We would like to compare the performance of three jockeys at a horse racing trac
ID: 3150660 • Letter: W
Question
We would like to compare the performance of three jockeys at a horse racing track. Placings for horses ridden by the three jockeys for a sample of races are summarized in the table below. (Assume that six horses run in each race.)
Placing
Jockey A
Jockey B
Jockey C
1st
5
16
12
2nd
10
9
16
3rd
13
17
11
4th
9
12
7
5th
16
10
14
6th
17
6
10
(a) Use JMP to conduct an appropriate hypothesis test at the 10% level of significance to determine whether the three jockeys are homogeneous with respect to their performance. Use the P-value method. Your contingency table should display only the observed cell counts, the expected cell counts and the cell chi-square values. Note thatJMP will do the calculations, but you must write the hypotheses and the conclusion yourself. You do not need to attach the JMP output to your assignment. You don't have to show the calculations of the test statistic and P-value. You can take these directly from the output.
Follows these instructions to get the proper JMP output:
Create three columns in a JMP data table. The first column should titled Jockey, the second column should be titled Placing and the third column should be titled Count.
In the first column, type Jockey A in the first six rows, Jockey B in the next six rows and Jockey C in the next six rows. In the second column, type 1st, 2nd, 3rd, 4th, 5th, 6th in the first six rows, and the same thing in the next six rows and the next six rows. In the third column, type the observed cell frequencies from the table above.
Go to Analyze > Fit Y by X. Choose Jockey as Y, Placing as X and Count as Freq. Click OK. A contingency table will appear. The values in the table are not the values we want. Under the red arrow beside Contingency Table, uncheck Total %, Col % andRow %, and leave Count checked. Now also check Expected and Cell Chi Square.
(b) Show how the expected cell count and the cell chi-square values are calculated for the number of races Jockey C finished in 5th Place.
(c) Provide an interpretation of the P-value of the test in (a).
(d) Which four cells contribute the most to the value of the test statistic?
(e) Suppose you had instead used the critical value method to conduct the test in (a). What would be the decision rule and the conclusion of the test?
(f) Explain what it would mean to make a Type I Error and a Type II Error for this hypothesis test.
Placing
Jockey A
Jockey B
Jockey C
1st
5
16
12
2nd
10
9
16
3rd
13
17
11
4th
9
12
7
5th
16
10
14
6th
17
6
10
Explanation / Answer
Anova: Single Factor SUMMARY Groups Count Sum Average Variance Jockey A 6 70 11.66667 20.66667 Jockey B 6 70 11.66667 17.86667 Jockey C 6 70 11.66667 9.866667 ANOVA Source of Variation SS df MS F P-value F crit Between Groups -2.8E-14 2 -1.4E-14 -8.8E-16 #NUM! 3.68232 Within Groups 242 15 16.13333 Total 242 17
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