# Solved (Free): Randomly selected nonfatal occupational injuries and illnesses are categorized according to the day of the week that they first occurred

#### ByDr. Raju Chaudhari

Apr 3, 2021

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.