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Practical Statistics Chi-Square Statistics. There are six statistics that will answer 90% of all questions! 1. Descriptive 2. Chi-square 3. Z-tests 4.

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Presentation on theme: "Practical Statistics Chi-Square Statistics. There are six statistics that will answer 90% of all questions! 1. Descriptive 2. Chi-square 3. Z-tests 4."— Presentation transcript:

1 Practical Statistics Chi-Square Statistics

2 There are six statistics that will answer 90% of all questions! 1. Descriptive 2. Chi-square 3. Z-tests 4. Comparison of Means 5. Correlation 6. Regression

3 Chi-square: Chi-square is a simple test for counts….. Which means: nominal data and… if some cases… Ordinal data

4 Chi-square: There are three types: 1. Test for population variance 2. Test of “goodness-of-fit” 3. Contingency table analysis Which is essentially a measure of association!

5 Chi-square: There are three types: 1. Test for population variance

6 Chi-square: There are three types: 1. Test for population variance 2. Test of “goodness-of-fit” Where o = frequency of actual observation, and e = frequency you expected to find

7 Coin thrown 100 times: Expect (e): heads = 50, tails = 50 Observed (o): heads = 40, tails = 60 Is this a “fair” coin?

8 According to marketing research, the clientele of a Monkey Shine Restaurant is made up of 30% Western businessmen, 30% women who stop in while shopping, 30% Chinese businessmen, and 10% tourists. A random sample of 600 customers at the Kowloon Monkey Shine found 150 Western businessmen, 190 Chinese businessmen, 100 tourists, and 65 women who were shopping. Is the clientele at this establishment different than the norm for this company?

9 TypePercentExpected 600 Observed 600 Western Business 30% 180 150 Chinese Business 30% 180 190 Women Shoppers 30% 180 160 Tourist10% 60 100

10 = 5.00 + 0.56 + 2.22 + 26.67 = 34.45 With (4-1) degrees of freedom

11 The chi-square distribution is highly skewed and dependent upon how many degrees of freedom (df) a problems has.

12 The chi-square for the restaurant problem was: Chi-square = 34.45, df = 3 By looking in a table, the critical value of Chi-square with df = 3 is 7.82. The probability that the researched frequency equals the frequency found in the MR project was p <.001. http://www.fourmilab.ch/rpkp/experiments/analysis/chiCalc.html

13 = 5.00 + 0.56 + 2.22 + 26.67 = 34.45 df = 3 By looking at the analysis, it is obvious that the largest contribution to chi-square came from the tourists. Hence, the Kowloon property is attracting more tourist than what would be expected at the Monkey Shine.

14 Chi-square: There are three types: 1. Test for population variance 2. Test of “goodness-of-fit” 3. Contingency table analysis Where o = frequency of actual observation, and e = frequency you expected to find

15 A contingency table is a table with numbers grouped by frequency.

16 A contingency table is a table with numbers grouped by frequency. Consider a study: There are three groups: brand loyal customers, regular buyers, and occasional buyers. Each is asked if they like the taste of new product over the old. They answer with a “yes” or a “no.”

17 A contingency table would look like this: YES NO Totals Loyal 50 40 90 Regular 60 40 100 Occasional 40 80 Total 150 120 270

18 A contingency table is a table with numbers grouped by frequency. All the numbers in the table are “observed” frequencies (o). So, what are the expected values?

19 YES NO Totals Loyal 50 40 90 Regular 60 40 100 Occasional 40 80 Total 150 120 270 The expected values (e) would be a random distribution of frequencies.

20 YES NO Totals Loyal 50 40 90 Regular 60 40 100 Occasional 40 80 Total 150 120 270 The expected values (e) would be a random distribution of frequencies. These can be calculated by multiplying the row frequency by the column frequency and dividing by the total number of observations.

21 YES NO Totals Loyal 50 40 90 Regular 60 40 100 Occasional 40 80 Total 150 120 270 For example, the expected values (e) of “loyal” and “yes” would be (150 X 90)/270 = 50

22 YES NO Totals Loyal 50 40 90 Regular 60 40 100 Occasional 40 80 Total 150 120 270 For example, the expected values (e) of “regular” And “no” would be (120 X 100)/270 = 44.4

23 The expected values (e) for the entire table would be: YES NO Totals Loyal 50.0 40.0 90 Regular 55.6 44.4 100 Occasional 44.4 35.6 80 Total 150 120 270

24 The chi-square value is calculated for every cell, and then summed over all the cells. YES NO Totals Loyal 50.0 40.0 90 Regular 55.6 44.4 100 Occasional 44.4 35.6 80 Total 150 120 270

25 The chi-square value is calculated for every cell: For Cell A: (50-50)^2/50 = 0 For Cell D: (40-44.4)^2/44.4 = 0.44 YES NO Totals Loyal A 50.0 40.0 90 Regular 55.6 D 44.4 100 Occasional 44.4 35.6 80 Total 150 120 270

26 The chi-square value is calculated for every cell: YES NO Totals Loyal 0 0 Regular.36.44 Occasional.44.55 Total

27 The chi-square value is calculated for every cell: Chi-square = 0 + 0 +.35 +.44 +.44 +.54 = 1.77 The df = (r-1)(c-1) = 1 X 2 = 2 YES NO Totals Loyal 0 0 Regular.35.44 Occasional.44.54 Total

28 A chi-square with a df = 2 has a critical value of 5.99, this chi-square = 1.77, so the results are nonsignificant. The probability = 0.4127. This means that the distribution is random, and there is no association between customer type and taste preference. http://www.fourmilab.ch/rpkp/experiments/analysis/chiCalc.html

29 A chi-square with a df = 2 has a critical value of 5.99, this chi-square = 1.77, so the results are nonsignificant. This means that the distribution is random, and there is no association between customer type and taste preference. Note : This type of chi-square is a test of association using nothing but counts (frequency); VERY useful in business research.

30 Service Encounter and Personality Normally, 60% of our shoppers are women. Is our sample correct? 0.6 X 271 = 163 women.4 X 271 = 109 men

31 Service Encounter and Personality Do men and women shop at different times?

32 Service Encounter and Personality Do men and women shop at different times?


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