We conduct a chi-square test with 6 degrees of freedom and find a test statistic of 20.32. The p-value for this test would be:

0.002429

Step-by-step 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

the answer is y < -3 or   y > 3

-3y > 7 + 2

-3 y > 9

y < -3

answer: ♡♡♡♡♡♡♡♡♡♡

step-by-step explanation: