Confidence interval help needed

Let θ denote the true ability of a student in a certain subject, representing the probability of astudent answering correctly an arbitrary question and let X1, . . . , Xn be the marks to n questions the student answered, taking values either 0 or 1. We model them as n i.i.d. Bernoulli(θ) randomvariables. Then, we approximate a student’s ability by its ML estimate.

a) We can approximate the distribution of

(θ(X) - θ) / ((θ(X)(1−θ(X)))/n)^0.5


(Where θ(X) is the Maximum likelihood estimate)

by a standard gaussian N(0,1) distribution, under distribution P_0. Using this result, construct the 95 percent confidence interval for θ.

b) How big should n be so that we can estimate the true ability within 0.01 marks, with probabilityat least 95%? Justify your answer.

I have no idea where to start here. Why are we approximating that scary looking distribution in the first place, and what does it mean by 'under distribution P₀ ? How do we start to construct an interval?

What it is saying here is that, if $\hat{\theta}$ is the maximum likelihood estimator of $\theta$, then

$\displaystyle \frac{\theta - \hat{\theta}}{\sqrt{\frac{\hat{ \theta }\left(1 - \hat{\theta}\right)}{n}}}$

is approximately distributed as a standard normal variable. I've changed your notation of $\theta(X)$ to the more common $\hat{\theta}$. The point here is that $\hat{\theta}$ is your estimate of $\theta$ and

$\displaystyle \sqrt{\frac{\hat{\theta}\left(1 - \hat{\theta}\right)}{n}}$

is an estimate of it's standard error. You can now use these facts to construct the 95% confidence interval in the usual way, that is $\hat{\theta}$ plus or minus 1.96 times that standard error.

The terminology "under distribution P0" is odd, and appears to be superfluous.

What it is saying here is that, if $\hat{\theta}$ is the maximum likelihood estimator of $\theta$, then

$\displaystyle \frac{\theta - \hat{\theta}}{\sqrt{\frac{\hat{ \theta }\left(1 - \hat{\theta}\right)}{n}}}$

is approximately distributed as a standard normal variable. I've changed your notation of $\theta(X)$ to the more common $\hat{\theta}$. The point here is that $\hat{\theta}$ is your estimate of $\theta$ and

$\displaystyle \sqrt{\frac{\hat{\theta}\left(1 - \hat{\theta}\right)}{n}}$

is an estimate of it's standard error. You can now use these facts to construct the 95% confidence interval in the usual way, that is $\hat{\theta}$ plus or minus 1.96 times that standard error.

The terminology "under distribution P0" is odd, and appears to be superfluous.

So, is my interval the estimator +/- 1.96 ? (Sorry for large text, I copied and pasted). Do I need to do anything else?

That's it.

Thanks. They didn't actually teach us how to do it. For the second part of the question, is it similar? Do I need to rearrange the interval for n?

Yes, you have to choose n so that the confidence interval has width 0.01.

My methods therefore is upper interval - lower interval <= 0.01. I rearranged for n and got 153664(theta)(1-theta)<= n. It doesn't seem right to have it in terms of my estimator, is it ok?

If you are not actually given any information that allows you to calculate an actual value of $\hat{\theta}$ then you need to think about what the "worst" value of $\hat{\theta}$ could be, from the point of view of the width of the confidence interval. As a hint, consider the graph of the function f(x)=x(1-x).