Sample size required to estimate proportion
The formula to estimate the sample size required to estimate the proportion is
$$ n =p*(1-p)\bigg(\frac{z}{E}\bigg)^2 $$
where,
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$p$ is the proportion of success,
-
$z$ is the
$Z_{\alpha/2}$(critical value of $Z$), -
$E$ is the margin of error.
Sample size calculator to estimate proportion
Use this calculator to find the minimum sample size required to estimate proportion $p$.
| Sample size to estimate proportion | ||
|---|---|---|
| Confidence Level ($1-\alpha$) | ||
| Proportion of success ($p$) | ||
| Margin of Error ($E$) | ||
| Results | ||
| Z value: | ||
| Required Sample Size : ($n$) | ||
How to use calculator to determine sample size to estimate proportion?
Step 1 - Select the confidence level $(1-\alpha)$
Step 2 - Enter the proportion of success
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 proportion Example 1
A survey of 1000 randomly selected registered U.S. voters found that 37% were in favor of the recent tax reform plan. A new survey is being proposed to estimate the true proportion in favor of the recent tax reform plan.
How many voters should be surveyed if the the goal is to estimate the proportion of voters within 0.05 with 90% confidence?
Solution
Given that the proportion of voters who favor the recent tax reform plan is $p =0.37$. The margin of error $E =0.05$. The confidence coefficient is $0.9$.
The critical value of $Z$ is $Z_{\alpha/2} = 1.64$.
The minimum sample size required to estimate the proportion voters who favor the recent tax reform plan is
$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.37(1-0.37)\bigg(\frac{1.64}{0.05}\bigg)^2\\ &=250.7783\\ &\approx 251. \end{aligned} $$
Thus, the sample of size $n=251$ will ensure that the $90$% confidence interval for the proportion voters who favor the recent tax reform plan will have a margin of error $0.05$.
Sample size to estimate proportion Example 2
Suppose we know that the unemployment rate has been about 18%. However, we wish to update our estimate in order to make an important decision about the national economic policy.
Accordingly, let us say we wish to be 98% confident that the new estimate of $p$ is within 0.01 of true $p$. What is the sample size needed to achieve this?
Solution
Given that the proportion of unemployment is $p =0.18$. The margin of error $E =0.01$. The confidence coefficient is $0.98$.
The critical value of $Z$ is $Z_{\alpha/2} = 2.33$.
The minimum sample size required to estimate the proportion of unemployment is
$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.18(1-0.18)\bigg(\frac{2.33}{0.01}\bigg)^2\\ &=8013.0564\\ &\approx 8014. \end{aligned} $$
Thus, the sample of size $n=8014$ will ensure that the $98$% confidence interval for the proportion of unemployment will have a margin of error $0.01$.
Sample size to estimate proportion Example 3
An insurance company estimates 45 percent of its claims have errors. The insurance company wants to estimate with 90 percent confidence the proportion of claims with errors. What sample size is needed if they wish to be within 5 percentage points of the actual?
Solution
Given that the proportion of claims with error is $p =0.45$. The margin of error $E =0.05$. The confidence coefficient is $0.9$.
The critical value of $Z$ is $Z_{\alpha/2} = 1.64$.
The minimum sample size required to estimate the proportion of claims with error is
$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.45(1-0.45)\bigg(\frac{1.64}{0.05}\bigg)^2\\ &=266.2704\\ &\approx 267. \end{aligned} $$
Thus, the sample of size $n=267$ will ensure that the $90$% confidence interval for the proportion of claims with error will have a margin of error $0.05$.
Sample size to estimate proportion Example 4
A television spots commentator wants to estimate the proportion of citizens who "follow professional football". What sample size should be obtained if he wants to be within 2 percentage points with 95% confidence if he uses an estimate of 54% obtained from a poll?
Solution
Given that the proportion of citizens who follow professional football is $p =0.54$. The margin of error $E =0.02$. The confidence coefficient is $0.95$.
The critical value of $Z$ is $Z_{\alpha/2} = 1.96$.
The minimum sample size required to estimate the proportion citizens who follow professional football is
$$ \begin{aligned} n&= p(1-p)\bigg(\frac{z}{E}\bigg)^2\\ &= 0.54(1-0.54)\bigg(\frac{1.96}{0.02}\bigg)^2\\ &=2385.6336\\ &\approx 2386. \end{aligned} $$
Thus, the sample of size $n=2386$ will ensure that the $95$% confidence interval for the proportion citizens who follow professional football will have a margin of error $0.02$.
Endnote
In this tutorial, you learned about how to determine the sample size required to estimate population proportion. You also learned how to solve numerical examples on sample size determination to estimate proportion.
To learn more about other calculators for the determination of sample size, please refer to the following tutorials:
Let me know in the comments if you have any questions on determination of sample size required to estimate the proportion and your thought on this article.
