In: Statistics and Probability
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 statisticΛ20 to test the claim that the probabilities of winning are the same in the different positions. Use a= 0.05. The results are based on 240 wins.
Starting Position | 1 | 2 | 3 | 4 | 5 | 6 |
Number of Wins | 44 | 36 | 33 | 50 | 32 | 45 |
A)Hypothesis
Ho :
Ha :
B) Critical Values= and decision to reject or fail to reject Ho
C) Conclusion about the problem statement.
A)
Ho: probabilities of winning are the same in the different positions
Ha: probabilities of winning are not the same in the different positions
B)
degree of freedom =categories-1= | 5 | |||
for 0.05 level and 5 df :crtiical value X2 = | 11.070 | |||
Decision rule: reject Ho if value of test statistic X2>11.07 |
applying chi square goodness of fit test: |
relative | observed | Expected | residual | Chi square | |
category | frequency(p) | Oi | Ei=total*p | R2i=(Oi-Ei)/√Ei | R2i=(Oi-Ei)2/Ei |
1 | 1/6 | 44 | 40.00 | 0.63 | 0.400 |
2 | 1/6 | 36 | 40.00 | -0.63 | 0.400 |
3 | 1/6 | 33 | 40.00 | -1.11 | 1.225 |
4 | 1/6 | 50 | 40.00 | 1.58 | 2.500 |
5 | 1/6 | 32 | 40.00 | -1.26 | 1.600 |
6 | 1/6 | 45 | 40.00 | 0.79 | 0.625 |
total | 1.000 | 240 | 240 | 6.7500 | |
test statistic X2 = | 6.750 |
since test statistic does not falls in rejection region we fail to reject null hypothesis |
C)
we do not have have sufficient evidence to conclude that probabilities of winning are not the same in the different positions |