10.97
chi-square test of independence can be used to assess this
10.98
c1 | c2 | c3 | sum | |||||
r1 | 17 | 113 | 13 | 143 | ||||
r2 | 7 | 115 | 21 | 143 | ||||
sum | 24 | 228 | 34 | 286 | ||||
Eij | 1 | 2 | 3 | |||||
expected | 1 | 12 | 114 | 17 | ||||
2 | 12 | 114 | 17 | |||||
0i | 17 | 113 | 13 | 7 | 115 | 21 | ||
Ei | 12 | 114 | 17 | 12 | 114 | 17 | ||
sum | ||||||||
(Oi-Ei)^2/Ei | 2.083333 | 0.008772 | 0.941176 | 2.083333 | 0.008772 | 0.941176471 | 6.066563 | |
critical value | 5.991465 | |||||||
p-value | 0.048157 |
since TS = 6.066563 > 5.99 (critical value at 0.05 level)
p-value = 0.048
we reject the null hypothesis
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