Sample size calculator to estimate mean

Sample size to estimate mean

In inferential statistics, knowing how many samples to take from the population is extremely important.

In this tutorial we will explore how to determine minimum sample size needed to estimate population mean with a given amount of error and desired confidence level.

Sample size to estimate mean

The formula to estimate the sample size required to estimate the mean is

 $$n =\bigg(\dfrac{z\sigma}{E}\bigg)^2$$
where $z$ is the $Z_{\alpha/2}$, $\sigma$ is the population standard deviation and $E$ is the margin of error.

Sample size Calculator to estimate mean

Use this calculator to find the minimum sample size required to estimate population mean $\mu$.

Sample Size to estimate mean
Confidence Level ($1-\alpha$)
Standard Deviation ($\sigma$)
Margin of Error ($E$)
Results
Z value:
Required Sample Size : ($n$)

How to use calculator to determine sample size to estimate mean?

Step 1 - Select the confidence level $(1-\alpha)$

Step 2 - Enter the standard deviation

Step 3 - Enter the margin of error $(E)$

Step 4 - Click on "Calculate" button to get the $Z$-value and the required sample size

Sample size to estimate mean Example 1

An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will she need to be 90% confident that her sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that $\sigma = 40$ seconds.

Solution

Given that the standard deviation $\sigma =40$, margin of error $E =15$. The confidence coefficient is $1-\alpha=0.9$. Thus $\alpha = 0.1$.

The critical value of $Z$ is $z=Z_{\alpha/2} = 1.64$.

The minimum sample size required to estimate the mean is

 \begin{aligned} n &= \bigg(\dfrac{z*\sigma}{E}\bigg)^2\\ & = \bigg(\dfrac{1.64*40}{15}\bigg)^2\\ & =19.126\\ &\approx 20. \end{aligned}
Thus, the sample of size $n=20$ will ensure that the $90$% confidence interval for the mean will have a margin of error $15$.

Sample size to estimate mean Example 2

An electrical firm manufactures light bulbs that have a length of life that is approximately normal with a standard deviation of 30 hours. How large a sample is needed if we wish to be 99% confident that our sample mean will be within 10 hours of the true mean?

Solution

Given that the standard deviation $\sigma =30$, margin of error $E =10$. The confidence coefficient is $1-\alpha=0.99$. Thus $\alpha = 0.01$.

The critical value of $Z$ is $z=Z_{\alpha/2} = 2.58$.

The minimum sample size required to estimate the mean is

 \begin{aligned} n &= \bigg(\dfrac{z*\sigma}{E}\bigg)^2\\ & = \bigg(\dfrac{2.58*30}{10}\bigg)^2\\ & =59.9076\\ &\approx 60. \end{aligned}
Thus, the sample of size $n=60$ will ensure that the $99$% confidence interval for the mean will have a margin of error $10$.

Sample size to estimate mean Example 3

How large a sample is needed to assess the population mean, with a standard deviation of \$17 with an allowable error of \$1.50 at 95 percent confidence?

Solution

Given that the standard deviation $\sigma =17$, margin of error $E =1.5$. The confidence coefficient is $1-\alpha=0.95$. Thus $\alpha = 0.05$.

The critical value of $Z$ is $z=Z_{\alpha/2} = 1.96$.

The minimum sample size required to estimate the mean is

 \begin{aligned} n &= \bigg(\dfrac{z*\sigma}{E}\bigg)^2\\ & = \bigg(\dfrac{1.96*17}{1.5}\bigg)^2\\ & =493.4322\\ &\approx 494. \end{aligned}

Thus, the sample of size $n=494$ will ensure that the $95$% confidence interval for the mean will have a margin of error $1.5$.

Endnote

In this tutorial, you learned about how to determine the sample size required to estimate population mean. You also learned how to solve numerical examples on sample size determination to estimate mean.