Let two independent random variables, $Y_1$ and $Y_2$ that have binomial distribution have parameters $n_1 = n_2 = 100$, $p_1$ and $p_2$, respectively, be observed to be equal to $y_1 = 50$ and $y_2 = 40$. Determine an approximate $90\%$ confidence interval for $p_1 - p_2$.

I'm pretty new to confidence intervals and wanted help on this problem from my book. I have tried to apply the following definition:

Let $X_1, X_2, \ldots, X_n$ be a sample on a random variable $X$, where $X$ has p.d.f. $f(x;\theta)$, where $\theta\in \Omega$. Let $0 < \alpha < 1$ be specified. Let $L = L(X_1, \ldots, X_n)$ and $U = U(X_1, \ldots, X_n)$ be two statistics. We say that $(L, U)$ is a $(1 - \alpha)100\%$ confidence interval for $\theta$ if $1 - \alpha = P_{\theta}(\theta \in (L, U))$.

However, I have been having trouble applying this definition directly. I can identify $\alpha = 0.1$ here, which means that I think we want $P(L < p_1 - p_2 < U) = 0.9$. Now I'm really not sure how to use this information to get the answer; I have no clue where the observed values would come into play.

Any help is appreciated.


The idea is that the sample averages $Y_1/n_1$ and $Y_2/n_2$ should be good approximations of the proportions $p_1$ and $p_2$ respectively. We can use the observed value $y_1/n_1 - y_2/n_2$ as a point estimate of $p_1 - p_2$, and get a confidence interval by figuring out the approximate distribution of $Y_1/n_1 - Y_2/n_2$.

Because of the Central Limit Theorem, we know that the distribution $\operatorname{Bin}(n, p)$ is approximately normal with mean $np$ and variance $np(1-p)$. Therefore since $Y_1$ and $Y_2$ are independent, the random variable $Y_1/n_1 - Y_2/n_2$ is approximately distributed like a normal random variable with mean $p_1 - p_2$ and variance $p_1(1-p_1)/n_1 + p_2(1-p_2)/n_2$. After standardizing this tells us that the test statistic $$ \frac{(Y_1/n_1 - Y_2/n_2) - (p_1 - p_2)}{\sqrt{p_1(1-p_1)/n_1 + p_2(1-p_2)/n_2}} $$ approximately follows a standard normal distribution. Using this, a little bit of algebraic manipulation will show that $$ \mathbb{P} \left( | (p_1 - p_2) - (Y_1/n_1 - Y_2/n_2)| \leq 1.645 \sqrt{p_1(1-p_1)/n_1 + p_2(1-p_2)/n_2} \right) = 90\% $$ (the number $1.645$ comes from the $95$th percentile of a standard normal distribution) So an approximate $90$% confidence interval would be $$ (Y_1/n_1 - Y_2/n_2) \pm 1.645 \sqrt{p_1(1-p_1)/n_1 + p_2(1-p_2)/n_2}. $$ The only problem is that $p_1$ and $p_2$ still appear in this expression, but they are unknown. One standard way to fix this is just to replace $p_1$ and $p_2$ by their point estimates $Y_1/n_1$ and $Y_2/n_2$ in the above expression, and you are still left with a pretty good approximation.

  • $\begingroup$ I'm confused about why you use $1.645$ in a $90\%$ confidence interval instead of $1.28$ if $\Phi(1.28) = 0.9$? $\endgroup$ – user882487 Mar 11 at 14:39
  • $\begingroup$ Because you want a two-sided confidence interval (I'm assuming). That means you need 5% probability in the left tail and 5% probability in the right tail, so you should take the left and right cutoffs to be the 5th and 95th percentile respectively $\endgroup$ – Adam Mar 11 at 14:43
  • $\begingroup$ Just to make sure I understand: Would it end up being $(50/100 - 40/100) \pm 1.645\sqrt{\cdots}$? And in the square-root, I'd substitute $p_i = Y_i/n_i$ $\endgroup$ – user882487 Mar 11 at 14:46
  • $\begingroup$ Looks right to me $\endgroup$ – Adam Mar 11 at 14:49

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