here we want to test the
null hypothesis H0: physical activity among Ohioans is similar
to the U.S. population
alternate hypothesis H0: physical activity among Ohioans is not
similar to the U.S. population
we use chi-square test and
chi-squar=sum((O-E)2/E)=379.48 with k-1=4-1=3 df and
critical chi-square(0.05,3)=7.81 is less than calculated
chi-squre=379.48, so we reject the null hypothesis and conclude
that distribution in physical activity among Ohioans is not similar
to the U.S. population.
following information has been generated
|
group |
Expected% (US) |
Observed%(Ohiaonas) |
Expected frequency(E) |
Observedfrequency(O) |
(O-E) |
(O-E)2/E |
|
donot meet physical activity |
45.9 |
50 |
2054.025 |
2237.5 |
183.475 |
16.39 |
|
strength |
3.5 |
5 |
156.625 |
223.75 |
67.125 |
28.77 |
|
aerobic |
29 |
35 |
1297.75 |
1566.25 |
268.5 |
55.55 |
|
both strength and aerobic |
21.6 |
10 |
966.6 |
447.5 |
-519.1 |
278.78 |
|
sum= |
100 |
100 |
4475 |
4475 |
0 |
379.48 |
|
|
|
|
|
|
critical chi-square(0.05,3)= |
7.81 |
