## CI for difference between two population proportions

In this tutorial we will discuss some examples on confidence interval for difference between two population proportions.

## Example 1

A random sample of 85 parts manufactured by machine A yields 10 defective and a random sample of 110 parts manufactured by machine B shows 28 defective.

Compute 95% confidence interval for difference between the proportions of defectives.

Are the two machine differ significantly with respect to the proportion of defectives?

### Solution

Given information

. | Machine A | Machine B |
---|---|---|

Sample size | $n_1=85$ | $n_2=110$ |

Observed no. of defectives | $X_1=10$ | $X_2=28$ |

We wish to determine $95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$.

### Step 1 Specify the confidence level

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

### Step 2 Given information

Given that $X_1 = 10$, $X_2 = 28$, $n_1 = 85$, $n_2 = 110$.

The estimate of the population proportions $p_1$ is $\hat{p}_1 =\frac{X_1}{n_1} =\frac{10}{85}=0.1176$ and the estimate of the population proportion $p_2$ is $\hat{p}_2 =\frac{X_2}{n_2} =\frac{28}{110}=0.2545$.

### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is
```
$$
\begin{aligned}
(\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E.
\end{aligned}
$$
```

where `$E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}$`

and `$Z_{\alpha/2}$`

is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus `$Z_{\alpha/2} = Z_{0.025} = 1.96$`

.

### Step 5 Compute the margin of error

The margin of error for the difference $(p_1-p_2)$ is
```
$$
\begin{aligned}
E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}\\
& = 1.96 \sqrt{\frac{0.1176*(1-0.1176)}{85}+\frac{0.2545*(1-0.2545)}{110}}\\
&= 0.1064.
\end{aligned}
$$
```

### Step 6 Determine the confidence interval

$95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is
```
$$
\begin{aligned}
(\hat{p}_1-\hat{p}_2) - E &\leq (p_1-p_2) \leq (\hat{p}_1-\hat{p}_2) + E\\
(0.1176-0.2545) - 0.1064 & \leq (p_1-p_2) \leq (0.1176-0.2545) + 0.1064\\
-0.2433 & \leq (p_1-p_2) \leq -0.0305
\end{aligned}
$$
```

Thus, $95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is $(-0.2433,-0.0305)$.

### Interpretation

We can be $95$% confident that the difference between two population proportions $(p_1-p_2)$ is between $-0.2433$ and $-0.0305$.

Because the $95$% confidence interval does not include the point zero, we conclude that at $0.05$ level of significance the two machine differ significantly with respect to the proportion of defectives.

## Example 2

In a recent survey of randomly selected adults 65 or older, 411 of 1012 men and 525 of 1062 women say they suffer from some form of arthritis.

a. Construct a 98% confidence interval for the difference between senior men and women who suffer from arthritis.

b. Does there appear to be a difference between senior men and women as far as suffering from arthritis?

### Solution

Given information

. | Men | Women |
---|---|---|

Sample size | $n_1=1012$ | $n_2=1062$ |

Suffering from arthritis | $X_1=411$ | $X_2=525$ |

We want to determine $98$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$.

### Step 1 Specify the confidence level

Confidence level is $1-\alpha = 0.98$. Thus, the level of significance is $\alpha = 0.02$.

### Step 2 Given information

Given that $X_1 = 411$, $X_2 = 525$, $n_1 = 1012$, $n_2 = 1062$.

The estimate of the population proportions $p_1$ is $\hat{p}_1 =\frac{X_1}{n_1} =\frac{411}{1012}=0.4061$ and the estimate of the population proportion $p_2$ is $\hat{p}_2 =\frac{X_2}{n_2} =\frac{525}{1062}=0.4944$.

### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is
```
$$
\begin{aligned}
(\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E.
\end{aligned}
$$
```

where `$E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}$`

and `$Z_{\alpha/2}$`

is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus `$Z_{\alpha/2} = Z_{0.01} = 2.33$`

.

### Step 5 Compute the margin of error

The margin of error for the difference $(p_1-p_2)$ is
```
$$
\begin{aligned}
E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}\\
& = 2.33 \sqrt{\frac{0.4061*(1-0.4061)}{1012}+\frac{0.4944*(1-0.4944)}{1062}}\\
&= 0.0507.
\end{aligned}
$$
```

### Step 6 Determine the confidence interval

$98$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is
```
$$
\begin{aligned}
(\hat{p}_1-\hat{p}_2) - E &\leq (p_1-p_2) \leq (\hat{p}_1-\hat{p}_2) + E\\
(0.4061-0.4944) - 0.0507 & \leq (p_1-p_2) \leq (0.4061-0.4944) + 0.0507\\
-0.1389 & \leq (p_1-p_2) \leq -0.0375
\end{aligned}
$$
```

Thus, $98$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is $(-0.1389,-0.0375)$.

### Interpretation

We can be $98$% confident that the difference between two population proportions $(p_1-p_2)$ is between $-0.1389$ and $-0.0375$.

Because the $98$% confidence interval does not include the point zero, we conclude that at $0.02$ level of significance there appears to be significant difference between the proportion of senior men and proportion of women who suffer from arthritis.

## Example 3

Two machines used in the same operation are to be compared. A random sample of 80 parts from the first machine yields 6 non-conforming ones. A random sample of 120 parts from the second machine shows 14 non-conforming ones.

Find a 95% confidence interval for difference in the proportion of non-conforming parts between the two machines.

## Solution

Given information

. | First Machine | Second Machine |
---|---|---|

Sample size | $n_1=80$ | $n_2=200$ |

no.of non-confirming | $X_1=6$ | $X_2=14$ |

We want to determine $95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$.

### Step 1 Specify the confidence level

Confidence level is $1-\alpha = 0.95$. Thus, the level of significance is $\alpha = 0.05$.

### Step 2 Given information

Given that $X_1 = 6$, $X_2 = 14$, $n_1 = 80$, $n_2 = 200$.

The estimate of the population proportions $p_1$ is $\hat{p}_1 =\frac{X_1}{n_1} =\frac{6}{80}=0.075$ and the estimate of the population proportion $p_2$ is $\hat{p}_2 =\frac{X_2}{n_2} =\frac{14}{200}=0.07$.

### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is
```
$$
\begin{aligned}
(\hat{p}_1-\hat{p}_2) - E \leq (p_1 -p_2) \leq (\hat{p}_1 -\hat{p}_2)+ E.
\end{aligned}
$$
```

where `$E = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}$`

and `$Z_{\alpha/2}$`

is the $Z$ value providing an area of $\alpha/2$ in the upper tail of the standard normal probability distribution.

### Step 4 Determine the critical value

The critical value of $Z$ for given level of significance is $Z_{\alpha/2}$.

Thus `$Z_{\alpha/2} = Z_{0.025} = 1.96$`

.

### Step 5 Compute the margin of error

The margin of error for the difference $(p_1-p_2)$ is
```
$$
\begin{aligned}
E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}_1*(1-\hat{p}_1)}{n_1}+\frac{\hat{p}_2*(1-\hat{p}_2)}{n_2}}\\
& = 1.96 \sqrt{\frac{0.075*(1-0.075)}{80}+\frac{0.07*(1-0.07)}{200}}\\
&= 0.0677.
\end{aligned}
$$
```

### Step 6 Determine the confidence interval

$95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is
```
$$
\begin{aligned}
(\hat{p}_1-\hat{p}_2) - E &\leq (p_1-p_2) \leq (\hat{p}_1-\hat{p}_2) + E\\
(0.075-0.07) - 0.0677 & \leq (p_1-p_2) \leq (0.075-0.07) + 0.0677\\
-0.0627 & \leq (p_1-p_2) \leq 0.0727
\end{aligned}
$$
```

Thus, $95$% confidence interval estimate of the difference between two population proportions $(p_1-p_2)$ is $(-0.0627,0.0727)$.

### Interpretation

We can be $95$% confident that the difference between two population proportions $(p_1-p_2)$ is between $-0.0627$ and $0.0727$.

Because the $95$% confidence interval include the point zero, we conclude that at $0.05$ level of significance there is no significant difference between the proportion of non-confirming parts between the two machines.