How to determine the 90% confidence interval for the population proportion of students who found the math test easy?

Question

Out of $120$ students, $85%$ found the math test easy. How to determine the $90%$ confidence interval for the population proportion of students who found the math test easy?

I am not sure about my computation but I tried to solve it in the following way.

The $85%$ of 120 students is 102 students.

Since the confidence level is 90%, there will be 92 students

My answer is not an interval but a fixed percentage which is $76.5%$.

Any comments and suggestions will be highly appreciated.

Answer 1

No, that is not correct.

You want to find the real percentage of students who find this test easy.

Your empirical percentage you got is $85/120 \approx 0.708$. So you could say, the percentage of students who find this test easy is 70.8%.

However, this could be just because you the first students who left the test, who might find the test easier as the other ones who are still writing. Or you could ask only the last people, who might find the test harder than others.

In other words, you have no way to check, if your empirical percentage is equal to the real one.

Thus, you need a model for the probability, e.g. each of the student has the same percentage of finding the test easy and they dont interact with each other. Then, you can now look at the empirical mean value and standard deviation and see what real percentage would agree with a probability of 90% with your empirical finding of 85 out of 120.

I dont want to spoil the whole computation here, but hopefully this helped you understand what you are trying to achieve.

Answer 2

The issue with your question is that there are several ways to construct a confidence interval for a binomial proportion.

Let $n$ be the total number of observations (in your case, $n = 120$ students observed). Let $x$ be the number of observations for which the event of interest occurred (here, $x = 85$ represents the number of students that found the test easy).

Now let $\hat p = x/n$ be the observed proportion of students that found the test easy.

Let $\alpha$ be chosen such that the confidence level of the interval is $100(1-\alpha)\%$. In your case, $\alpha = 0.10$ for a $90\%$ confidence interval.

Let $z_{\alpha/2}$ represent the upper $\alpha/2$ percentile of the standard normal distribution; that is to say, it is a $z$-score for which $$\Pr[Z > z_{\alpha/2}] = \alpha/2$$ where $Z$ is a standard normal random variable. In your case, $z_{0.05} \approx 1.645$ because for $Z \sim \operatorname{Normal}(0,1)$, $\Pr[Z > 1.645] \approx 0.05$.

With all this, there is the Wald interval:

$$\hat p \pm z^*_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}.$$

There is the Wilson score interval:

$$\frac{1}{1+z_{\alpha/2}^2/n} \left(\hat p + \frac{z_{\alpha/2}^2}{2n}\right) \pm \frac{z_{\alpha/2}}{1 + z_{\alpha/2}^2/n} \sqrt{\frac{\hat p (1 - \hat p)}{n} + \frac{z_{\alpha/2}^2}{4n^2}}.$$

Then there is also the exact Clopper-Pearson interval, whose formula involves the evaluation of the $F$-ratio distribution or incomplete beta function, or a binomial sum.

The Wikipedia article also includes the Agresti-Coull interval and arcsine interval. The first two are the most popular in use. Which one do you want to use?