Partial preview of the text
Download Stats notes chap 4 so and more Quizzes Statistics in PDF only on Docsity!
> ig-to eorge 16 ght ship am Pitt e ninister lin was slumnist Uframer ion 1. Note that row and column headings do not count in dd columns. A contingency table is designated as an R X C (rd R= 2and ; hence, this table is a 2X 3 canting is called a cell and is designated by its row and column a frequency of 80 is designated as C, 2, or row 1, colt Column 1 Column 2 The degrees of freedom for any contingency table are (rows — 1) times (columns — 1); that is, df. = (R — 1)(C — 1). In this case, (2 — 1) — 1) = ()Q) = 2. The reason for this formula for d.f. is that all the expected values except one are free to vary in each row and in each column. Using the previous table, you can compute the expected frequencies for each block (or cell), as shown next. 1, Find the sum of cach row and cach column, and find the grand total, as shown. Prefer new Prefer old No Group procedure procedure preference Total Row I sum ~ Nurses 100 80 20 200 Row 2 sum Doctors +50 +120 =30 200 Total 150 200 30 400 Column 1 sum Column 2 sum Column 3 sum Grand total nv . For each cell, multiply the corresponding row sum by the column sum and divide by the grand total, to get the expected value: row sum X column sum Expected value = “2 Sm ~ column sum grand total For example, for C,,, the expected value, denoted by £1, is (refer to the previous tables) _ (200/200) _ =—00 7 100 For each cell, the expected values are computed as follows: _ (200150) _ _ (200)(200) _ _ (20050) _ Ey = agg 2 75 Bia = agp = 00 Eis = Gog = 25 _ (200)(150) _ (200)(200) _ _ (200150) _ 21 = —a0g 75 Exa= Gog — = 100 Ey = yg = 25 The expected values can now be placed in the corresponding cells along with the observed values, as shown. Prefernew Prefer old Group procedure Procedure Nopreference _Total Nurses 100 (75) 80 (100) 20 (25) 200 Doctors 50 (75) 120 (100) 30 (25) 200 Total 150, 200 50 400 1-7 11 Other Chi-Square Tests The rationale for the computation of the expected frequencies for a contingency table uses proportions. For C,,, a total of 150 out af 400 people prefer the new procedure. And since there are 200 nurses, you would expect, if the null hypothesis were true, (150/400)(200), or 75, of the nurses to be in favor of the new procedure. The formula for the test value for the independence test is the same as the one used for the goodness-of-fit test. It is (0 — EP a 
