Answer:
c) Yes, because the p-value = 0.0172
Step-by-step explanation:
The following table is obtained:
Categories    Observed(fo)     Expected (fe)     (fo-fe)²/fe
NW Oregon     3109      4357*0.727=3167.539    1.082
SW Oregon     902        4357*0.207=901.899      0
Central Oregon   244      4357*0.048=209.136     5.812
Eastern Oregon   102      4357*0.028=121.996      3.277
Sum = Â Â Â Â Â Â Â Â Â Â 4357 Â Â Â Â 4357 Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 10.171
(1) Null and Alternative Hypotheses
The following null and alternative hypotheses need to be tested:
H0​:p1​=0.727,p2​=0.207,p3​=0.048,p4​=0.028
Ha​: Some of the population proportions differ from the values stated in the null hypothesis
This corresponds to a Chi-Square test for Goodness of Fit.
(2) Rejection Region
Based on the information provided, the significance level is α=0.05, the number of degrees of freedom is df=4−1=3, so then the rejection region for this test is R={χ2:χ2>7.815}.
(3) Test Statistics
The Chi-Squared statistic is computed as follows:
[tex]X^2=\sum^n_{i=1}\frac{(O_i-E_i)^2}{y} \\\\= 1.082+0+5.812 +3.277 = 10.171[/tex]
(4) Decision about the null hypothesis
Since it is observed that
[tex]X^2 = 10.171 > X_c^2 = 7.815[/tex]
it is then concluded that the null hypothesis is rejected.
(5) Conclusion
It is concluded that the null hypothesis H_o is rejected. Therefore, there is enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.05 significance level.
