Each observation in a random sample of 100 bicycle accidents resulting in death was classified according to the day of the week on which the accident occurred. Data consistent with information given on the website are given in the following table.

Day | Mon | Tues | Wed | Thurs | Fri | Sat | Sun |
---|---|---|---|---|---|---|---|

Frequency | 14 | 13 | 12 | 15 | 14 | 17 | 15 |

Based on these data,is it reasonable to conclude that the proportion of accidents is not the same for all days of the week?

#### Solution

The observed data is

Day | Obs. Freq.$(O)$ | Prop. |
---|---|---|

Mon | 14 | 0.1429 |

Tues | 13 | 0.1429 |

Wed | 12 | 0.1429 |

Thurs | 15 | 0.1429 |

Fri | 14 | 0.1429 |

Sat | 17 | 0.1429 |

Sun | 15 | 0.1429 |

##### Step 1 Setup the hypothesis

The null and alternative hypothesis are as follows:

`$H_0:p_{Mon} =\cdots = p_{Sat} =p_{Sun}=\frac{1}{7}$`

$H_1:$ At least one of the proportion is different from `$\frac{1}{7}$`

.

##### 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 $df=k-1=7-1 =6$.

The critical value of $\chi^2$ for $df=6$ and $\alpha=0.05$ level of significance is $\chi^2 =12.5916$.

##### 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

` $$ \begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 100*0.1429\\ &=&14.29. \end{eqnarray*} $$ `

Day | Obs. Freq.$(O)$ | Prop. $p_i$ | Expe.Freq.$(E)$ | $(O-E)^2/E$ |
---|---|---|---|---|

Mon | 14 | 0.1429 | 14.29 | 0.006 |

Tues | 13 | 0.1429 | 14.29 | 0.116 |

Wed | 12 | 0.1429 | 14.29 | 0.367 |

Thurs | 15 | 0.1429 | 14.29 | 0.035 |

Fri | 14 | 0.1429 | 14.29 | 0.006 |

Sat | 17 | 0.1429 | 14.29 | 0.514 |

Sun | 15 | 0.1429 | 14.29 | 0.035 |

The test statistic is

` $$ \begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(14-14.29)^2}{14.29}+\cdots + \frac{(15-14.29)^2}{14.29}\\ &=& 0.006 +\cdots + 0.035\\ &=& 1.079. \end{eqnarray*} $$ `

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

The test statistic is $\chi^2 =1.079$ which falls $outside$ the critical region bounded by the critical value $12.5916$, we $\textit{fail to reject}$ the null hypothesis.

**OR**

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

The p-value is $P(\chi^2_{6}>1.079) =0.98244$.

As the p-value $0.9824$ is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

Thusthe proportion of accidents is same for all days of the week.

**Interpretation : ** The goodness of fit test suggest that the proportion of accidents is same for all days of the week.

#### Further Reading

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