Confidence Interval for Mean
Use this calculator to compute the confidence interval for population mean when the population standard deviation is known.
| Confidence Interval Calculator for mean | |
|---|---|
| Sample Size ($n$) | |
| Sample Mean ($\overline{x}$) | |
| Population Standard Deviation ($\sigma$) | |
| Confidence Level ($1-\alpha$) | |
| Results | |
| Standard Error of Mean: | |
| Z Critical Value : ($Z$) | |
| 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 mean $\overline{X}$
Step 2 - Enter the population standard deviation $\sigma$
Step 3 - Select the confidence level
Step 4 - Click on Calculate button to get the required confidence interval for mean when population variance is known
Step 5 - Gives the result for SE of mean
Step 6 - Gives the $Z$-critical value and margin of error $E$
Step 7 - Gives the lower and upper confidence limits for the differencce between means when population variances are known.
Confidence interval for mean Example 1
Among those who take the Graduate Record Examination (GRE), 67 persons are randomly selected. This sample group has a mean score of 558 on the quantitative portion of the GRE. The population standard deviation is 139 (from the Educational Testing Service). Estimate at the 98 % level the mean quantitative score on the GRE for all persons taking the test.
Solution
Given that sample size $n = 67$, sample mean $\overline{X}= 558$ and population standard deviation $\sigma = 139$.
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 sample size $n =67$, sample mean $\overline{X}=558$ and population standard deviation is $\sigma = 139$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval for the population mean $\mu$ is
$$ \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned} $$
where $E = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$, 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 mean is
$$ \begin{aligned} E & = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\\ & = 2.33 \frac{139}{\sqrt{67}} \\ & = 39.567. \end{aligned} $$
Step 6 Determine the confidence interval
$98$ % confidence interval estimate for population mean is
$$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 558 - 39.567 & \leq \mu \leq 558 + 39.567\\ 518.433 & \leq \mu \leq 597.567. \end{aligned} $$
Thus, $98$% confidence interval estimate for population mean is $(518.433,597.567)$.
Interpretation
We are $98$% confident that the true population mean quantitative score on the GRE test is between $518.433$ and $597.567$.
Confidence interval for mean Example 2
A sample of 81 observations is taken from a normal distribution with a standard deviation of 5. The sample mean is 40. Determine the 95% confidence interval for the population mean.
Solution
Given that sample size $n = 81$, sample mean $\overline{X}= 40$ and population standard deviation $\sigma = 5$.
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 sample size $n =81$, sample mean $\overline{X}=40$ and population standard deviation is $\sigma = 5$.
Step 3 Specify the formula
$100(1-\alpha)$% confidence interval for the population mean $\mu$ is
$$ \begin{aligned} \overline{X} - E \leq \mu \leq \overline{X} + E \end{aligned} $$
where $E = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}$, 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 mean is
$$ \begin{aligned} E & = Z_{\alpha/2} \frac{\sigma}{\sqrt{n}}\\ & = 1.96 \frac{5}{\sqrt{81}} \\ & = 1.089. \end{aligned} $$
Step 6 Determine the confidence interval
$95$% confidence interval estimate for population mean is
$$ \begin{aligned} \overline{X} - E & \leq \mu \leq \overline{X} + E\\ 40 - 1.089 & \leq \mu \leq 40 + 1.089\\ 38.911 & \leq \mu \leq 41.089. \end{aligned} $$
Thus, $95$% confidence interval estimate for population mean is $(38.911,41.089)$.
Interpretation
We are $95$% confident that the true population mean is between $38.911$ and $41.089$.
Conclusion
In this tutorial, you learned about how to calculate confidence interval for the mean when population variance is known. You also learned about the step by step examples to construct desired confidence interval for the mean when population variance is known.
To learn more about interval estimation and construction of confidence interval, please refer to the following tutorials:
CI for mean when the population variance is known
Let me know in the comments if you have any questions on confidence interval for mean when the population variance is known and your thought on this article.
