Monday, Dec. 2 Chi-square Goodness of Fit Chi-square Test of Independence: Two Variables.

gg yy yg yyyggggy 25%25% 25% 25%

Pea Color freq Observed freq Expected Yellow Green TOTAL

Pea Color freq Observed freq Expected Yellow Green TOTAL 22 =  (f o - f e ) 2 fefe i=1 k Chi Square Goodness of Fit d.f. = k - 1, where k = number of categories of in the variable.

“… the general level of agreement between Mendel’s expectations and his reported results shows that it is closer than would be expected in the best of several thousand repetitions. The data have evidently been sophisticated systematically, and after examining various possibilities, I have no doubt that Mendel was deceived by a gardening assistant, who knew only too well what his principal expected from each trial made…” -- R. A. Fisher

Pea Color freq Observed freq Expected Yellow Green TOTAL 22 =  (f o - f e ) 2 fefe i=1 k Chi Square Goodness of Fit d.f. = k - 1, where k = number of categories of in the variable.

Peas to Kids: Another Example Goodness of Fit At my children’s school science fair last year, where participation was voluntary but strongly encouraged, I counted about 60 boys and 40 girls who had submitted entries. Since I expect a ratio of 50:50 if there were no gender preference for submission, is this observation deviant, beyond chance level?

BoysGirls Expected:5050 Observed:6040

BoysGirls Expected:5050 Observed:6040 22 =  (f o - f e ) 2 fefe i=1 k

BoysGirls Expected:5050 Observed:6040 22 =  (f o - f e ) 2 fefe i=1 k For each of k categories, square the difference between the observed and the expected frequency, divide by the expected frequency, and sum over all k categories.

BoysGirls Expected:5050 Observed:6040 22 =  (f o - f e ) 2 fefe i=1 k For each of k categories, square the difference between the observed and the expected frequency, divide by the expected frequency, and sum over all k categories. (60-50) 2 (40-50) = 4.00=

BoysGirls Expected:5050 Observed:6040 22 =  (f o - f e ) 2 fefe i=1 k For each of k categories, square the difference between the observed and the expected frequency, divide by the expected frequency, and sum over all k categories. (60-50) 2 (40-50) = 4.00= This value, chi-square, will be distributed with known probability values, where the degrees of freedom is a function of the number of categories (not n). In this one-variable case, d.f. = k - 1.

BoysGirls Expected:5050 Observed:6040 22 =  (f o - f e ) 2 fefe i=1 k For each of k categories, square the difference between the observed and the expected frequency, divide by the expected frequency, and sum over all k categories. (60-50) 2 (40-50) = 4.00= This value, chi-square, will be distributed with known probability values, where the degrees of freedom is a function of the number of categories (not n). In this one-variable case, d.f. = k - 1. Critical value of chi-square at  =.05, d.f.=1 is 3.84, so reject H 0.

Chi-square Test of Independence Are two nominal level variables related or independent from each other? Is race related to SES, or are they independent?

Lo Hi SES WhiteBlack

Row n x Column n Total n The expected frequency of any given cell is Lo Hi SES WhiteBlack

22 = (f o - f e ) 2 fefe  r=1 r  c=1 c At d.f. = (r - 1)(c - 1)

Row n x Column n Total n The expected frequency of any given cell is (15x28)/47(15x19)/47 (32x28)/47(32x19)/47

Row n x Column n Total n The expected frequency of any given cell is (15x28)/47(15x19)/47 (32x28)/47(32x19)/

22 = (f o - f e ) 2 fefe  r=1 r  c=1 c Please calculate:

Important assumptions: Independent observations. Observations are mutually exclusive. Expected frequencies should be reasonably large: d.f. 1, at least 5 d.f. 2, >2 d.f. >3, if all expected frequencies but one are greater than or equal to 5 and if the one that is not is at least equal to 1.