindness.

(a) Find a 92% confidence interval for the true proportion of men from the sampled population that have this type of color blindness. (b) Using the results from the above mentioned survey, how many men should be sampled to estimate the true proportion of men with this type of color blindness to within 2% with 98% confidence? (c) If no previous estimate of the sample proportion is available, how large of a sample should be used in (b)?

I have already answered (a). However, I am at a complete loss as to how to answer (b) or (c). For one thing, I am not quite sure what (b) is even asking. Advice on this question would be greatly appreciated. Thanks!

Without giving all the answers, I think the following outline will be helpful.

Let $X$ be the number of color-blind men in a sample of $n.$ Then $X \sim \mathsf{Binom}(n, \theta),$ where $\theta$ is the population proportion of color-bindness. The MLE and MME of $\theta$ is $\hat \theta = X/n.$ In the study mentioned $\hat \theta = 24/107 = 0.2243.$

(a) A traditional 92% CI for $\theta$ is of the form $$\hat \theta \pm 1.74\sqrt{\frac{\hat \theta(1-\hat\theta)}{n}},$$ where 1.74 cuts 4% of the area from the upper tail of the standard normal distribution. In this formula, the '(estimated) standard error' is $\sqrt{\frac{\hat \theta(1-\hat\theta)}{n}}$ and the 92% 'margin of error' is $1.75\sqrt{\frac{\hat \theta(1-\hat\theta)}{n}}.$

(b) I will leave it to you to find the 98% margin of error $M$ and then let $M = .02$ and solve for $n$ with $\hat \theta = 0.2243.$

(c) Because $\theta(1-\theta)$ is at its maximum when $\theta = 1/2,$ use that value in $M$ when solving for $n.$

Notes:(1) When finding $n$ as in (b) and (c) you will not generally get an integer value. In that case, it is customary to round up to the nearest integer value.

(2) The displayed formula in (a) depends on two approximations: (i) the normal approximation to the binomial and (ii) the approximation of $\hat \theta$ to $\theta$ in the standard error. Especially for $n$ as small as $n = 107$ the latter approximation is questionable. Other forms of CIs are more accurate, including those due to Wilson and Agresti, which you can explore in statistics texts and on the Internet. My guess is that you are expected to use the formula I provided.

• Thanks for your answer! Just for clarification purposes, what do MLE and MME stand for? – ImRellyBadAtMath Oct 26 '17 at 16:34
• actually never mind, i solved it. Thank you for your help! It was greatly appreciated – ImRellyBadAtMath Oct 26 '17 at 17:25
• Maximum Likelihood Estimation, Method of Moments estimation – BruceET Oct 26 '17 at 22:43