Confidence intervals for the difference of two means

Confidence intervals for the difference of two means

A confidence interval for the difference of two means estimates how different two population means are, based on sample data. This is useful when comparing two groups — such as teachers versus police officers, a treatment group versus a control group, or one school versus another — and you want to quantify the gap between their averages.

Sampling distribution of ${\bar{x}}_1-{\bar{x}}_2$

Recall the following facts about the sampling distribution of ${\bar{x}}_1-{\bar{x}}_2$:

1. The set of all differences of sample means is approximately normally distributed.

2. The mean of the set of differences of sample means equals ${\mu }_1-{\mu }_2$, which is the same as the difference of the population means.

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

$$\begin{array}{rl}{\sigma }_{{\bar{x}}_1-{\bar{x}}_2}& \approx \sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}\end{array}$$

Conditions for inference

Before finding a confidence interval or conducting a t-test, make sure these conditions are met:

1. The samples are random (ideally simple random samples).
2. The samples are taken independently of one another.
3. If sampling without replacement, each sample is less than $10\%$ of its respective population.

Confidence interval form

Definitions

Standard error of ${\bar{x}}_1-{\bar{x}}_2$

An estimate of the standard deviation of the sampling distribution of ${\bar{x}}_1-{\bar{x}}_2$.

$$\begin{array}{rl}\text{SE}& =\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\end{array}$$

Margin of error ($\text{ME}$)

The amount we add and subtract from the point estimate to form the confidence interval.

$$\begin{array}{rl}\text{ME}& =t^{\ast }\cdot \text{SE}\end{array}$$

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

$$\begin{array}{r}({\bar{x}}_1-{\bar{x}}_2)\pm t^{\ast }\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\end{array}$$

Practice problem

Example:

A sociologist surveys high school teachers and police officers about their planned retirement age.

• $28$ randomly selected high school teachers have a sample mean planned retirement age of $64.5$ and a sample standard deviation of $4.1$ years.
• $30$ randomly selected police officers have a sample mean planned retirement age of $58.2$ and a sample standard deviation of $5.3$ years.

Part a.

Construct a $95\%$ confidence interval for the difference in population means ${\mu }_1-{\mu }_2$, where ${\mu }_1$ is the population mean planned retirement age of teachers and ${\mu }_2$ is the population mean planned retirement age of police officers.

From the problem:

$$\begin{array}{rl}{n}_1& =28\\ {\bar{x}}_1& =64.5\\ s_1& =4.1\\ n_2& =30\\ {\bar{x}}_2& =58.2\\ s_2& =5.3\end{array}$$

Compute the point estimate:

$$\begin{array}{rl}{\bar{x}}_1-{\bar{x}}_2& =64.5-58.2\\ & =6.3\text{ years}\end{array}$$

Compute the standard error:

$$\begin{array}{rl}\text{SE}& =\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ & =\sqrt{\frac{4.1^2}{28}+\frac{5.3^2}{30}}\\ & =\sqrt{\frac{16.81}{28}+\frac{28.09}{30}}\\ & =\sqrt{0.6004+0.9363}\\ & \approx 1.2396\end{array}$$

Find the critical value for a $95\%$ confidence interval:

$$\begin{array}{rl}{t}^{\ast }=t_{0.975,54}& \approx 2.005\end{array}$$

Compute the margin of error:

$$\begin{array}{rl}\text{ME}& =t^{\ast }\cdot \text{SE}\\ & =2.005\cdot 1.2396\\ & \approx 2.485\end{array}$$

Construct the interval:

$$\begin{array}{rl}({\bar{x}}_1-{\bar{x}}_2)\pm \text{ME}& =6.3\pm 2.485\end{array}$$

Solution:

Part b.

Based on your interval, is there convincing evidence that the average teacher’s planned retirement age is more than $5$ years greater than the average police officer’s planned retirement age?

To answer this, check whether the entire confidence interval $(3.815,8.785)$ lies above $5$.

Solution: