## Confidence interval calculator for difference of means

Use this calculator to determine the confidence interval for difference between means when variances are known.

Confidence Interval Calculator for Difference of means | ||
---|---|---|

Sample 1 | Sample 2 | |

Sample Mean | ||

Sample Size | ||

Population Standard Deviation | ||

Confidence Level ($1-\alpha$) | ||

Results |
||

Standard Error of Diff. of Means: | ||

Z-critical value: ($Z_{\alpha/2}$) | ||

Margin of Error: ($E$) | ||

Lower Confidence Limits: | ||

Upper Confidence Limits: | ||

## How to use the calculator to calculate CI for difference between means?

Step 1 - Enter the sample means for sample 1 and sample 2

Step 2 - Enter the sample sizes for sample 1 and sample 2

Step 3 - Enter the population standard deviations for sample 1 and sample 2

Step 4 - Select the confidence level

Step 5 - Click on Calculate button to get the required confidence interval for difference between means

Step 6 - Gives the result for SE of difference of means

Step 7 - Gives the $Z$-critical value and margin of error $E$

Step 8 - Gives the lower and upper confidence limits for the differencce between means when population variances are known.

## CI for difference between means (Variances are known) Examples

In this tutorial we will discuss some examples on confidence interval for difference in means when the population variances are known. To know more about the theory refer confidence interval for difference in means when the population variances are known.

## Confidence Interval for means 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 $-0.3943$.

## Confidence Interval for means 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 $3.55$.

## Conclusion

In this tutorial, you learned about how to calculate confidence interval for the difference between two means when population variances are known. You also learned about the step by step examples to construct desired confidence interval for the difference between two means when population variances are known.

To learn more about interval estimation and construction of confidence interval, please refer to the following tutorials:

CI for difference between two means when the population variances are known

Let me know in the comments if you have any questions on **confidence interval for difference between two means when the population variances are known** and your thought on this article.