Confidence intervals for difference of two proportions

Confidence intervals for the difference of two proportions

A confidence interval here gives a range of reasonable values for the true difference in population proportions, $p_1-p_2$, using information from two samples. In many real comparisons — such as School A versus School B, a treatment group versus a control group, or men versus women — you usually want more than “Are the sample proportions different?” You want to estimate how different the underlying population proportions might be.

Notation

Definitions

${\hat{p}}_1,{\hat{p}}_2$

The sample proportions from samples $1$ and $2$.

$p_1,p_2$

The true population proportions for groups $1$ and $2$.

$n_1,n_2$

The sample sizes of samples $1$ and $2$.

${\sigma }_d$

The standard deviation of the distribution of differences between two sample proportions.

Conditions and sampling distribution

Important pre-requisite information:

1. The set of all differences of sample proportions is approximately normally distributed due to the central limit theorem. The central limit theorem requirements state that $n_1{\hat{p}}_1$, $n_1(1-{\hat{p}}_1)$, $n_2{\hat{p}}_2$, and $n_2(1-{\hat{p}}_2)$ should all be at least $10$. Both samples should ideally be simple random and must be taken independently of one another.

2. The mean of the sampling distribution of ${\hat{p}}_1-{\hat{p}}_2$ is $p_1-p_2$.

3. The standard deviation of the set of differences of sample proportions is:

$$\begin{array}{rl}{\sigma }_d& ={\sigma }_{{\hat{p}}_1-{\hat{p}}_2}\\ & \approx \sqrt{\frac{p_1(1-p_1)}{n_1}+\frac{p_2(1-p_2)}{n_2}}\end{array}$$

Confidence interval form

A confidence interval for the difference of two proportions is the estimate $\pm$ the margin of error:

$$\begin{array}{r}({\hat{p}}_1-{\hat{p}}_2)\pm z^{\ast }\cdot \text{SE}\end{array}$$

If the confidence interval does not include $0$, that’s evidence of a statistically significant difference between the two groups. A difference of $0$ means “no difference,” and if $0$ is not in the interval, it isn’t a plausible value for the true difference based on the data.

Practice problem

Example:

One of the Texas school districts wants to find out if there is a difference in the proportion of students who support a new dress code policy which is aimed at making the summer heat more manageable for students. A simple random sample of $120$ students from School A found that $78$ support the policy. A separate simple random sample of $150$ students from School B found that $81$ support the policy. Assume that each sample size is less than $10\%$ of its school’s student population.

Part a.

What are the sample proportions of students who support the policy at each school?

From the problem:

• School A: $78$ out of $120$ students support the policy.
• School B: $81$ out of $150$ students support the policy.

Set up the sample proportions:

$$\begin{array}{rl}{\hat{p}}_A& =\frac{78}{120}\\ {\hat{p}}_B& =\frac{81}{150}\end{array}$$

Solution:

Part b.

What is the standard deviation for the difference in sample proportions?

The standard deviation of the difference in sample proportions is estimated by the standard error:

$$\begin{array}{rl}\text{SE}& =\sqrt{\frac{{\hat{p}}_A(1-{\hat{p}}_A)}{n_A}+\frac{{\hat{p}}_B(1-{\hat{p}}_B)}{n_B}}\end{array}$$

From Part a. and the problem:

$$\begin{array}{rl}{\hat{p}}_A& =0.65\\ {\hat{p}}_B& =0.54\\ n_A& =120\\ n_B& =150\end{array}$$

Substitute into the formula:

$$\begin{array}{rl}\text{SE}& =\sqrt{\frac{0.65(0.35)}{120}+\frac{0.54(0.46)}{150}}\\ & =\sqrt{\frac{0.2275}{120}+\frac{0.2484}{150}}\\ & =\sqrt{0.001896+0.001656}\\ & =\sqrt{0.003552}\end{array}$$

Solution:

Part c.

Construct a $95\%$ confidence interval for the true difference in proportions $p_A-p_B$.

For a $95\%$ confidence interval, the critical value is:

$$\begin{array}{rl}{z}^{\ast }& =1.96\end{array}$$

The confidence interval has the form:

$$\begin{array}{rl}({\hat{p}}_A-{\hat{p}}_B)\pm z^{\ast }\cdot \text{SE}& =(0.65-0.54)\pm 1.96\cdot 0.0596\\ & =0.11\pm 0.1168\end{array}$$

Solution:

Part d.

Interpret the confidence interval in context.

Solution:

Part e.

Is there convincing evidence of a difference in support of the policy between School A and School B? Why or why not?

Solution: