Question

25. (9.2)Take a random sample of 100 trees in each quadrant. Here are the data l21 Quadrant Count lQl 18 03 39

Answer 1

(1)

If the trees are randomly distributed then we expected 100/4 = 25 trees in each quadrant

(2)

The following table is obtained:

Categories	Observed	Expected	(fo-fe)2/fe
Category 1	18	100*0.25=25	(18-25)2/25 = 1.96
Category 2	22	100*0.25=25	(22-25)2/25 = 0.36
Category 3	39	100*0.25=25	(39-25)2/25 = 7.84
Category 4	21	100*0.25=25	(21-25)2/25 = 0.64
Sum =	100	100	10.8

(1) Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$\dpi{100} H_0: p_1 = 0.25, p_2 = 0.25, p_3 = 0.25, p_4 = 0.25$

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 $\dpi{100} R = \{\chi^2: \chi^2 > 7.815\}$

(3) Test Statistics

The Chi-Squared statistic is computed as follows:

$\dpi{100} \chi^2 = \sum_{i=1}^n \frac{(O_i-E_i)^2}{E_i} = 1.96+0.36+7.84+0.64 = 10.8$

(4) Decision about the null hypothesis

Since it is observed that $\dpi{100} \chi^2 = 10.8 > \chi_c^2 = 7.815$ , it is then concluded that the null hypothesis is rejected.

(5) Conclusion

It is concluded that the null hypothesis Ho 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.