In: Statistics and Probability
Post Position | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
Wins | 19 | 14 | 11 | 15 | 15 | 7 | 8 | 12 | 5 | 11 |
The table below lists the frequency of wins for different post positions in the Kentucky Derby horse race.
Use a 0.05 significance level to test the claim that the likelihood of winning is the same for the different post positions.
What is the critical value (the X2 value)? [Round to the nearest thousandths place]
total frequency= 117
expected frequncy,E = expected proportions*total
frequency
observed frequencey, O | expected proportion | expected frequency,E | (O-E) | (O-E)²/E | |
19 | 0.100 | 11.70 | 7.30 | 4.555 | |
14 | 0.100 | 11.70 | 2.30 | 0.452 | |
11 | 0.100 | 11.70 | -0.70 | 0.042 | |
15 | 0.100 | 11.70 | 3.30 | 0.931 | |
15 | 0.100 | 11.70 | 3.30 | 0.931 | |
7 | 0.100 | 11.700 | -4.70 | 1.888 | |
8 | 0.100 | 11.700 | -3.70 | 1.170 | |
12 | 0.100 | 11.700 | 0.30 | 0.008 | |
5 | 0.100 | 11.700 | -6.70 | 3.837 | |
11 | 0.100 | 11.700 | -0.70 | 0.042 |
chi square test statistic,X² = Σ(O-E)²/E =
13.855
level of significance, α= 0.05
Degree of freedom=k-1= 10 -
1 = 9
Critical value = 16.919 [
Excel function: =chisq.inv.rt(α,df) ]
Decision: test stat < critical , Do not reject
Ho