Z-test calculator for two proportions examples

$Z$-Test for two proportions

In this tutorial we will discuss $z$-test calculator for testing two population proportions with step by step numerical examples on $Z$-test for testing two population proportions.

Z test calculator for two population proportions

The $Z$-test calculator for testing two population proportions makes it easy to calculate the test statistic, $Z$ critical value and the $p$-value given the sample information, level of significance and the type of alternative hypothesis (i.e. left-tailed, right-tailed or two-tailed.)

Z test Calculator for two proportions
Sample 1 Sample 2
Sample size
No. of Successes
Level of Significance ($\alpha$)
Tail Left tailed
Right tailed
Two tailed
Results
sample proportions:
pooled estimate of proportion:
Standard Error of Diff. of prop.:
Test Statistics Z:
Z-critical value(s):
p-value:

How to use $z$-test calculator for testing two proportions?

Step 1 - Enter the sample size for first sample $n_1$ and second sample $n_2$

Step 2 - Enter the no. of successes for first sample $X_1$ and second sample $X_2$

Step 3 - Enter the level of significance $\alpha$

Step 4 - Select the alternative hypothesis (left-tailed / right-tailed / two-tailed)

Step 5 - Click on "Calculate" button to get the result

$Z$-Test for two proportions Example 1

A survey indicate that of 900 women randomly sampled, 345 use smart-phones. For the men, 450 of the 1025 who were randomly sampled use smartphones. Test whether a percentage of women who uses smartphone is less than men. Use $\alpha=0.05$.

Solution

Given that among $n_1 = 900$ women $X_1= 345$ women use smartphones and among $n_2=1025$ men $X_2=450$ men use smartphones.

The sample proportions of women and men who use smartphones are respectively
$\hat{p}_1=\frac{X_1}{n_1}=\frac{345}{900}=0.383$ and
$\hat{p}_2=\frac{X_2}{n_2}=\frac{450}{1025}=0.439$.

The pooled estimate of sample proportion is
$\hat{p} =\frac{X_1+X_2}{n_1+n_2}=\frac{345+450}{900+1025} =0.413$

Step 1 State the hypothesis testing problem

We wish to test the hypothesis that percentage of women who use smartphones is less as compared to percentage of men who use smartphones, (i.e., $p_1<p_2$).

So the hypothesis testing problem can be setup as

$H_0 : p_1 = p_2$ against $H_1 : p_1 < p_2$ ($\textit{left-tailed}$)

Step 2 Define test statistic

The test statistic for testing above hypothesis testing problem is

 \begin{aligned} Z & =\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}. \end{aligned}
The test statistic $Z$ follows standard normal distribution $N(0,1)$.

Step 3 Specify the level of significance $\alpha$

The significance level is $\alpha = 0.05$.

Step 4 Determine the critical value

As the alternative hypothesis is $\textit{left-tailed}$, the critical value of $Z$ $\text{ is }$ $\text{-1.64}$ (From Normal Statistical Table).

The rejection region (i.e. critical region) is $\text{Z < -1.64}$.

Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} Z_{obs}&= \frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}\\ &= \frac{(0.383-0.439)-0}{\sqrt{\frac{0.413*(1-0.413)}{900}+\frac{0.413*(1-0.413)}{1025}}}\\ &= -2.476 \end{aligned}

The rejection region (i.e. critical region) is $\text{Z < -1.64}$. The test statistic is $Z_{obs} =-2.476$ which falls $inside$ the critical region, we $\textit{reject}$ the null hypothesis.

OR

Step 6 Decision ($p$-value approach)

The test is $\text{left-tailed}$ test, so the p-value is the area to the $\text{left}$ of the test statistic ($Z_{obs}=-2.476$) is p-value = $0.0066$.

The p-value is $0.0066$ which is $\textit{less than}$ the significance level of $\alpha = 0.05$, we $\textit{reject}$ the null hypothesis.

Interpretation

There is enough evidence to conclude that percetage of women who use smartphones is less than the percentage of men who use smartphones.

$Z$-Test for two proportions Example 2

A simple random sample of front-seat occupants involved in car crashes is obtained. Among 2823 occupants not wearing seat belts, 31 were killed. Among 7765 occupants wearing seat belts, 16 were killed. The claim is that the fatality rate is higher for those not wearing seat belts. Are Seat Belts Effective? Use $\alpha = 0.01$.

Solution

Let $p_1$ be the proportion of occupants not wearing seat belts and $p_2$ be the proportion of occupants wearing seat belts.

Given that number of front-seat occupants not wearing seat belt $n_1 = 2823$ among them $X_1= 31$ were killed and the number of front-seat occupants wearing seat belt $n_2=7765$ among them $X_2=16$ were killed.

The estimated sample proportions are
$\hat{p}_1=\frac{X_1}{n_1}=\frac{31}{2823}=0.011$.

$\hat{p}_2=\frac{X_2}{n_2}=\frac{16}{7765}=0.002$.

The pooled estimate of sample proportion is
$\hat{p} =\frac{X_1+X_2}{n_1+n_2}=\frac{31+16}{2823+7765} =0.004$

The claim that the fatality rate is higher for those not wearing seat belts can be expressed as $p_1 > p_2$.

Step 1 State the hypothesis testing problem

The hypothesis testing problem is

$H_0 : p_1 = p_2$ against $H_1 : p_1 > p_2$ ($\textit{right-tailed}$)

Step 2 Define test statistic

The test statistic for testing above hypothesis testing problem is

 \begin{aligned} Z & =\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}. \end{aligned}

The test statistic $Z$ follows standard normal distribution $N(0,1)$.

Step 3 Specify the level of significance

The significance level is $\alpha = 0.01$.

Step 4 Determine the critical value

As the alternative hypothesis is $\textit{right-tailed}$, the critical value of $Z$ $\text{ is }$ $\text{2.33}$ (From Normal Statistical Table).

The rejection region (i.e. critical region) is $\text{Z > 2.33}$.

Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} Z_{obs}&= \frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}\\ &= \frac{(0.011-0.002)-0}{\sqrt{\frac{0.004*(1-0.004)}{2823}+\frac{0.004*(1-0.004)}{7765}}}\\ &= 6.106 \end{aligned}

The rejection region (i.e. critical region) is $\text{Z > 2.33}$. The test statistic is $Z_{obs} =6.106$ which falls $inside$ the critical region, we $\textit{reject}$ the null hypothesis.

OR

Step 6 Decision ($p$-value approach)

The test is $\text{right-tailed}$ test, so the p-value is the area to the $\text{right}$ of the test statistic ($Z_{obs}=6.106$) is p-value = $0$.

The p-value is $0$ which is $\textit{less than}$ the significance level of $\alpha = 0.01$, we $\textit{reject}$ the null hypothesis.

Interpretation

There is enough evidence to support the claim that the fatality rate is higher for those not wearing seat belts.

Thus we conclude that the use of seat belts appears to be effective in saving lives.

$Z$-Test for two proportions Example 3

Two machines used in the same operation are to be compared. A random sample of 80 parts from the first machine yields 6 non-conforming ones. A random sample of 120 parts from the second machine shows 14 non-conforming ones.

Are the two machine differ significantly with respect to the proportion of non-confirming? Use $\alpha= 0.05$.

Solution

Given information

. First Machine Second Machine
Sample size $n_1=80$ $n_2=200$
no.of non-confirming $X_1=6$ $X_2=14$

The sample proportions are
$\hat{p}_1=\frac{X_1}{n_1}=\frac{6}{80}=0.075$.

$\hat{p}_2=\frac{X_2}{n_2}=\frac{14}{200}=0.07$.

The pooled estimate of sample proportion is
$\hat{p} =\frac{X_1+X_2}{n_1+n_2}=\frac{6+14}{80+200} =0.071$

Step 1 State the hypothesis testing problem

The hypothesis testing problem is $H_0:$ Two machines do not differ significantly with respect to proportion of non-confirming against $H_1:$ Two machines differ significantly with respect to proportion of non-confirming.

i.e., $H_0 : p_1 = p_2$ against $H_1 : p_1 \neq p_2$ ($\textit{two-tailed}$)

Step 2 Define test statistic

The test statistic for testing above hypothesis testing problem is

 \begin{aligned} Z & =\frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}. \end{aligned}

The test statistic $Z$ follows standard normal distribution $N(0,1)$.

Step 3 Specify the level of significance $\alpha$

The significance level is $\alpha = 0.05$.

Step 4 Determine the critical value

As the alternative hypothesis is $\textit{two-tailed}$, the critical value of $Z$ $\text{ are }$ $\text{-1.96 and 1.96}$ (From Normal Statistical Table).

The rejection region (i.e. critical region) is $\text{Z < -1.96 or Z > 1.96}$.

Step 5 Computation

The test statistic under the null hypothesis is

 \begin{aligned} Z_{obs}&= \frac{(\hat{p}_1-\hat{p}_2)-(p_1-p_2)}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n_1}+\frac{\hat{p}(1-\hat{p})}{n_2}}}\\ &= \frac{(0.075-0.07)-0}{\sqrt{\frac{0.071*(1-0.071)}{80}+\frac{0.071*(1-0.071)}{200}}}\\ &= 0.147 \end{aligned}

The rejection region (i.e. critical region) is $\text{Z < -1.96 or Z > 1.96}$. The test statistic is $Z_{obs} =0.147$ which falls $outside$ the critical region, we $\textit{fail to reject}$ the null hypothesis.

OR

Decision ($p$-value approach)

The test is $\text{two-tailed}$ test, so the p-value is the area to the $\text{extreme}$ of the test statistic ($Z_{obs}=0.147$) is p-value = $0.8833$.

The p-value is $0.8833$ which is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

Interpretation

There is no sufficient evidence to support the alternative hypothesis. Thus we conclude that the two machine do not differ significantly with respect to the proportion of non-confirming.

Endnote

In this tutorial, you learned the about how to solve numerical examples on $Z$-test for testing two population proportions. You also learned about the step by step procedure to apply $Z$-test for testing two population proportions and how to use $Z$-test calculator for testing two population proportions to get the value of test statistic, p-value, and z-critical value.

Let me know in the comments if you have any questions on $Z$-test calculator for two proportions with examples and your thought on this article.