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

(a) reject $H_0$at $\alpha =0.05$; there is strong evidence that the trees are randomly distributed.

(b) reject $H_0$at $\alpha =0.05$; there is not strong evidence that the trees are randomly distributed.

(c) reject $H_0$at $\alpha =0.05$; there is strong evidence that the trees are not randomly distributed.

(d) fail to reject $H_0$at $\alpha =0.05$; there is not strong evidence that the trees are randomly distributed.

(e) fail to reject $H_0$at $\alpha =0.05$; there is strong evidence that the trees are randomly distributed.

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:

$0.0129.$Table is

$P$-value is

Level of significance $5\%$

The null and alternative hypotheses are:

$H_0:p_1=p_2=p_3=p_4=0.24$

$H_a:$ At least one of the $p_i$ is incorrect

The calculation for test statistic could be done as:

The test statistic is:

${\chi }^2=\sum \frac{(O-E{)}^2}{E}$

$=\frac{(18-25{)}^2}{25}+\frac{(22-25{)}^2}{25}+\frac{(39-25{)}^2}{25}+\frac{(21-25{)}^2}{25}$

$=10.8$

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).