How to find a confidence interval for difference in response rates?

Question

I found this question on a past exam:

I know how to find a confidence interval with one mean: $\bar x \pm z_{\alpha/2} \frac{\sigma}{\sqrt n}$

and I also know how to do it with two means however that also requires a sample standard variation, so I'm not sure how to even start this question.

I though maybe: $\bar x_1- \bar x_2 = 0$

however I'm not sure, any help would be appreciated thank you.

Answer 1

You are asked to find a confidence interval for the difference of two population proportions.

Consider a general problem in this connection:

Assume $X_1\sim \mathsf{bin}(n_1,p_1)$ and $X_2\sim\mathsf{bin}(n_2,p_2)$ are independent sample observations. To find a confidence interval (C.I.) for $p_1-p_2$.

You can adopt the following methods to solve your particular problem.

A usual approach is to construct an asymptotic Wald-type C.I. by assuming $n_1,n_2$ to be large enough.

Let $\hat p_i=\frac{X_i}{n_i}$ be the observed proportion of success, $i=1,2$.

Then we have the pivot $$\frac{\hat p_1-\hat p_2-(p_1-p_2)}{\sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}}\stackrel{L}\longrightarrow N(0,1)$$

An asymptotic C.I. of $p_1-p_2$ with confidence coefficient approximately $1-\alpha$ based on the above pivot has confidence limits $$\hat p_1-\hat p_2\mp z_{\alpha/2}\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2}}$$

, where $z_{\alpha/2}$ is such that $P(Z>z_{\alpha/2})=\frac{\alpha}{2}$ with $Z\sim N(0,1)$.

Using the data given, Minitab software gives the following result:

Sample   X    N  Sample p
1       35  100  0.350000
2       65  100  0.650000

Difference = p (1) - p (2)
Estimate for difference:  -0.3
95% CI for difference:  (-0.432207, -0.167793)

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Another approach is to use a variance stabilising transformation on $\hat p_i$ and find a Bonferroni confidence region for $(p_1,p_2)$.

Here we have the pivots

$$2\sqrt{n_i}\left(\sin^{-1}\sqrt{\hat p_i}-\sin^{-1}\sqrt{p_i}\right)\stackrel{L}\longrightarrow N(0,1)\quad,\,i=1,2$$

It can be verified that a large sample C.I. for $p_i$ with confidence coefficient approximately $1-\frac{\alpha}{2}$ has confidence limits $$\sin^2\left(\sin^{-1}\sqrt{\hat p_i}\mp \frac{z_{\alpha/4}}{2\sqrt{n_i}}\right)\quad,\,i=1,2$$

So for some $L_i,U_i$ we have approximately $$P\left[L_i\le p_i\le U_i\right]= 1-\frac{\alpha}{2}\quad,\,i=1,2$$

Therefore,

\begin{align} P\left[\underbrace{L_1\le p_1\le U_1}_{A_1}\,,\,\underbrace{L_2\le p_2\le U_2}_{A_2}\right]&=P(A_1\cap A_2) \\&\ge P(A_1)+P(A_2)-1 \\&=1-\alpha \end{align}

This is the same as saying $$P\left[L_1-U_2\le p_1-p_2\le U_1-L_2\right]\ge 1-\alpha$$

That is, $[L_1-U_2,U_1-L_2]$ is an asymptotic C.I. for $p_1-p_2$ with confidence coefficient at least $1-\alpha$.

Answer 2

I am a beginner learning about confidence intervals in my book. So if my answer is wrong please let me know.

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Let $p_1 =$ true response rate of the blank cover

Let $p_2 =$ true response rate of the skydiver cover

We want to know what $a$ and $b$ are for $P(a \le p_2 - p_1 \le b) = .95$

Letting $X \sim Bin(n,p_1)$ and $Y \sim Bin(n,p_2)$ and using the Normal approximation to the Binomial,

$$ Z_1 = \frac{X-100p_1}{\sqrt{100(.35)(.65)}} \dot{\sim} N(0,1) $$

$$ Z_2 = \frac{Y-100p_2}{\sqrt{100(.65)(.35)}} \dot{\sim} N(0,1) $$

so $Z_2 - Z_1 \, \dot{\sim} N(0,2)$ and therefore

$$ \frac{Y-100p_2 -X + 100p_1}{\sqrt{200(.65)(.35)}} \dot{\sim} N(0,1) $$

So we have

$$P \left(-1.96 \le \frac{Y-100p_2 -X + 100p_1}{\sqrt{200(.65)(.35)}} \le 1.96 \right) = .95$$

Note that $\hat{p_1} = X/n$ so $X = n\hat{p_1}$ therefore

$$P \left(-1.96 \le \frac{65-100p_2 - 35 + 100p_1}{\sqrt{200(.65)(.35)}} \le 1.96 \right) = .95$$

$$P( .1677 \le p_2 - p_1 \le .4322) = .95$$

So in theory with 95 percent confidence the difference between the two response rates is between $.1677$ and $.4322$ so it seems like the sky diver cover is for sure better since $p_2 - p_1$ isn't likely to be a negative number. Let me know what you think of this answer.