Decide which chi-square test (goodness-of fit, homogeneity, or independence) would be most appropriate for the given situation. A car insurance company performed a study to determine whether an association exists between age and the frequency of car accidents. They obtained the following sample data.
Age Group
Under 25 25-45 Over 45 Total
Number of accidents in past 3 years 0 74 90 84 248
1 19 8 12 39
1 7 2 4 13
Total 100 100 100 300
A. Test for Homogeneity.
B. Test for Independence.
C. Test for Goodness-of-fit.

C. Test for Goodness-of-fit.

Step-by-step explanation:

C. Test for Goodness-of-fit would be most appropriate for the given situation.

A. Test Of  Homogeneity.

The value of q is large when the sample variances differ greatly and is zero when all variances are zero . Sample variances do not differ greatly in the given question.

B. Test for Independence.

The chi square  is used to test the hypothesis about the independence of two variables each of which is classified into number of attributes. They are not classified into attributes.

C. Test for Goodness-of-fit.

The chi square test is applicable when the cell probabilities depend upon unknown parameters provided that the unknown parameters are replaced with  their estimates and provided that one degree of freedom is deducted for each parameter estimated.

i = 2 / - 5

step-by-step explanation:

evaluation of ( 2 - 4 ) ( p + q ) when p = 2 and q = 5 i, by putting values of p and q,                                                                    

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !