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Chapter 14:

Chapter 14 Chi Square - 2

Chi Square:

Chi Square Chi Square is a non-parametric statistic used to test the null hypothesis. It is used for nominal data. It is equivalent to the F test that we used for single factor and factorial analysis.

… Chi Square:

… Chi Square Nominal data puts each participant in a category. Categories are best when mutually exclusive and exhaustive. This means that each and every participant fits in one and only one category. Chi Square looks at frequencies in mutually exclusive and exhaustive categories into which participants are assigned after a single measurement.

Expected frequencies and the null hypothesis ...:

Expected frequencies and the null hypothesis ... Chi Square compares the expected frequencies in categories to the observed frequencies in categories. “Expected frequencies”are the frequencies in each cell predicted by the null hypothesis

… Expected frequencies and the null hypothesis ...:

… Expected frequencies and the null hypothesis ... The null hypothesis: H0: fo = fe There is no difference between the observed frequency and the frequency predicted (expected) by the null. The experimental hypothesis: H1: fo  fe The observed frequency differs significantly from the frequency predicted (expected) by the null.

Calculating 2:

Calculating 2 Calculate the deviations of the observed from the expected. For each cell: Square the deviations. Divide the squared deviations by the expected value.

Calculating 2:

Calculating 2 Add ‘em up. Then, look up 2 in Chi Square Table df = k - 1 (one sample 2) OR df= (Columns-1) * (Rows-1) (2 or more samples)

Slide8:

df 1 2 3 4 5 6 7 8 .05 3.84 5.99 5.82 9.49 11.07 12.59 14.07 15.51 .01 6.63 9.21 11.34 13.28 15.09 16.81 18.48 20.09 df 9 10 11 12 13 14 15 16 .05 16.92 18.31 19.68 21.03 22.36 23.68 25.00 26.30 .01 21.67 23.21 24.72 26.22 27.69 29.14 30.58 32.00 df 17 18 19 20 21 22 23 24 .05 27.59 28.87 30.14 31.41 32.67 33.92 35.17 36.42 .01 33.41 34.81 36.19 37.57 38.93 40.29 41.64 42.98 df 25 26 27 28 29 30 .05 37.65 38.89 40.14 41.34 42.56 43.77 .01 44.31 45.64 46.96 48.28 49.59 50.89 Critical values of 2

Slide9:

Slide10:

Slide11:

Example:

Example If there were 5 degrees of freedom, how big would 2 have to be for significance at the .05 level?

Slide13:

Using the 2 table.:

Using the 2 table. If there were 2 degrees of freedom, how big would 2 have to be for significance at the .05 level? Note: Unlike most other tables you have seen, the critical values for Chi Square get larger as df increase. This is because you are summing over more cells, each of which usually contributes to the total observed value of chi square.

Slide15:

One sample example: Party: 75% male, 25% femaleThere are 40 swimmers. Since 75% of people at party are male, 75% of swimmers should be male. So expected value for males is .750 X 40 = 30. For women it is .250 x 40 = 10.00:

One sample example: Party: 75% male, 25% female There are 40 swimmers. Since 75% of people at party are male, 75% of swimmers should be male. So expected value for males is .750 X 40 = 30. For women it is .250 x 40 = 10.00 Male Female Observed 20 20 Expected 30 10 O-E -10 10 (O-E)2 100 100 (O-E)2/E 3.33 10 df = k-1 = 2-1 = 1

Slide17:

2 sample example:

2 sample example Freshman and sophomores who like horror movies. Likes horror films Dislikes horror films 150 200 100 50

Slide19:

There are 500 altogether. 200 (or a proportion of .400 like horror movies, 300 (.600) dislike horror films. (Proportions appear in parentheses in the margins.) Multiplying by the proportion in the “likes horror films” row by the number in the “Freshman” column yield the following expected frequency for the first cell. The formula is: Expected Frequency = (Proprowncol). (EF appears in parentheses in each cell.) Likes horror films Dislikes horror films 150 200 (150) 100 (150) 50 (100) 200 (.400) 300 (.600) 250 250 500 (100)

Computing 2 :

Computing 2 Fresh Likes Fresh Dislikes Soph Likes Soph Dislikes Observed 150 100 50 200 Expected 100 150 100 150 df = (C-1)(R-1) = (2-1)(2-1) = 1

Slide21:

The only (slightly)hard part is computing expected frequencies:

The only (slightly)hard part is computing expected frequencies In one sample case, multiply n by a hypothetical proportion based on the null hypothesis that frequencies will be random.

Simple Example - 100 teenagers listen to radio stations:

Simple Example - 100 teenagers listen to radio stations H1: Some stations are more popular with teenagers than others. H0: Radio station do not differ in popularity with teenagers. Expected frequencies are the frequencies predicted by the null hypothesis. In this case, the problem is simple because the null predicts an equal proportion of teenagers will prefer each of the four radio stations. Is the observed significantly different from the expected?

Slide24:

Observed Expected df = k-1 = (4-1) = 3 2(3, n=100) = 20.00, p<.01 Station 1 Station 2 Station 3 Station 4 40 30 20 10 25 25 25 25 15 5 -5 15 225 25 25 225 9.00 1.00 1.00 9.00 Differential popularity of Radio station among teenagers

The only (slightly)hard part is computing expected frequencies:

The only (slightly)hard part is computing expected frequencies In the multi-sample case, multiply proportion in row by numbers in each column to obtain EF in each cell.

A 3 x 4 Chi Square:

A 3 x 4 Chi Square Women, stress, and seating preferences. (and perimeter vs. interior, front vs. back Very Stressed Females Moderately Stressed Females Control Group Females Front Front Back Back Perim Inter Perim Inter 10 60 15 35 30 50 70 5 10 15 15 25 20 300 60 30 150 100 100 100

Expected frequencies:

Expected frequencies Women, stress, and perimeter versus interior seating preferences. Very Stressed Females Moderately Stressed Females Control Group Females 10 60 15 35 30 50 70 5 10 15 15 25 20 300 60 30 150 100 100 100 (20) (20) (20) Front Front Back Back Perim Inter Perim Inter

Column 2:

Column 2 Women, stress, and perimeter versus interior seating preferences. Very Stressed Females Moderately Stressed Females Control Group Females 10 60 15 35 30 50 70 5 10 15 15 25 20 300 60 30 150 100 100 100 (20) (20) (20) (50) (50) (50) Front Front Back Back Perim Inter Perim Inter

Column 3:

Column 3 Women, stress, and perimeter versus interior seating preferences. Very Stressed Females Moderately Stressed Females Control Group Females 10 60 15 35 30 50 70 5 10 15 15 25 20 300 60 30 150 100 100 100 (20) (20) (20) (50) (50) (50) (10) (10) (10) Front Front Back Back Perim Inter Perim Inter

All the expected frequencies:

All the expected frequencies Women, stress, and perimeter versus interior seating preferences. Very Stressed Females Moderately Stressed Females Control Group Females 10 60 15 35 30 50 70 5 10 15 15 25 20 300 60 30 150 100 100 100 (20) (20) (20) (50) (50) (50) (10) (10) (10) (20) (20) (20) Front Front Back Back Perim Inter Perim Inter

Slide31:

FrontP FrontI BackP BackI Observed 10 70 5 15 Expected 20 50 10 20 df = (C-1)(R-1) = (4-1)(3-1) = 6 Very Stressed FrontP FrontI BackP BackI 15 50 10 25 20 50 10 20 -5 0 0 5 25 0 0 25 1.25 0.00 0.00 1.25 Moderately Stressed FrontP FrontI BackP BackI 35 30 15 20 20 50 10 20 15 -20 5 0 225 400 25 0 11.25 8.00 2.50 0.00 Control Group

Slide32:

Slide33:

FrontP FrontI BackP BackI Observed 10 70 5 15 Expected 20 50 10 20 O-E -10 20 -5 -5 (O-E)2 100 400 25 25 (O-E)2/E 5.00 8.00 2.50 1.25 2 = 41.00 df = (C-1)(R-1) = (4-1)(3-1) = 6 Very Stressed FrontP FrontI BackP BackI 15 50 10 25 20 50 10 20 -5 0 0 5 25 0 0 25 1.25 0.00 0.00 1.25 Moderately Stressed FrontP FrontI BackP BackI 35 30 15 20 20 50 10 20 15 -20 5 0 225 400 25 0 11.25 8.00 2.50 0.00 Control Group Very stressed women avoid the perimeter and prefer the front interior. The control group prefers the perimeter and avoids the front interior.

Summary: Different Ways of Computing the Frequencies Predicted by the Null Hypothesis:

Summary: Different Ways of Computing the Frequencies Predicted by the Null Hypothesis One sample Expect subjects to be distributed equally in each cell. OR Expect subjects to be distributed proportionally in each cell. OR Expect subjects to be distributed in each cell based on prior knowledge, such as, previous research. Multi-sample Expect subjects in different conditions to be distributed similarly to each other. Find the proportion in each row and multiply by the number in each column to do so.

Conclusion - Chi Square:

Conclusion - Chi Square Chi Square is a non-parametric statistic,used for nominal data. It is equivalent to the F test that we used for single factor and factorial analysis. Chi Square compares the expected frequencies in categories to the observed frequencies in categories.

… Conclusion - Chi Square:

… Conclusion - Chi Square The null hypothesis: H0: fo = fe There is no difference between the observed frequency and frequency predicted by the null hypothesis. The experimental hypothesis: H1: fo  fe The observed frequency differs significantly from the frequency expected by the null hypothesis.

Ch14slideswebsitefin al7222006

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Chapter 14:

Chi Square:

… Chi Square:

Expected frequencies and the null hypothesis ...:

… Expected frequencies and the null hypothesis ...:

Calculating 2:

Calculating 2:

Slide8:

Slide9:

Slide10:

Slide11:

Example:

Slide13:

Using the 2 table.:

Slide15:

One sample example: Party: 75% male, 25% female There are 40 swimmers. Since 75% of people at party are male, 75% of swimmers should be male. So expected value for males is .750 X 40 = 30. For women it is .250 x 40 = 10.00:

Slide17:

2 sample example:

Slide19:

Computing 2 :

Slide21:

The only (slightly)hard part is computing expected frequencies:

Simple Example - 100 teenagers listen to radio stations:

Slide24:

The only (slightly)hard part is computing expected frequencies:

A 3 x 4 Chi Square:

Expected frequencies:

Column 2:

Column 3:

All the expected frequencies:

Slide31:

Slide32:

Slide33:

Summary: Different Ways of Computing the Frequencies Predicted by the Null Hypothesis:

Conclusion - Chi Square:

… Conclusion - Chi Square:

The end. Hope you found the slides helpful! RK:

One sample example: Party: 75% male, 25% femaleThere are 40 swimmers. Since 75% of people at party are male, 75% of swimmers should be male. So expected value for males is .750 X 40 = 30. For women it is .250 x 40 = 10.00:

The end. Hope you found the slides helpful!RK: