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

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.

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

Solution

The observed data is

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 $df=k-1=4-1 =3$.

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

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

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.

Further Reading

• Statistics
• Descriptive Statistics
• Probability Theory
• Probability Distribution
• Hypothesis Testing
• Confidence interval
• Sample size determination
• Non-parametric Tests
• Correlation Regression
• Statistics Calculators