Answer: 0.002429
Explanation:
Given that:
Degree of freedom = 6
Value of Chisquare test statistic = 20.32
The P value of the Chisquare distribution with a test statistic of 20.32 and a degree of freedom of 6 can be obtained using a Chisquare p value calculator.
P(20.32, 6), at 0.05 the degree of freedom obtained is :
0.002429
This p value shows that result is significant at p < 0.05.
As the P value obtained is lesser than the significance level.
= 0.002429