Chi-Square and Nonparametric Techniques Q560: Experimental Methods in Cognitive Science Lecture 14

The Tolman/Hull Debate (1946) Can animals create “cognitive maps” or are they purely stimulus-response learners? This was a huge debate in the 40s between Clark Hull (S-R) and Edward Tolman (cognition). In a classic experiment, Tolman trained rats to run down an alley into a circular arena. From there, they were trained to go straight across to another alley which turned right to a goal box After training, he put them in a new maze that also contained alleys at other angles, one heading directly towards the goal box. The rats had a choice to take the alley they had learned, or a novel one that pointed in the direction of the goal

If Hull was right, the rat learned an S-R sequence during learning, and would continue this response If Tolman was right, the rat learned a cognitive map “goal on the right” and would enter the alley on the right Goal B A C D Start

If Hull was right, the rat learned an S-R sequence during learning, and would continue this response If Tolman was right, the rat learned a cognitive map “goal on the right” and would enter the alley on the right Goal Alley Chosen B A C A B C D Observed Expected 4 8 5 15 D Who was right? Start

Chi-Square Test Our first problem is that each rat has only provided us with an alley preference…this is a nomial scale (frequency) not an interval or ratio scale, so we cannot compute means or variances So, we use a Chi-Square statistic: comparing observed frequencies to those expected by chance Why divide by expected freq? Relative discrepancy from expected: Party example

Expected Frequencies A B C D 25% A B C D 8 If the null hypothesis is true, alley choices should be random…no preference. So with 4 options: p(option) = .25 A B C D 25% For each option, , where p = proportion if null is true, and n is sample size (n=32 rats) A B C D 8 This is our chance model

Chi-Square Test As with our other statistical tests, we compare to a chance distribution for Chi-square. Like the F-distribution, we actually have a family of distributions depending on df (Table on p. 697) df = C -1 (where C = # of columns)...why?

The shape can change quite a bit depending on df

For Tolman’s rats: df = C - 1 = 4-1 = 3 For df = 3, X2-crit = 7.81 Note: df are based on the number of “options” rather than sample size (as with our previous tests). Hence, as the number of options increases, the chance of a difference due purely to chance also increases. So, the critical value for Chi-square actually increases with larger df rather than decreasing (as with t and F) Alley Chosen A B C D Observed Expected 4 8 5 15

Alley Chosen X2-crit = 7.81 X2-obt = 9.25 A B C D 4 8 5 15 Observed Expected 4 8 5 15 X2-crit = 7.81 X2-obt = 9.25

Steps in Hypothesis Testing State hypotheses (No preference/preference) Determine critical values (chance model) Calculate Chi-sqaure statistic Decision and conclusions

An Example Coke Pepsi RC Cola 13 9 8 We’ll settle the age-old debate of whether people can actually detect their favorite cola based solely on taste. For 30 coke-lovers, I blindfold them, and have them sample 3 colas…is there a true difference, or are these preference differences explainable by chance? Coke Pepsi RC Cola 13 9 8

Step 1: Hypotheses Null: There are no preferences: The population is divided evenly among the brands Alternate: There are preferences: The population is not divided evenly among the brands Note: This is non-parametric, so we don’t need to talk about parameters (our “code symbols”)

Step 2: Chance Model Coke Pepsi RC Cola Observed 13 9 8 Expected 10 df = C -1 = 3 -1 = 2, set  = .05 For df = 2, X2-crit = 5.99 Next, we determine the frequencies we would expect if the null hypothesis were true ....is this 1- or 2-tailed? Coke Pepsi RC Cola Observed 13 9 8 Expected 10

Step 3: Calculate Chi-Square Coke Pepsi RC Cola Observed 13 9 8 Expected 10

Step 4: Decision and Conclusion Conclude that the preferences are evenly divided among the colas when the logos are removed.

An SPSS example w/ RPS

The Chi-Square Test of Independence This is the nominal equivalent of a factorial ANOVA: we have two variables, and we want to test whether there is a relationship between them. We present the data in a matrix. If there is no relationship, then the observed frequencies should not differ as a function of the second variable

Littering 17 28 49 Not Littering 73 62 41 Cialdini et al. (1990) examined whether a person’s tendency to litter depended on the amount of litter already in the area. People were handed a flyer as they entered an amusement park. The entrance area had already been prepared with either no litter, a small amount of litter, or a lot of litter lying on the ground. The frequency data are: Amount of Litter Small Amount Large Amount None Littering 17 28 49 Not Littering 73 62 41 Is there a relationship between tendency to litter and litter already in the area?

Littering 17 28 49 Not Littering 73 62 41 94 176 90 90 90 n = 270 Amount of Litter Small Amount Large Amount None Littering 17 28 49 Not Littering 73 62 41 94 176 90 90 90 n = 270 Where: df = (R-1)(C-1) = (3-1)(2-1) = 2

Litt (Obs): Expected: 17 31.33 28 49 No Litt (Obs): 73 58.67 62 41 94 Amount of Litter Small Amount Large Amount None Litt (Obs): Expected: 17 31.33 28 49 No Litt (Obs): 73 58.67 62 41 94 176 90 90 90 n = 270 Where:

There is a relationship between tendency to litter and litter already in the area

Likelihood Ratios Note that the likelihood ratio is chi-square distributed with k-1 degrees of freedom: In model testing, the likelihood ratio is used to determine the likelihood of Model A generating the data over Model B. In the chi-square case, we are determining the likelihood that the observed data are generated by a chance model.