# Solved (Free): 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

#### ByDr. Raju Chaudhari

Apr 3, 2021

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.