Lecture 43 Sections 14.4 – 14.5 Mon, Nov 26, 2007 Test of Homogeneity Lecture 43 Sections 14.4 – 14.5 Mon, Nov 26, 2007

Homogeneous Populations Two distributions are called homogeneous if they exhibit the same proportions within the same categories.

Examples If two colleges’ student bodies are each 55% female and 45% male, then the distributions are homogeneous. If a teacher teaches two sections of Statistics using two different teaching methods and the two sections have the same grade distribution, then they are homogeneous.

Example Suppose a teacher teaches two sections of Statistics and uses two different teaching methods. At the end of the semester, he gives both sections the same final exam and he compares the grade distributions. He wants to know if the differences that he observes are significant.

Example Does there appear to be a difference? Or are the two populations homogeneous? A B C D F Method I 5 7 36 17 Method II 11 18

The Test of Homogeneity The null hypothesis is that the populations are homogeneous. The alternative hypothesis is that the populations are not homogeneous. H0: The populations are homogeneous. H1: The populations are not homogeneous.

Independence Two variables are independent if the value of one has no bearing on the value of the other.

Mendel’s Experiments In Mendel’s experiments, Mendel observed 75% yellow seeds, 25% green seeds. 75% smooth seeds, 25% wrinkled seeds. Because color and texture were independent, he also observed 9/16 yellow and smooth 3/16 yellow and wrinkled 3/16 green and smooth 1/16 green and wrinkled

Mendel’s Experiments That is, he observed the same ratios within categories that he observed for the totals. Smooth Wrinkled Yellow 9 3 Green 1

Mendel’s Experiments That is, he observed the same ratios within categories that he observed for the totals. Smooth Wrinkled Yellow 9 3 Green 1 3 : 1 Ratio

Mendel’s Experiments That is, he observed the same ratios within categories that he observed for the totals. Smooth Wrinkled Yellow 9 3 Green 1 3 : 1 Ratio

Mendel’s Experiments That is, he observed the same ratios within categories that he observed for the totals. Smooth Wrinkled Yellow 9 3 Green 1 3 : 1 Ratio

Mendel’s Experiments That is, he observed the same ratios within categories that he observed for the totals. Smooth Wrinkled Yellow 9 3 Green 1 3 : 1 Ratio

Mendel’s Experiments Had the traits not been independent, he might have observed something different. Smooth Wrinkled Yellow 11 1 Green 3

Example Suppose a university researcher suspects that a student’s SAT-M score is related to his performance in Statistics. At the end of the semester, he compares each student’s grade to his SAT-M score for all Statistics classes at that university. He wants to know whether the student’s with the higher SAT-M scores got the higher grades.

Example Does there appear to be a difference between the rows? Or are the rows independent? Grade A B C D F 400 – 500 7 8 16 20 21 500 – 600 13 28 32 22 600 – 700 23 10 9 700 – 800 14 5 SAT-M

The Test of Independence The null hypothesis is that the variables are independent. The alternative hypothesis is that the variables are not independent. H0: The variables are independent. H1: The variables are not independent.

An Example of Homogeneity We will work the homogeneity example of the two teaching methods. A B C D F Method I 5 7 36 17 Method II 11 18

The Test Statistic The test statistic is the chi-square statistic, computed as The question now is, how do we compute the expected counts?

Expected Counts Under the assumption of homogeneity or independence (H0), the rows should exhibit the same proportions. We can get the best estimate of those proportions by pooling the rows. That is, find the column totals then compute the column proportions from them.

Row and Column Proportions B C D F Method I 5 7 36 17 Method II 11 18

Row and Column Proportions B C D F Method I 5 7 36 17 Method II 11 18 Col Total 12 54 24

Row and Column Proportions B C D F Method I 5 7 36 17 Method II 11 18 Col Total 12 54 24 10% 15% 45% 20%

Expected Counts Similarly, the columns should exhibit the same proportions, so we can get the best estimate by pooling the columns. That is, find the row totals and then compute the row proportions from them.

Row and Column Proportions B C D F Method I 5 7 36 17 Method II 11 18 Col Total 12 54 24 10% 15% 45% 20%

Row and Column Proportions B C D F Row Total Method I 5 7 36 17 72 Method II 11 18 48 Col Total 12 54 24 10% 15% 45% 20%

Row and Column Proportions B C D F Row Total Method I 5 7 36 17 72 60% Method II 11 18 48 40% Col Total 12 54 24 10% 15% 45% 20%

Row and Column Proportions B C D F Row Total Method I 5 7 36 17 72 60% Method II 11 18 48 40% Col Total 12 54 24 120 10% 15% 45% 20% Grand Total

Row and Column Totals A B C D F Total Method I 5 7 36 17 Method II 11 18 12 54 24 120

Expected Counts Now apply the appropriate row and column proportions to each cell to get the expected count. In the upper-left cell, according to the row and column proportions, it should contain 60% of 10% of 120.

Expected Counts That is, the expected count is 0.60 0.10  120 = 7.2 This is

Expected Counts Therefore, the quick formula is

Expected Counts Apply that formula to each cell to find the expected counts and add them to the table. A B C D F Method I 5 (7.2) 7 (10.8) 36 (32.4) 17 (14.4) Method II (4.8) 11 18 (21.6) (9.6)

The Test Statistic Now compute 2 in the usual way.

df = (no. of rows – 1)  (no. of cols – 1). Degrees of Freedom The number of degrees of freedom is df = (no. of rows – 1)  (no. of cols – 1). In our example, df = (2 – 1)  (5 – 1) = 4. To find the p-value, calculate 2cdf(7.2106, E99, 4) = 0.1252. At the 5% level of significance, the differences are not statistically significant.

TI-83 – Test of Homogeneity or Independence The tables in these examples are not lists, so we can’t use the lists in the TI-83. Instead, the tables are matrices. The TI-83 can handle matrices.

TI-83 – Test of Homogeneity or Independence Enter the observed counts into a matrix. Press MATRIX. Select EDIT. Use the arrow keys to select the matrix to edit, say [A]. Press ENTER to edit that matrix. Enter the number of rows and columns. (Press ENTER to advance.) Enter the observed counts in the cells. Press 2nd Quit to exit the matrix editor.

TI-83 – Test of Homogeneity or Independence Perform the test of homogeneity. Select STATS > TESTS > 2-Test… Press ENTER. Enter the name of the matrix of observed counts. Enter the name (e.g., [E]) of a matrix for the expected counts. These will be computed for you by the TI-83. Select Calculate.

TI-83 – Test of Homogeneity or Independence The window displays The title “2-Test”. The value of 2. The p-value. The number of degrees of freedom. See the matrix of expected counts. Press MATRIX. Select matrix [E]. Press ENTER.

TI-83 – Example Work the previous example, using the TI-83.

Example Is the Math SAT score independent from the grade in Statistics? Grade A B C D F 400 - 500 7 8 16 20 21 500 – 600 13 28 32 22 600 – 700 23 10 9 700 - 800 14 5 SAT-M

Expected Counts A B C D F 400 - 500 7 (8.64) 8 (17.28) 16 (20.16) 20 (14.40) 21 (11.52) 500 – 600 13 (12.96) 28 (25.92) 32 (30.24) 22 (21.60) 600 – 700 23 10 9 700 - 800 (5.76) 14 (13.44) (9.60) 5 (7.68)

The Test Statistic The value of 2 is 23.7603.

df = (no. of rows – 1)  (no. of cols – 1). Degrees of Freedom The degrees of freedom are the same as before df = (no. of rows – 1)  (no. of cols – 1). In our example, df = (4 – 1)  (5 – 1) = 12.

The p-value To find the p-value, calculate 2cdf(23.7603, E99, 12) = 0.0219. The results are significant at the 5% level.

TI-83 – Test of Independence The test for independence on the TI-83 is identical to the test for homogeneity.