A physician claims that 13.6% of all pregnant females smoke during their pregnancy. A random sample of 400 pregnant women revealed that 60 of them are smoking while pregnant. Test the physician's claim at the 0.05 level of significance.

Solution

Given that $n = 400$, $X= 60$.

The sample proportion is

$$\hat{p}=\frac{X}{n}=\frac{60}{400}=0.15$$.

Step 1 Hypothesis Testing Problem

The hypothesis testing problem is
$H_0 : p = 0.136$ against $H_1 : p \neq 0.136$ ($\text{two-tailed}$)

Step 2 Test Statistic

The test statistic for testing above hypothesis testing problem is

$$ \begin{aligned} Z & = \frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}} \end{aligned} $$
which follows $N(0,1)$ distribution.

Step 3 Significance Level

The significance level is $\alpha = 0.05$.

Step 4 Critical values

As the alternative hypothesis is $\textit{two-tailed}$, the critical value of $Z$ $\text{ are }$ $\text{-1.96 and 1.96}$.

Z-critical region two-tailed
Z-critical region two-tailed

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

Step 5 Computation

The test statistic is

$$ \begin{aligned} Z & = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\\ &= \frac{0.15-0.136}{\sqrt{\frac{0.136* (1-0.136)}{400}}}\\ & =0.817 \end{aligned} $$

Step 6 Decision (Traditional approach)

The test statistic is $Z =0.817$ which falls $outside$ the critical region, we $\text{fail to reject}$ the null hypothesis.

OR

Step 6 Decision ($p$-value approach)

This is a $\text{two-tailed}$ test, so the p-value is the area to the left of the test statistic ($Z=0.817$). Thus the $p$-value = $P(Z < 0.817) =0.414$.

The p-value is $0.414$ which is $\text{greater than}$ the significance level of $\alpha = 0.05$, we $\text{fail to reject}$ the null hypothesis.

A physicians claim is accepted at 0.05 level of significance. That is 13.6% of all pregnant females smoke during their pregnancy.

Further Reading