Answer
Step 1:
$H_0$ : The schools and the results of students in a basic fitness test are independent from each other.
$H_1$: The schools and the results of students in a basic fitness test are dependent upon each other.
Step 2:
Since α=0.05, the critical value using Table G with (2-1)(4-1) = (1)(3) =3 degrees of freedom is 7.815.
Step 3:
Expected Value:
$E_1,1$ = $\frac{(167)(120)}{(480)}$ = 41.75
$E_1,2$ = $\frac{(167)(120)}{(480)}$ = 41.75
$E_1,3$ = $\frac{(167)(120)}{(480)}$= 41.75
$E_1,4$ = $\frac{(167)(120)}{(480)}$ = 41.75
$E_2,1$ = $\frac{(313)(120)}{(480)}$ = 78.25
$E_2,2$ = $\frac{(313)(120)}{(480)}$ = 78.25
$E_2,3$ = $\frac{(313)(120)}{(480)}$ =78.25
$E_2,4$ = $\frac{(313)(120)}{(480)}$ = 78.25
Test Value :
χ2 = Σ $\frac{(O-E)^{2}}{E}$
=
$\frac{(49-41.75)^{2}}{41.75}$ + $\frac{(38-41.75)^{2}}{41.75}$ + $\frac{(46-41.75)^{2}}{41.75}$ + $\frac{(34-41.75)^{2}}{41.75}$ + $\frac{(71-78.25)^{2}}{78.25}$ + $\frac{(82-78.25)^{2}}{78.25}$ + $\frac{(74-78.25)^{2}}{78.25}$ + $\frac{(86-78.25)^{2}}{78.25}$ +
=1.259+0.337+0.433+1.439+0.672+0.18+0.231+0.768
=5.317
Step 4:
Since 5.317 < 7.815, the decision is to reject the null hypothesis.
Step 5:
There is not enough evidence to claim that the schools and the results of students in a basic fitness test are dependent upon each other.
