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# 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 $(0.1388,0.3012)$.