# Solved (Free): A random sample of 160 car crashes are selected and categorized by age. The results are listed below. The age distribution of drivers

#### ByDr. Raju Chaudhari

Apr 3, 2021

A random sample of 160 car crashes are selected and categorized by age. The results are listed below. The age distribution of drivers for the given categories is 18% for the under 26 group, 39% for the 26 - 45 group, 31% for the 46 - 65 group, and 12% for the group over 65.

Age Under 26 26-45 46-65 Over 65
Drivers 66 39 25 30

Test the claim that all ages have crash rates proportional to their driving rates. Use $\alpha = 0.05$.

#### Solution

The observed data is

Age Obs. Freq.$(O)$ Prop.
Under 26 66 0.18
26-45 39 0.39
46-65 25 0.31
Over 65 30 0.12
##### Step 1 Setup the hypothesis

The null and alternative hypothesis are as follows:

$H_0:p_{1}=0.18, p_{2} =0.39, p_{3} =0.31, p_{4}= 0.12$
(i.e., All ages have crash rate proportional to their driving rates.)

$H_1:$ The crash rate is not proportional to ther diving rates.

##### Step 2 Test statistic

The test statistic for testing above hypothesis is

 $$\begin{equation*} \chi^2= \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ \end{equation*}$$

##### Step 3 Level of Significance

The level of significance is $\alpha =0.05$.

##### Step 4 Critical value of $\chi^2$

The level of significance is $\alpha =0.05$. Degrees of freedom $df=k-1=4-1 =3$. chi-square critical region

The critical value of $\chi^2$ for $df=3$ and $\alpha=0.05$ level of significance is $\chi^2 =7.8147$.

##### Step 5 Test Statistic

The expected frequencies can be calculated as

 $$\begin{equation*} E_{i} =N*p_i \end{equation*}$$

For example, $E_{1}$ is given by

 $$\begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 160*0.18\\ &=&28.8. \end{eqnarray*}$$

Age Obs. Freq.$(O)$ Prop. $p_i$ Expe.Freq.$(E)$ $(O-E)^2/E$
Under 26 66 0.18 28.8 48.05
26-45 39 0.39 62.4 8.775
46-65 25 0.31 49.6 12.201
Over 65 30 0.12 19.2 6.075

The test statistic is

 $$\begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(66-28.8)^2}{28.8}+\cdots + \frac{(30-19.2)^2}{19.2}\\ &=& 48.05 +\cdots + 6.075\\ &=& 75.101. \end{eqnarray*}$$

##### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =75.101$ which falls $inside$ the critical region bounded by the critical value $7.8147$, we $\textit{reject}$ the null hypothesis.

OR

##### Step 6 Decision ($p$-value approach)

The p-value is $P(\chi^2_{3}>75.101) =0$.

As the p-value $0$ is $\textit{less than}$ the significance level of $\alpha = 0.05$, we $\textit{reject}$ the null hypothesis.

There is sufficient evidence to reject the claim that all ages have crash rates proportional to their driving rates.