# Solved-Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest t

#### ByDr. Raju Chaudhari

Oct 7, 2020

Many track runners believe that they have a better chance of winning if they start in the inside lane that is closest to the field. For the data below, the lane closest to the field is Lane 1, the next lane is Lane 2, and so on until the outermost lane, Lane 6. The data lists the number of wins for track runners in the different starting positions.

Test the claim that the number of wins is uniformly distributed across the different starting positions. The results are based on 240 wins.

Starting Position 1 2 3 4 5 6
Number of wins 32 36 33 45 50 44

#### Solution

The observed data is

position Obs. Freq.$(O)$ Prop.
1 32 0.1667
2 36 0.1667
3 33 0.1667
4 45 0.1667
5 50 0.1667
6 44 0.1667

#### Step 1 The null and alternative hypothesis are as follows:

$H_0:$ The number of wins is uniformly distributed across the different starting positions,

i.e., $H_0: p_1=p_2=p_3=p_4 = p_5=p_6=\frac{1}{6}=0.1666667$

$H_1:$ The number of wins is not uniformly distributed across the different starting positions.

#### 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 $\alpha =0.01$.

#### Step 4 Critical value of $\chi^2$

The level of significance is $\alpha =0.01$. Degrees of freedom $df=k-1=6-1 =5$.

The critical value of $\chi^2$ for $df=5$ and $\alpha=0.01$ level of significance is $\chi^2 =15.0863$.

#### 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

 $$\begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 240*0.1667\\ &=&40.008. \end{eqnarray*}$$

position Obs. Freq.$(O)$ Prop. $p_i$ Expe.Freq.$(E)$ $(O-E)^2/E$
1 32 0.1667 40.008 1.603
2 36 0.1667 40.008 0.402
3 33 0.1667 40.008 1.228
4 45 0.1667 40.008 0.623
5 50 0.1667 40.008 2.496
6 44 0.1667 40.008 0.398

The test statistic is

 $$\begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(32-40.008)^2}{40.008}+\cdots + \frac{(44-40.01)^2}{40.01}\\ &=& 1.603 +\cdots + 0.398\\ &=& 6.75. \end{eqnarray*}$$

#### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =6.75$ which falls $outside$ the critical region bounded by the critical value $15.0863$, we $\textit{fail to reject}$ the null hypothesis.

OR

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

The p-value is $P(\chi^2_{5}>6.75) =0.23991$.

As the p-value $0.2399$ is $\textit{greater than}$ the significance level of $\alpha = 0.01$, we $\textit{fail to reject}$ the null hypothesis.

We conclude that the number of wins is uniformly distributed across the different starting positions.