Answer
Step 1:
$H_0$ : The size of the population is independent from the age.
$H_1$: The size of the population is dependent upon the age.
Step 2:
Since α=0.05, the critical value using Table G with (2-1)(6-1) = (1)(5) =5 degrees of freedom is 11.071.
Step 3:
Expected Value:
$E_1,1$ = $\frac{(11079)(1461)}{(22335)}$ = 724.71
$E_1,2$ = $\frac{(11079)(4244)}{(22335)}$ = 2105.18
$E_1,3$ = $\frac{(11079)(1167)}{(22335)}$= 578.88
$E_14$ = $\frac{(11079)(6874)}{(22335)}$ = 3409.76
$E_1,5$ = $\frac{(11079)(5189)}{(22335)}$ = 2573.94
$E_1,6$ = $\frac{(11079)(3400)}{(22335)}$ = 1686.53
$E_2,1$ = $\frac{(11256)(1461)}{(22335)}$ = 736.29
$E_2,2$ = $\frac{(11256)(4244)}{(22335)}$ = 2138.82
$E_2,3$ = $\frac{(11256)(1167)}{(22335)}$ = 588.12
$E_2,4$ = $\frac{(11256)(6874)}{(22335)}$ = 3464.24
$E_2,5$ = $\frac{(11256)(5189)}{(22335)}$ = 2615.06
$E_2,6$ = $\frac{(11256)(3400)}{(22335)}$ = 1713.47
Test Value :
χ2 = Σ $\frac{(O-E)^{2}}{E}$
=
$\frac{(721-724.71)^{2}}{724.71}$ + $\frac{(2140-2105.18)^{2}}{2105.18}$ + $\frac{(102-578.88)^{2}}{578.88}$ + $\frac{(3515-3409.76)^{2}}{3409.76}$ + $\frac{(2702-2573.94)^{2}}{2573.94}$ +$\frac{(1899-1686.53)^{2}}{1686.53}$ + $\frac{(740-736.29)^{2}}{236.29}$ + $\frac{(2104-2138.82)^{2}}{2138.82}$ + $\frac{(1065-588.12)^{2}}{588.12}$ + $\frac{(3359-3464.24)^{2}}{3464.24}$ + $\frac{(2487-2615.06)^{2}}{2615.06}$ + $\frac{(1501-1713.47)^{2}}{1713.47}$
=0.019+0.575+392.849+3.248+6.371+26.768+0.019+0.567+386.671+3.197+6.271+26.347
=852.902
Step 4:
Since 852.902 > 11.071, the decision is to reject the null hypothesis.
Step 5:
There is enough evidence to claim that the size of the population is dependent upon the age.
