Randomly selected nonfatal occupational injuries and illnesses are categorized according to the day of the week that they first occurred, and the results are listed below (based on data from the Bureau of Labor Statistics).

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

No. of Births | 23 | 23 | 21 | 21 | 09 |

Use a 0.05 significance level to test the claim that such injuries and illnesses occur with equal frequency on the different days of the week.

#### Solution

The observed data is

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

Mon | 23 | 0.2 |

Tues | 23 | 0.2 |

Wed | 21 | 0.2 |

Thurs | 21 | 0.2 |

Fri | 9 | 0.2 |

##### Step 1 Setup the hypothesis

The null and alternative hypothesis are as follows:

`$H_0:p_{Mon} =\cdots = p_{Fri} =\frac{1}{5}$`

$H_1:$ At least one of the proportion is different from `$\frac{1}{5}$`

.

##### 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 $df=k-1=5-1 =4$.

The critical value of $\chi^2$ for $df=4$ and $\alpha=0.05$ level of significance is $\chi^2 =9.4877$.

##### 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

` $$ \begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 97*0.2\\ &=&19.4. \end{eqnarray*} $$ `

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

Mon | 23 | 0.2 | 19.4 | 0.668 |

Tues | 23 | 0.2 | 19.4 | 0.668 |

Wed | 21 | 0.2 | 19.4 | 0.132 |

Thurs | 21 | 0.2 | 19.4 | 0.132 |

Fri | 9 | 0.2 | 19.4 | 5.575 |

The test statistic is

` $$ \begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(23-19.4)^2}{19.4}+\cdots + \frac{(9-19.4)^2}{19.4}\\ &=& 0.668 +\cdots + 5.575\\ &=& 7.175. \end{eqnarray*} $$ `

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

The test statistic is $\chi^2 =7.175$ which falls $outside$ the critical region bounded by the critical value $9.4877$, we $\textit{fail to reject}$ the null hypothesis.

**OR**

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

The p-value is $P(\chi^2_{4}>7.175) =0.12692$.

As the p-value $0.1269$ is $\textit{greater than}$ the significance level of $\alpha = 0.05$, we $\textit{fail to reject}$ the null hypothesis.

Thus the claim that such injuries and illnesses occur with equal frequency on the different 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