A researcher claims the proportion of auto accidents that involve teenage drivers is less than 20%. ABC Insurance Company checks police records on 200 randomly selected auto accidents and notes that teenagers were at the wheel in 32 of them. Assume the company wants to use a 0.05 significance level to test the researcher's claim.

(a) Identify the null hypothesis and the alternative hypothesis.
(b) Determine the test statistic. Show all work; writing the correct test statistic, without supporting work, will receive no credit.
(c) Determine the P-value for this test. Show all work; writing the correct P-value, without supporting work, will receive no credit.
(d) Is there sufficient evidence to support the researcher's claim that the proportion of auto accidents that involve teenage drivers is less than 20%? Explain.

Solution

Given that $n = 200$. The sample proportion is $\hat{p}=\frac{X}{n}= \frac{32}{200} =0.16$.

(a) Null and alternative hypothesis :

The hypothesis testing problem is
$H_0 : p = 0.2$ against $H_1 : p < 0.2$ ($\text{left-tailed}$)

(b) 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.

The test statistic is
$$ \begin{aligned} Z & = \frac{\hat{p}-p}{\sqrt{\frac{p(1-p)}{n}}}\\ &= \frac{0.16-0.2}{\sqrt{\frac{0.2* (1-0.2)}{200}}}\\ & =-1.414 \end{aligned} $$

(c) The $p$-value for this test is

This is a $\text{left-tailed}$ test, so the p-value is the area to the left of the test statistic ($Z=-1.414$). Thus the $p$-value = $P(Z<-1.414) =0.0786$.

(d) The p-value is $0.0786$ which is $\text{greater than}$ the significance level of $\alpha = 0.05$, we $\text{fail to reject}$ the null hypothesis.

There is no sufficient evidence to support the researcher's claim that the proportion of auto accidents that involve teenage drivers is less than 20%.

Further Reading