## CI for difference between two population means (variances are known)

In this tutorial we will discuss some examples on confidence interval for difference in means when the population variances are known.

## Example 1

Two kinds of thread are being compared for tensile strength. Fourty pieces of each type of thread are tested under similar conditions. Brand A has an average tensile strength of 81.6 kilograms with a standard deviation of 4.5 kilograms, while brand B had an average tensile strength of 84.5 kilograms with a standard deviation of 5.1 kilograms.

Construct a 98% confidence interval for the difference of the population means.

### Solution

Given that $n_1 = 40$, $\overline{x}_1 =81.6$, $\sigma_1 = 4.5$, $n_2 =40$, $\overline{x}_2 =84.5$ and $\sigma_2 = 5.1$.

We wish to determine $98$% confidence interval estimate for the difference $(\mu_1-\mu_2)$.

### Step 1 Specify the confidence level $(1-\alpha)$

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

### Step 2 Given information

Given that $n_1 = 40$, $\overline{x}_1= 81.6$, $\sigma_1 = 4.5$, $n_2 = 40$, $\overline{x}_2= 84.5$, $\sigma_2 = 5.1$.

### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval estimate for the difference `$(\mu_1-\mu_2)$`

is
`$$ \begin{aligned} (\overline{x}_1 -\overline{x}_2)- E & \leq (\mu_1-\mu_2) \leq (\overline{x}_1 -\overline{x}_2) + E. \end{aligned} $$`

where `$E = Z_{\alpha/2}\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$`

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 difference of means $\mu_1-\mu_2$ is
`$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\ & = 2.33 \sqrt{\frac{4.5^2}{40}+\frac{5.1^2}{40}}\\ & = 2.506. \end{aligned} $$`

### Step 6 Determine the confidence interval

$98$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is
`$$ \begin{aligned} (\overline{x}_1 -\overline{x}_2)- E & \leq \mu_1-\mu_2 \leq (\overline{x}_1 -\overline{x}_2) + E\\ (81.6-84.5) - 2.506 & \leq \mu_1-\mu_2 \leq (81.6-84.5) + 2.506\\ -5.406 & \leq \mu_1-\mu_2 \leq -0.394. \end{aligned} $$`

Thus, $98$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is $(-5.406,-0.394)$.

### Interpretation

We Can be $98$% confident that the true difference in the mean tensile strength between the two brands is between $-5.4057$ and $-0.3943$.

## Example 2

Two groups of students (Mathematics majors and Computer majors) are given a problem solving task on logical reasoning. The results about the logical reasong scores are as follows:

. | Mathematics majors | Computer majors |
---|---|---|

Sample size | 38 | 42 |

Sample mean | 82.4 | 80.5 |

Population standard deviation | 3.9 | 3.6 |

Find the 95% confidence interval of the true difference in mean lo.

### Solution

Given that $n_1 = 38$, $\overline{x}_1 =82.4$, $\sigma_1 = 3.9$, $n_2 =42$, $\overline{x}_2 =80.5$ and $\sigma_2 = 3.6$.

Here we wish to determine $95$% confidence interval estimate for the difference $(\mu_1-\mu_2)$.

### Step 1 Specify the confidence level $(1-\alpha)$

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

### Step 2 Given information

Given that $n_1 = 38$, $\overline{x}_1= 82.4$, $\sigma_1 = 3.9$, $n_2 = 42$, $\overline{x}_2= 80.5$, $\sigma_2 = 3.6$.

### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval estimate for the difference `$(\mu_1-\mu_2)$`

is
`$$ \begin{aligned} (\overline{x}_1 -\overline{x}_2)- E & \leq (\mu_1-\mu_2) \leq (\overline{x}_2 -\overline{x}_2) + E. \end{aligned} $$`

where `$E = Z_{\alpha/2}\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$`

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 difference of means $\mu_1-\mu_2$ is
`$$ \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}\\ & = 1.96 \sqrt{\frac{3.9^2}{38}+\frac{3.6^2}{42}}\\ & = 1.65. \end{aligned} $$`

### Step 6 Determine the confidence interval

$95$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is
`$$ \begin{aligned} (\overline{x}_1 -\overline{x}_2)- E & \leq \mu_1-\mu_2 \leq (\overline{x}_1 -\overline{x}_2) + E\\ (82.4-80.5) - 1.65 & \leq \mu_1-\mu_2 \leq (82.4-80.5) + 1.65\\ 0.25 & \leq \mu_1-\mu_2 \leq 3.55. \end{aligned} $$`

Thus, $95$% confidence interval estimate for the difference $(\mu_1-\mu_2)$ is $(0.25,3.55)$.

### Interpretation

We can be $95$% confident that the true difference in logical reasoning task by mathematics majors and computer majors is between $0.25$ and $3.55$.