Amateur triathletes who competed in the 2015 Lake Placid Ironman were classified by gender and age with the following results:
18 to 34 |
35 to 49 |
50 to 79 |
|
Men |
590 |
735 |
168 |
Women |
165 |
140 |
120 |
Do these data provide sufficient evidence at the 0.01 level of
significance to conclude that age and gender are independent among
amateur Ironman triathletes?
Use the Critical Value Method of Testing.
In your work space below, you will need to have -
1. The null hypothesis, Ho (2 points)
2. The alternative hypothesis, H1 (2 points)
3. The expected frequencies for each table value (6
points)
4. The test statistic (9 points)
5. The critical value (2 points)
6. The decision to accept Ho or reject Ho (4 points)
1) null hypothesis: Ho: age and gender are independent among amateur Ironman triathletes
2)The alternative hypothesis, H1 age and gender are not independent among amateur Ironman triathletes
3) fr expected frequency =row total*column total/grand total
hence below is expected frequencies table:
Ei=?row*?column/?total | 18-34 | 35-49 | 50-79 |
men | 587.703 | 681.113 | 224.184 |
women | 167.297 | 193.887 | 63.816 |
4)applying chi square test of independence:
chi square | =(Oi-Ei)2/Ei | 18-34 | 35-49 | 50-79 | Total |
men | 0.01 | 4.26 | 14.08 | 18.35 | |
women | 0.03 | 14.977 | 49.464 | 64.47 | |
Total | 0.04 | 19.24 | 63.54 | 82.824 |
chi square test statsitic =82.824
5)
degree of freedom =(row-1)*(column-1)=(2-1)*(3-1)=2
for 2 df and 0.01 level critical value =9.210
6)
as test statsitic is higher than critical value we reject null hypothesis
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