Question

The table below lists the frequency of wins for different post positions in a horse race. A post position of 1 is closest to the inside rail, so that horse has the shortest distance to run. (Because the number of horses varies from year to year, only the first 10 post positions are included.) Use a 0.05 significance level to test the claim that the likelihood of winning is the same for the different post positions. Based on the result, should bettors consider the post position of a horse race? 5 8 Post Position Wins 10 18 Determine the null and alternative hypotheses. Calculate the test statistic, x2 x2(Round to three decimal places as needed.) Calculate the P-value. P-value(Round to four decimal places as needed.) What is the conclusion for this hypothesis test? A. Fail to reject Ho. There is sufficient evidence to warrant rejection of the claim that the likelihood of winning is the same for the different post positions. B. Reject Ho- There is sufficient evidence to warrant rejection of the claim that the likelihood of winning is the same for the different post positions. C. Reject Ho- There is insufficient evidence to warrant rejection of the claim that the likelihood of winning is the same for the different post positions. D. Fail to reject Ho. There is insufficient evidence to warrant rejection of the claim that the likelihood of winning is the same for the different post positions Based on the result, should bettors consider the post position of a horse race? O Yes

Answer 1

SOLUTION:

From given data,

Determine the null and alternative hypothesis:

Here, we would be using chi square test

H₀: All horses are equally likely to win
H₁: Horses are not equally likely to win

Post Position	Wins (Observed)	Expected (E)	(O-E)2/E
1	18	10.9	(18-10.9)2/10.9 = 4.6247
2	14	10.9	(14-10.9)2/10.9 = 0.8816
3	10	10.9	(10-10.9)2/10.9 = 0.0743
4	15	10.9	(15-10.9)2/10.9 = 1.5422
5	13	10.9	(13-10.9)2/10.9 = 0.4045
6	6	10.9	(6-10.9)2/10.9 = 2.2027
7	7	10.9	(7-10.9)2/10.9 = 1.3954
8	11	10.9	(11-10.9)2/10.9 = 0.0009
9	4	10.9	(4-10.9)2/10.9 = 4.3678
10	11	10.9	(11-10.9)2/10.9 = 0.0009

Average = $\Sigma$ O / n = (18+14+10+15+13+6+7+11+4+11) / 10 = 109/10 = 10.9

Test statistics:

$\chi ^{2}$ = $\Sigma$ (O-E)²/E = 4.6247+0.8816+0.0743+1.5422+0.4045+2.2027+1.3954+0.0009+4.3678+0.0009

$\chi ^{2}$ = $\Sigma$ (O-E)²/E = 15.495

Degree of freedom (df)

df = n-1 = 10 -1 = 9

P-value :

p-value = 0.0782

Significance level = $\alpha$ = 0.05

What is the conclusion for this hypothesis test.

p-value = 0.0782 > Significance level = $\alpha$ = 0.05

option (B) is correct,

Reject $H_{0}$ , There is sufficient evidence to warrant rejection of the claim that the likelihood of winning is the same for the different post positions.

Please thumbs-up / vote up this answer if it was helpful. In case of any problem, please comment below. I will surely help. Down-votes are permanent and not notified to us, so we can't help in that case