Null hypothesis: Ho: Sampling data is similar to the distribution that would exist if the teams are evenly matched
Alternate hypothesis: Ha: Sampling data is not similar to the distribution that would exist if the teams are evenly matched
degree of freedom =categories-1= | 3 | |||
for 0.05 level and 3 df :crtiical value X2 = | 7.8147 | |||
Decision rule: reject Ho if value of test statistic X2>7.815 |
applying chi square goodness of fit test: |
relative | observed | Expected | Chi square | ||
category | frequency(p) | Oi | Ei=total*p | R2i=(Oi-Ei)2/Ei | |
4 | 0.13 | 16 | 10.50 | 2.881 | |
5 | 0.25 | 17 | 21.00 | 0.762 | |
6 | 0.31 | 19 | 26.25 | 2.002 | |
7 | 0.31 | 32 | 26.25 | 1.260 | |
total | 1.000 | 84 | 84 | 6.9048 | |
test statistic X2 = | 6.905 |
since test statistic does not falls in rejection region we fail to reject null hypothesis |
we do not have have sufficient evidence to conclude that data support the distribution that would exist if the teams are evenly matched |
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