Many track hurdlers 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 hurdlers in the different starting positions. Calculate the chi-square test statistic $\chi ^ { 2 }$ to test the claim that the probabilities of winning are the same in the different positions. Use $\alpha = 0.05$. The results are based on 240 wins.
$\begin{array}{c|cccccc}
\text { Starting Position } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Number of Wins } & 50 & 36 & 45 & 44 & 32 & 33
\end{array}$

A) $6.750$
B) $9.326$
C) $12.592$
D) $15.541$

Question

Many track hurdlers 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 hurdlers in the different starting positions. Calculate the chi-square test statistic $\chi ^ { 2 }$ to test the claim that the probabilities of winning are the same in the different positions. Use $\alpha = 0.05$. The results are based on 240 wins.
$\begin{array}{c|cccccc}
\text { Starting Position } & 1 & 2 & 3 & 4 & 5 & 6 \
\hline \text { Number of Wins } & 50 & 36 & 45 & 44 & 32 & 33
\end{array}$

A) $6.750$
B) $9.326$
C) $12.592$
D) $15.541$

Quizplus · Accepted Answer

The chi-square test statistic is calculated as:$$\chi^2 = \sum_{i=1}^k \frac{(O_i - E_i)^2}{E_i}$$where $O_i$ is the observed frequency, $E_i$ is the expected frequency, and $k$ is the number of categories.In this case, the expected frequency for each category is 40 (since there are 240 wins and 6 categories).Therefore, the chi-square test statistic is:$$\chi^2 = \frac{(50 - 40)^2}{40} + \frac{(36 - 40)^2}{40} + \frac{(45 - 40)^2}{40} + \frac{(44 - 40)^2}{40} + \frac{(32 - 40)^2}{40} + \frac{(33 - 40)^2}{40} = 6.750$$The degrees of freedom for the chi-square test is $k - 1 = 5$.Using a chi-square distribution table with 5 degrees of freedom and $\alpha = 0.05$, we find that the critical value is 11.070.Since the chi-square test statistic (6.750) is less than the critical value (11.070), we fail to reject the null hypothesis that the probabilities of winning are the same in the different positions.

Many Track Hurdlers Believe That They Have a Better Chance χ2\chi ^ { 2 }χ2

Many Track Hurdlers Believe That They Have a Better Chance $\chi ^ { 2 }$