# Confidence Interval for Population Proportion Calculator

## Confidence Interval for Population Proportion Calculator

Confidence interval for population proportion calculator computes estimate of proportion,standard error of proportion,Z-critical value,margin of error,lower and upper confidence limits based on sample size,number of successes and confidence level input values.

Confidence Interval for population proportion calculator
Sample Size ($n$)
Number of successes ($k$)
Confidence Level ($1-\alpha$)
Results
Estimate of proportion: ($\hat{p}$)
Standard Error of proportion: ($SE$)
Z-critical value: ($Z_{\alpha/2}$)
Margin of Error: ($E$)
Lower Confidence Limits:
Upper Confidence Limits:

## Confidence Interval For Proportion Theory

Confidence interval can be used to estimate the population parameter with the help of an interval with some degree of confidence. One such a parameter that can be estimated is a population proportion.

We will discuss step by step procedure to construct a confidence interval for population proportion.

Let $X$ be the observed number of individuals possessing certain attributes (number of successes) in a random sample of size $n$ from a large population with population proportion $p$. Then $\hat{p}=\frac{X}{n}$ be the observed proportion of successes.

Let $C=1-\alpha$ be the confidence coefficient. We wish to construct $100(1-\alpha)$% confidence interval estimate of a population proportion $p$.

The margin of error for proportion is

 \begin{aligned} E = Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \end{aligned}

where $Z_{\alpha/2}$ is the table value from normal statistical table.

$100(1-\alpha)$% confidence interval for population proportion is

 \begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned}

## Assumptions

a. $np\geq 10$ and $n(1-p)\geq 10$.
b. The sample is a random sample.

## Step by step procedure

Step by step procedure to find the confidence interval for proportion is as follows :

#### Step 2 Given information

Specify the given information, sample size $n$, observed number of successes $X$. The estimate of population proportion of success is $\hat{p} =\frac{X}{n}$.

#### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval to estimate the population proportion is

 \begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned}

where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$.

#### Step 4 Determine the critical value

Find the critical value $Z_{\alpha/2}$ from the normal statistical table that corresponds to the desired confidence level.

#### Step 5 Compute the margin of error

The margin of error for proportion is

 \begin{aligned} E = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}} \end{aligned}

#### Step 6 Determine the confidence interval

$100(1-\alpha)\%$ confidence interval estimate for population proportion is

 \begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E \end{aligned}

Equivalently, $100(1-\alpha)\%$ confidence interval estimate of population proportion is $\hat{p} \pm E$ or $(\hat{p} -E, \hat{p} +E)$.

Thus $100(1-\alpha)$% confidence interval estimate of population proportion $p$ is

 \begin{aligned} \bigg(\hat{p}-Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}, \hat{p}+Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\bigg). \end{aligned}

## Confidence interval for Population Proportion Examples

We will discuss some numerical examples to understand how to construct a confidence interval for population proportion.

### Example 1

In a simple random sample of 50 animated children's movies, 19 show tobacco use by main characters. Construct a 90% confidence interval to estimate the proportion of all animated children's movies that show tobacco use by main characters. Assume the population is normally distributed.

#### Solution

Given that sample size $n = 50$, observed $X = 19$.

Thus the sample proportion of animated children's movies that show tobacco use by main characters is
$\hat{p}=\frac{X}{n}=\frac{19}{50}=0.38$.

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

Confidence level is $1-\alpha = 0.9$. Thus, the level of significance is $\alpha = 0.1$.

#### Step 2 Given information

Given that sample size $n =50$, observed number of animated children's movies that show tobacco use by main characters is $X=19$.

The estimate of the proportion of success is $\hat{p} =\frac{X}{n} =\frac{19}{50}=0.38$.

#### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for population proportion is

 \begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned}

where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ 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 $Z_{\alpha/2} = Z_{0.05} = 1.64$.

#### Step 5 Compute the margin of error

The margin of error for proportions is

 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.64 \sqrt{\frac{0.38*(1-0.38)}{50}}\\ & =0.113. \end{aligned}

#### Step 6 Determine the confidence interval

$90$% confidence interval estimate for population proportion is

 \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.38 - 0.113 & \leq p \leq 0.38 + 0.113\\ 0.2674 & \leq p \leq 0.4926. \end{aligned}

Thus, $90$% confidence interval estimate for population proportion $p$ is $(0.2674,0.4926)$.

### Example 2

A marketing research firm wishes to estimate the proportion of adults who are planning to buy a new car in the next 6 months. A simple random sample of 100 adults led to 22 who were planning to buy a new car in the next 6 months.

Compute the 95% confidence interval for the proportion of adults who are planning to buy a new car in the next 6 months

#### Solution

Given that sample size $n = 100$, observed $X = 22$.

Thus the sample proportion adults who are planning to buy a new car in the next 6 months is
$\hat{p}=\frac{X}{n}=\frac{22}{100}=0.22$.

#### 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 $n =100$, adults who are planning to buy a new car in the next 6 months is $X=22$.

The estimate of the proportion of adults who are planning to buy a new car in the next 6 months is $\hat{p} =\frac{X}{n} =\frac{22}{100}=0.22$.

#### Step 3 Specify the formula

$100(1-\alpha)$% confidence interval for population proportion is

 \begin{aligned} \hat{p} - E \leq p \leq \hat{p} + E. \end{aligned}

where $E=Z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$ 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 proportions is

 \begin{aligned} E & = Z_{\alpha/2} \sqrt{\frac{\hat{p}*(1-\hat{p})}{n}}\\ & = 1.96 \sqrt{\frac{0.22*(1-0.22)}{100}}\\ & =0.081. \end{aligned}

#### Step 6 Determine the confidence interval

$95$% confidence interval estimate for population proportion is

 \begin{aligned} \hat{p} - E & \leq p \leq \hat{p} + E\\ 0.22 - 0.081 & \leq p \leq 0.22 + 0.081\\ 0.1388 & \leq p \leq 0.3012. \end{aligned}

Thus, $95$% confidence interval estimate for population proportion $p$ is $(0.1388,0.3012)$. VRCBuzz co-founder and passionate about making every day the greatest day of life. Raju is nerd at heart with a background in Statistics. Raju looks after overseeing day to day operations as well as focusing on strategic planning and growth of VRCBuzz products and services. Raju has more than 25 years of experience in Teaching fields. He gain energy by helping people to reach their goal and motivate to align to their passion. Raju holds a Ph.D. degree in Statistics. Raju loves to spend his leisure time on reading and implementing AI and machine learning concepts using statistical models.