ChiSquare Test for Qualitative Data


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1 ChiSquare Test for Qualitative Data For qualitative data (measured on a nominal scale) * Observations MUST be independent  No more than one measurement per subject * Sample size must be large enough  Expected frequencies must be 5
2 Chisquare distribution Critical Values Table on page 537 in your book!
3 X rollercoaster right here in California
4 Goodness of Fit χ 1 variable H 0 : observed & expected frequencies do not differ Steps: Calculate expected frequencies Compute χ Compare to critical value df # categories  1 (fofe) fe Observed frequency Expected frequency
5 Example: Goodness of Fit χ Married Single Separated Divorced Widowed Total Sample (N 100) f o expected freq. f e % Is the marital status of our sample representative of the population? Statistical Hypotheses: H 0 f o s (observed frequencies) conform to f e s (expected) H 1 the sample differs from the expected frequencies Decision rule: α.05; df 51 4; critical χ 9.49 Calculate test statistic: (*expected frequencies should not below 5 in any cell!) ( fo fe) χ fe (50 55) ( 1) (8 9) (18 10) ( 5) χ χ
6 Getting the Critical Value
7 Example: Goodness of Fit χ Observed statistical test value: χ (4) 8.81, p >.05 Make a decision & interpret  Retain H 0 because 8.81 < The sample does not significantly differ from the population, with regard to marital status
8 Another Example Rated G Rated PG13 Rated NC17 Sample (N 4) f o expected freq. f e Is there an association between sexy advertising and buying more products? Statistical Hypotheses: H 0 there is no association between sexy advertising and purchases; H 1 there is an association between advertising and purchases Decision rule: α.05; df 31 ; critical χ 5.99 Calculate statistic: (remember: expected frequencies should not below 5 in any cell!) ( fo fe) χ fe (5 8) (5 8) (14 8) χ χ
9 Another Example Observed statistical test value: χ () 6.75, p <.05 Make a decision & interpret Reject H 0 because 6.75 > 5.99 Sex sells!
10 Practice! Goodness of Fit χ Lets say you roll a 6sided dice 10 times. You would EXPECT that each side would come up 1/6 of the time (i.e., 0 times) f o Now your friend gets his own 6sided dice and rolls it 10 times. You would have the same EXPECTED frequency here, right? f o Calculate a goodness of fit χ for both you and your friend, and determine whether one of you has a weighted dice, at α.05. Don t forget to calculate df to get the critical χ value! Is one of the dice suspect?
11 Your 10 Rolls Dice Obs. Exp. OE (OE) (O E) E
12 Friend s 10 Rolls Dice Obs. Exp. OE (OE) (O E) E
13 df & critical value l df #categories 1 5 l Critical χ 11.07
14 Practice: Goodness of Fit χ You: χ (OE) E 1.4 NOT SIGNIFICANT Friend: χ (OE) E 85 SIGNIFICANT Is your friend using a weighted dice?
15 χ Test for Independence Tests the association between categorical variables Do the frequencies you actually observe differ from the expected frequencies by more than chance alone? Statistical hypotheses: Steps: H 0 : the variables are independent (i.e. no association) H 1 : the variables are not independent Calculate expected frequency of each cell Compute χ Compare to critical value df (# rows 1) x (# columns 1) Observed frequency (fofe) fe Expected frequency
16 Example: χ Test for Independence Is there an association between gender and vegetarianism? Statistical Hypotheses: Vegetarian NonVegetarian Total: Male Female Total: H 0 : gender and food preference are independent H 1 : gender and food preference are associated/ not independent Decision rule: α.05 df (# rows 1) x (# columns 1) à (1) x (1) 1 Critical χ 3.841
17 Next step: calculate the expected frequency of each cell Vegetarian NonVegetarian Total: Male fe 70 x Female fe 130 x x Total: fe fe 130 x expected frequency of each cell row total x column total grand total
18 Now put it into the table Sample (N 00) Male Veg Male NonVeg Female Veg Female NonVeg f o expected freq. f e χ χ ( fo f fe (10 1) 1 e) (60 49) + 49 (50 39) + 39 (80 91) + 91 χ
19 Example: χ Test for Independence Observed statistical test value: χ (1) 1.66, p <.05 Make a decision & interpret Reject H 0 and accept H 1 because 1.66 > 3.84 Gender is related to food preference!
20 Practice! Is there an association between cat ownership (yes/no) and life success (yes/no)? You survey 100 people Successful Not Successful Total: Cat No Cat Total: 100 Don t forget to get your row and column totals And follow the steps of hypothesis testing: Statistical Hypothesis Decision Rule Calculate Test Statistic Make a Decision & Interpret
21 Successful Not Successful Total: Cat No Cat Total: Statistical Hypotheses: H 0 : cat ownership and life success are independent H 1 : cat ownership and life success are related Decision rule: α.05 df (# rows 1) x (# columns 1) à (1) x (1) 1 Critical χ 3.841
22 Successful Not Successful Total: Cat fe 75 x No Cat fe 5 x x Total: fe fe 5 x Sample (N 100) Cat, Success No cat, Success Cat, No success f o expected freq. f e No cat, No Success
23 Sample (N 100) Cat, Success No cat, Success Cat, No success f o expected freq. f e No cat, No Success χ ( fo fe) χ fe ( ) ( ) ( ) (10 6.5) χ
24 Observed statistical test value: χ (1) 4.00, p <.05 Make a decision & interpret Reject H 0 because 4.00 > 3.84 Cat ownership is related to life success!
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