Psy B07 Chapter 6Slide 1 CATEGORICAL DATA & χ 2. Psy B07 Chapter 6Slide 2 A Quick Look Back  Reminder about hypothesis testing: 1) Assume what you believe.

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Psy B07 Chapter 6Slide 1 CATEGORICAL DATA & χ 2

Psy B07 Chapter 6Slide 2 A Quick Look Back  Reminder about hypothesis testing: 1) Assume what you believe (H 1 ) is wrong.  Construct H 0 and accept it as a default. 2) Show that some event is of sufficiently low probability given H 0 ***. 3) Reject H 0. *** In order to do this, we need to know the distribution associated with H 0, because we use that distribution as the basis for our probability calculation.

Psy B07 Chapter 6Slide 3 z-score  Use when we have acquired some data set, then want to ask questions concerning the probability of certain specific data values (e.g., do certain values seem extreme?).  In this case, the distribution associated with H 0 is described by X and S 2 because the data points reflect a continuous variable that is normally distributed.

Psy B07 Chapter 6Slide 4 Chi-square ( χ 2 ) test  The Chi-square test is a general purpose test for use with discrete variables.  It has a number of uses, including the detection of bizarre outcomes given some a priori probability for binomial situation, and for multinomial situations.

Psy B07 Chapter 6Slide 5 Chi-square ( χ 2 ) test  In addition, it allows us to go beyond questions of bizarreness, and move into the question of whether pairs of variables are related. For example:  It does so by mapping the discreet variables unto a continuous distribution assuming H 0, the chi-square distribution.

Psy B07 Chapter 6Slide 6 The chi-square distribution  Let’s reconsider a simple binomial problem. Say, we have a batter who hits.300 [i.e., P(Hit)=0.30], and we want to know whether it is abnormal for him to go 6 for 10 (i.e., 6 hits in 10 at bats).  We could do this using the binomial stuff that I did not cover in Chapter 5 (and for which you are not responsible)  But we can also do it with a chi-square test

Psy B07 Chapter 6Slide 7 The way of the chi 2  We can put our values into a contingency table as follows:  Then consider the distribution of the following formula given H 0 :

Psy B07 Chapter 6Slide 8 The way of the chi 2

Psy B07 Chapter 6Slide 9 The way of the chi 2

Psy B07 Chapter 6Slide 10 The way of the chi 2 In-Class Example:  Note that while the observed values are discreet, the derived score is continuous.  If we calculated enough of these derived scores, we could plot a frequency distribution which would be a chi-square distribution with 1 degree of freedom or  2 (1).  Given this distribution and appropriate tables, we can then find the probability associated with any particular  2 value.

Psy B07 Chapter 6Slide 11 The way of the chi 2 Continuing the Baseball Example: So if the probability of obtaining a  2 of 4.29 or greater is less than , then the observed outcome can be considered bizarre (i.e., the result of something other than a.300 hitter getting lucky). So if the probability of obtaining a  2 of 4.29 or greater is less than , then the observed outcome can be considered bizarre (i.e., the result of something other than a.300 hitter getting lucky).

Psy B07 Chapter 6Slide 12 The way of the chi 2  Just like the t-test, chi 2 distribution is based on degrees of freedom  Thus, since our obtained  2 value of 4.29 is greater than 3.84, we can reject H 0 and assume that hitting 6 of 10 reflects more than just chance performance.

Psy B07 Chapter 6Slide 13 The way of the chi 2 Going a Step Further:  Suppose we complicate the previous example by taking walks and hit by pitches into account. That is, suppose the average batter gets a hit with a probability of 0.28, gets walked with a probability of.08, gets hit by a pitch (HBP) with a probability of.02, and gets out the rest of the time.

Psy B07 Chapter 6Slide 14 The way of the chi 2  Now we ask, can you reject H 0 (that this batter is typical of the average batter) given the following outcomes from 50 at bats? 1) Calculate expected values (Np). 2) Calculate  2 obtained. 3) Figure out the appropriate df (C-1). 4) Find  2 critical and compare  2 obtained to it.

Psy B07 Chapter 6Slide 15 The way of the chi 2

Psy B07 Chapter 6Slide 16 Two types of chi 2 tests  So far, all the tests have been to assess whether some observation or set of observations seems out-of-line with some expected distribution. This is also known as the goodness-of-fit chi-square test  However, the logic of the chi-square test can be extended to examine the issue of whether two variables are independent (i.e., not systematically related) or dependent (i.e., systematically related).

Psy B07 Chapter 6Slide 17 χ 2 test for independence  Consider the following data set again:  Are the variables of gender and opinion concerning the legalization of marijuana independent?

Psy B07 Chapter 6Slide 18 χ 2 test for independence

Psy B07 Chapter 6Slide 19 χ 2 test for independence  If these two variables are independent, then by the multiplicative law, we expect that:

Psy B07 Chapter 6Slide 20 χ 2 test for independence  If we do this for all four cells, we get:

Psy B07 Chapter 6Slide 21 χ 2 test for independence  Are the observed values different enough from the expected values to reject the notion that the differences are due to chance variation?

Psy B07 Chapter 6Slide 22 χ 2 test for independence  The df associated with 2 variable contingency tables can be calculated using the formula:  where C is the number of columns and R is the number of rows.

Psy B07 Chapter 6Slide 23 χ 2 test for independence  Thus, to finish our previous example, the  2 critical with alpha equal.05 and 1 df equals 3.84. Since our  2 is not bigger than that (i.e., 3.6) we cannot reject H 0.

Psy B07 Chapter 6Slide 24 Assumptions of χ 2 Independence of observations:  Chi-square analyses are only valid when the actual observations within the cells are independent.  This independence of observations is different from the issue of whether the variables are independent, that is what the chi-square is testing.

Psy B07 Chapter 6Slide 25 Assumptions of χ 2 Independence of observations:  You know your observations are not independent when the grand total is larger than the number of subjects.  Example: The activity level of 5 rats was tested over 4 days, producing these values:

Psy B07 Chapter 6Slide 26 Assumptions of χ 2 Normality:  Use of the chi-square distribution for finding critical values assumes that the expected values (i.e., Np) are normally distributed.  This assumption breaks down when the expected values are small (specifically, the distribution of Np becomes more and more positively skewed as Np gets small).

Psy B07 Chapter 6Slide 27 Assumptions of χ 2 Normality:  Thus, one should be cautious using the chi-square test when the expected values are small.  How small? This is debatable but if expected values are as low as 5, you should be worried.

Psy B07 Chapter 6Slide 28 Assumptions of χ 2 Inclusion of Non-Occurrences:  The chi-square test assumes that all outcomes (occurrences and non- occurrences) are considered in the contingency table.  As an example of a failure to include a non-occurrence, see page 160 of the text.

Psy B07 Chapter 6Slide 29 A tale of tails  We only reject H 0 when values of  2 are larger than  2 obtained.  This suggests that the  2 test is always one-tailed and, in terms of the rejection region, it is.  In a different sense, however, the test is actually multiple tailed.

Psy B07 Chapter 6Slide 30 A tale of tails  Reconsider the following “marking scheme” example:  If we do not specify how we expect the results to fall out then any outcome with a high enough  2 obtained can be used to reject H 0.  However, if we specify our outcome, we are allowed to increase our alpha - in the example we can increase alpha to 0.30 if we specified the exact ordering (in advance) that was observed.

Psy B07 Chapter 6Slide 31 Measures of Association  The chi-square test only tells us whether two variables are independent, it does not say anything about the magnitude of the dependency if one is found to exist.  Stealing from the book, consider the following two cases, both of which produce a significant  2 obtained, but which imply different strengths of relation:

Psy B07 Chapter 6Slide 32 Measures of Association

Psy B07 Chapter 6Slide 33 Measures of Association  There are a number of ways to quantify the strength of a relation (see sections in the text on the contingency coefficient, Phi, & Odds Ratios), but the two most relevant to psychologists are Cramer’s Phi and Cohen’s Kappa.

Psy B07 Chapter 6Slide 34 Measures of Association  Cramer’s Phi ( φ c) can be used with any contingency table and is calculated as:  Values of range from 0 to 1. The values the tables on the previous page are 0.12 and 0.60 respectively, indicating a much stronger relation in the second example.

Psy B07 Chapter 6Slide 35 Measures of Association  Often, in psychology, we will ask some “judge” to categorize things into specific categories.  For example, imagine a beer brewing competition where we asked a judge to categorize beers as Yucky, OK, or Yummy.  Obviously, we are eventually interested in knowing something about the beers after they are categorized.

Psy B07 Chapter 6Slide 36 Measures of Association  However, one issue that arises is the judges abilities to tell the difference between the beers.  One way around this is to get two judges and show that a given beer is reliably rated across the judges (i.e., that both judges tend to categorize things in a similar way).

Psy B07 Chapter 6Slide 37 Measures of Association  Such a finding would suggest that the judges are sensitive to some underlying quality of the beers as opposed to just guessing.

Psy B07 Chapter 6Slide 38 Measures of Association  Note that if you just looked at the proportion of decisions that me and Judge 2 agreed on, it looks like we are doing OK: P(Agree)=21/30 = 0.70 or 70%

Psy B07 Chapter 6Slide 39 Measures of Association  There is a problem here, however, because both judges are biased to judge a beer as OK such that even if they were guessing, the agreement would seem high because both would guess OK on a lot of trials and would therefore agree a lot.

Psy B07 Chapter 6Slide 40 Measures of Association  Such a finding would suggest that the judges are sensitive to some underlying quality of the beers as opposed to just guessing.

Psy B07 Chapter 6Slide 41

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