Q.21

Question

The P-value for a chi-square goodness-of-fit test is 0.0129. The correct conclusion is

(a) reject H0 at α=0.05; there is strong evidence that the trees are randomly distributed.

(b) reject H0 at α=0.05; there is not strong evidence that the trees are randomly distributed.

(c) reject H0 at α=0.05; there is strong evidence that the trees are not randomly distributed.

(d) fail to reject H0 at α=0.05; there is not strong evidence that the trees are randomly distributed.

(e) fail to reject H0 at α=0.05; there is strong evidence that the trees are randomly distributed.

Step-by-Step Solution

Verified
Answer

Correct option is 

(c) reject H0 at α=0.05; there is strong evidence that the trees are not randomly distributed.

1Step 1: Given information

Given in the question that,  Researchers wondered whether the trees in a longleaf pine forest in Georgia are randomly distributed. To find out, they divided the forest into four equal quadrants. Then the researchers took a random sample of 100 trees and counted the number in each quadrant. Here are their data: 

2Step 2: Explanation

0.0129 .Table is

Quadratic Count

1
18
2
22
3
39
4
21

P-value is

Level of significance 5%

The null and alternative hypotheses are:

H0:p1=p2=p3=p4=0.24

Ha: At least one of the pi is incorrect

3Step 3: Calculation for test statistic

The calculation for test statistic could be done as:

Observed value(O)Expected value(E)

18


0.25(100)=25
22
0.25(100)=25
39
0.25(100)=25
21
0.25(100)=25

The test statistic is:

χ2=(O-E)2E

=(18-25)225+(22-25)225+(39-25)225+(21-25)225

=10.8

The P-value is 0.0129

The P-value is below the significance level. Thus, the null hypothesis is rejected. Thus at 5% significance level there is sufficient evidence to favor the claim about the trees.

Hence, the correct option is (c).