Answer:
The Chi-square test of independence
Step-by-step explanation:
The answer to this question is the Chi-square test of independence because it is what is used to determine the existence of an association between nominal/categorical variables. This test is non parametric in nature and it can also be called the chi square test of association.
The data is tabulated. each row is a representation of the category for one of the variables and each column is for the other variable.
The summation of negative 5 n minus 1, from n equals 3 to 12 can be
expressed as
n = 3/12
substitute the value of n to the equatio
-5n – 1
-5( 3/12 ) -1
-9/4
There is an app that will help for these types of questions called, "Showmath."
16 is correct
and so is negitive 8
