Question

4. When calculating a one-sided p-value for Fisher's exact test, the focus is on the upper left-hand cell of the table, typically denoted as nui but is denoted as k in the table below. Explanatory Response Yes No Level 1 k Ri- k R Level 2 C-k R-C1 + R2 CCT To find the p-value, for each value of k (k = 111,111 +1,...,min(Ri,C) or k = 0,1,...,111 for k >nul and k < 111, respectively), the proportion of possible randomizations leading to that value is R!R2!C!C! T!k!(R -k)!(C -k)!(R2-C1 + k)! Using the hypergeometric probability distribution, verify this equation. Hint: The hypergeometric proba- bility distribution is typically written in the following way: N N2 In-r where N is the total number in the first group, N2 is the total number in the second group, N = Ni + N2, is the number of successes in the first group, and ner is the number of successes in the second group. (20 points)

Answer 1

lutio GPven dato denoled yrtally Yes NO R-k k R Ci-k Ra & k (KEnn, Value mh (e,.ca ko,1.. k2n and ken i 5 a TIk (R):(G): (-4x)! hypergeomaleobabdstehuon total nunaber enenst Pa second P& total humber 1

Pg.No'ra P numbes &Succeses Pn st Succene n Seeon d umber n-x ohese, k nx R k N = T Sub () ( R.-K GD CI! C k (ci-k) - R,+t) R. joste shere, Re -Ck CRik

No3 Ci! RI!R2 Ci! k (Crk)B) (&-) Cal R,IR