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In an interview $63\%$ of $100$ randomly chosen people gave an positive answer. $p$ is this amount from all people. What is the maximum likelihood estimation of $p$? Calculate the wald interval of $p$!

I understand it so, that $p = 0.63 $. I also want to give the steps of the deviation of the formula of the likelihood. So I tried to start with this formula for the likelihood: $L(p)=\prod \limits_{i=1}^{n}p(x_i, p) = \prod \limits_{i=1}^{n}P(X_i=x_i) = \prod \limits_{i=1}^{100}P(X_i =0.63)$

$\hat{p} = \frac{1}{n} * \sum\limits_{i=1}^{n} x_i = \frac{1}{100} * 63 = 0.63$

And for the wald interval:

$\hat{p} \pm Z_{1-\frac{\alpha}{2}} * \sqrt{\frac{\hat{p} (1-\hat{p} )}{n}}$

When I use $0.63$ for $\hat{p}$ I get the interval $[0.54; 0.72]$.

So I don't really understand how to get from the formula of $L(p)$ to the value of $\hat{p}$. I'm also not sure if this is already my final result for the likelihood. I got this with the help of a book, and there seem to be some steps missing. I'm also not sure if this is in general the right way of solving it and if I used the right formula and values for the Wald interval.

BruceET
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mrs fourier
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1 Answers1

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You are generally on the right track, but you have not shown how to get from your first displayed equation to your second one. For details of that, please see page 5 of these notes, one of the many other similar derivations available online, or on this site, where the development is the same but the notation is quite different (so I am reluctant to call your Question a duplicate).

The outline of these steps is to find the logarithm $\ell(p)$ of $L(p),$ and to find the derivative $\frac{\partial}{\partial p}\ell(p)$ of $\ell(p)$, then finally to set the partial derivative equal to $0$ and solve for $p$ in terms of $X$ and $n$. As shown in the first link, the solution is $\hat p = X/n.$ (In addition, you should check to make sure that you have found a relative maximum, not minimum or saddle-point by setting the derivative equal to zero.)

The particular situation for your Question is shown graphically below. The curve is the likelihood function $L(p)$ and the red line indicates that its maximum is at $p = .63.$

enter image description here

If $X \sim \mathsf{Binom}(n = 100, p),$ then the maximum likelihood estimator (MLE) $\hat p$ of $p$ is $\hat p = X/n$ $= 63/100$ $= 0.63$ in your problem, as you say.

The 95% Wald confidence interval (CI) for $p$ is of the form $\hat p \pm 1.96\sqrt{\frac{\hat p(1-\hat p)}{n}},$ where 1.96 cuts 2.5% from the upper tail of the standard normal distribution and $\hat p$ is the MLE of $p.$

In your problem, this CI computes to $(0.535, 0.725),$ which matches your computation.

Note: The Wald interval is derived using asymptotic arguments, and so it works best for large $n$. In particular, the Wald interval makes two approximations: (a) the normal approximation to the binomial distribution and (b) the estimation of the standard error of $\hat p$ which is $\text{SE} = \sqrt{\frac{p(1-p)}{n}}$ by $\widehat{\text{SE}} = \sqrt{\frac{\hat p(1-\hat p)}{n}}.$ The combination of these two approximations causes the actual coverage probability of the Wald interval to be considerably different from 95% (often less than 95%) for $n < 500$ or $1000.$ The Wikipedia article on 'binomial distribution' discusses other styles of CIs for binomial $p$ that have more dependable coverage probabilities.

BruceET
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