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

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.

#### Further Reading

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