Chi Square Tests

## 12  Chi Square Tests

The chi-squared distribution allows for statistical tests of categorical data. Among these tests are those for goodness of fit and independence.

### 12.1  The chi-squared distribution

The c2-distribution (chi-squared) is the distribution of the sum of squared normal random variables. Let Zi be i.i.d. normal(0,1) random numbers, and set
c2 =
 n å i=1
Zi2
Then c2 has the chi-squared distribution with n degrees of freedom.

The shape of the distribution depends upon the degrees of freedom. These diagrams (figures 48 and 49) illustrate 100 random samples for 5 d.f. and 50 d.f.

> x = rchisq(100,5);y=rchisq(100,50)
> simple.eda(x);simple.eda(y)


Figure 48: c2 data for 5 degrees of freedom

Figure 49: c2 data for 50 degrees of freedom

Notice for a small number of degrees of freedom it is very skewed. However, as the number gets large the distribution begins to look normal. (Can you guess the mean and standard deviation?)

### 12.2  Chi-squared goodness of fit tests

A goodness of fit test checks to see if the data came from some specified population. The chi-squared goodness of fit test allows one to test if categorical data corresponds to a model where the data is chosen from the categories according to some specified set of probabilities. For dice rolling, the 6 categories (faces) would be assumed to be equally likely. For a letter distribution, the assumption would be that some categories are more likely than other.

Example: Is the die fair?
If we toss a die 150 times and find that we have the following distribution of rolls is the die fair?
 face 1 2 3 4 5 6 Number of rolls 22 21 22 27 22 36
Of course, you suspect that if the die is fair, the probability of each face should be the same or 1/6. In 150 rolls then you would expect each face to have about 25 appearances. Yet the 6 appears 36 times. Is this coincidence or perhaps something else?

The key to answering this question is to look at how far off the data is from the expected. If we call fi the frequency of category i, and ei the expected count of category i, then the c2 statistic is defined to be
c2 =
 n å i=1
(fi - ei)2
ei

Intuitively this is large if there is a big discrepancy between the actual frequencies and the expected frequencies, and small if not.
Statistical inference is based on the assumption that none of the expected counts is smaller than 1 and most (80%) are bigger than 5. As well, the data must be independent and identically distributed -- that is multinomial with some specified probability distribution.
If these assumptions are satisfied, then the c2 statistic is approximately c2 distributed with n-1 degrees of freedom. The null hypothesis is that the probabilities are as specified, against the alternative that some are not.

Notice for our data, the categories all have enough entries and the assumption that the individual entries are multinomial follows from the dice rolls being independent.

R has a built in test for this type of problem. To use it we need to specify the actual frequencies, the assumed probabilities and the necessary language to get the result we want. In this case -- goodness of fit -- the usage is very simple

> freq = c(22,21,22,27,22,36)
# specify probabilities, (uniform, like this, is default though)
> probs = c(1,1,1,1,1,1)/6  # or use rep(1/6,6)
> chisq.test(freq,p=probs)

Chi-squared test for given probabilities

data:  freq
X-squared = 6.72, df = 5, p-value = 0.2423

The formal hypothesis test assumes the null hypothesis is that each category i has probability pi (in our example each pi = 1/6) against the alternative that at least one category doesn't have this specified probability.

As we see, the value of c2 is 6.72 and the degrees of freedom are 6-1=5. The calculated p-value is 0.2423 so we have no reason to reject the hypothesis that the die is fair.

Example: Letter distributions
The letter distribution of the 5 most popular letters in the English language is known to be approximately 13
 letter E T N R O freq. 29 21 17 17 16
That is when either E,T,N,R,O appear, on average 29 times out of 100 it is an E and not the other 4. This information is useful in cryptography to break some basic secret codes. Suppose a text is analyzed and the number of E,T,N,R and O's are counted. The following distribution is found
 letter E T N R O freq. 100 110 80 55 14
Do a chi-square goodness of fit hypothesis test to see if the letter proportions for this text are pE=.29, pT=.21, pN=.17, pR=.17, pO=.16 or are different.

The solution is just slightly more difficult, as the probabilities need to be specified. Since the assumptions of the chi-squared test require independence of each letter, this is not quite appropriate, but supposing it is we get

> x = c(100,110,80,55,14)
> probs = c(29, 21, 17, 17, 16)/100
> chisq.test(x,p=probs)
Chi-squared test for given probabilities

data:  x
X-squared = 55.3955, df = 4, p-value = 2.685e-11

This indicates that this text is unlikely to be written in English.

Some Extra Insight: Why the cs?
What makes the statistic have the c2 distribution? If we assume that fi - ei = Zi
( ei )
 1/2
. That is the error is somewhat proportional to the square root of the expected number, then if Zi are normal with mean 0 and variance 1, then the statistic is exactly c2. For the multinomial distribution, one needs to verify, that asymptotically, the differences from the expected counts are roughly this large.

### 12.3  Chi-squared tests of independence

The same statistic can also be used to study if two rows in a contingency table are independent''. That is, the null hypothesis is that the rows are independent and the alternative hypothesis is that they are not independent.

For example, suppose you find the following data on the severity of a crash tabulated for the cases where the passenger had a seat belt, or did not:
 Injury Level None minimal minor major Seat Belt Yes 12,813 647 359 42 No 65,963 4,000 2,642 303
Are the two rows independent, or does the seat belt make a difference? Again the chi-squared statistic makes an appearance. But, what are the expected counts? Under a null hypothesis assumption of independence, we can use the marginal probabilities to calculate the expected counts. For example
P(none and yes) = P(none)P(yes)
which is estimated by the proportion of none'' (the column sum divided by n) and the proportion of yes: (the row sum divided by n). The expected frequency for this cell is then this product times n. Or after simplifying, the row sum times the column sum divided by n. We need to do this for each entry. Better to let the computer do so. Here it is quite simple.

> yesbelt = c(12813,647,359,42)
> nobelt = c(65963,4000,2642,303)
> chisq.test(data.frame(yesbelt,nobelt))

Pearson's Chi-squared test

data:  data.frame(yesbelt, nobelt)
X-squared = 59.224, df = 3, p-value = 8.61e-13

This tests the null hypothesis that the two rows are independent against the alternative that they are not. In this example, the extremely small p-value leads us to believe the two rows are not independent (we reject).

Notice, we needed to make a data frame of the two values. Alternatively, one can just combine the two vectors as rows using rbind.

### 12.4  Chi-squared tests for homogeneity

The test for independence checked to see if the rows are independent, a test for homogeneity, tests to see if the rows come from the same distribution or appear to come from different distributions. Intuitively, the proportions in each category should be about the same if the rows are from the same distribution. The chi-square statistic will again help us decide what it means to be close'' to the same.

Example: A difference in distributions?
The test for homogeneity tests categorical data to see if the rows come from different distributions. How good is it? Let's see by taking data from different distributions and seeing how it does.

We can easily roll a die using the sample command. Let's roll a fair one, and a biased one and see if the chi-square test can decide the difference.

First to roll the fair die 200 times and the biased one 100 times and then tabulate:


> die.fair = sample(1:6,200,p=c(1,1,1,1,1,1)/6,replace=T)
> die.bias = sample(1:6,100,p=c(.5,.5,1,1,1,2)/6,replace=T)
> res.fair = table(die.fair);res.bias = table(die.bias)
> rbind(res.fair,res.bias)
1  2  3  4  5  6
res.fair 38 26 26 34 31 45
res.bias 12  4 17 17 18 32

Do these appear to be from the same distribution? We see that the biased coin has more sixes and far fewer twos than we should expect. So it clearly doesn't look so. The chi-square test for homogeneity does a similar analysis as the chi-square test for independence. For each cell it computes an expected amount and then uses this to compare to the frequency. What should be expected numbers be?

Consider how many 2's the fair die should roll in 200 rolls. The expected number would be 200 times the probability of rolling a 1. This we don't know, but if we assume that the two rows of numbers are from the same distribution, then the marginal proportions give an estimate. The marginal total is 30/300 = (26 + 4)/300 = 1/10. So we expect 200(1/10) = 20. And we had 26.

As before, we add up all of these differences squared and scale by the expected number to get a statistic:
c2 = å
(fi - ei)2
ei

Under the null hypothesis that both sets of data come from the same distribution (homogeneity) and a proper sample, this has the chi-squared distribution with (2-1)(6-1)=5 degrees of freedom. That is the number of rows minus 1 times the number of columns minus 1.

The heavy lifting is done for us as follows with the chisq.test function.

> chisq.test(rbind(res.fair,res.bias))
Pearson's Chi-squared test
data:  rbind(res.fair, res.bias)
X-squared = 10.7034, df = 5, p-value = 0.05759

Notice the small p-value, but by some standards we still accept the null in this numeric example.
If you wish to see some of the intermediate steps you may. The result of the test contains more information than is printed. As an illustration, if we wanted just the expected counts we can ask with the exp value of the test


> chisq.test(rbind(res.fair,res.bias))[['exp']]
1  2        3  4        5        6
res.fair 33.33333 20 28.66667 34 32.66667 51.33333
res.bias 16.66667 10 14.33333 17 16.33333 25.66667


### 12.5  Problems

12.1
In an effort to increase student retention, many colleges have tried block programs. Suppose 100 students are broken into two groups of 50 at random. One half are in a block program, the other half not. The number of years in attendance is then measured. We wish to test if the block program makes a difference in retention. The data is:
 Program 1 yr 2 yr. 3 yr 4yr 5+ yrs. Non-Block 18 15 5 8 4 Block 10 5 7 18 10
Do a test of hypothesis to decide if there is a difference between the two types of programs in terms of retention.

12.2
A survey of drivers was taken to see if they had been in an accident during the previous year, and if so was it a minor or major accident. The results are tabulated by age group:

 Accident Type AGE None minor major under 18 67 10 5 18-25 42 6 5 26-40 75 8 4 40-65 56 4 6 over 65 57 15 1
Do a chi-squared hypothesis test of homogeneity to see if there is difference in distributions based on age.

12.3
A fish survey is done to see if the proportion of fish types is consistent with previous years. Suppose, the 3 types of fish recorded: parrotfish, grouper, tang are historically in a 5:3:4 proportion and in a survey the following counts are found
 Type of Fish Parrotfish Grouper Tang observed 53 22 49
Do a test of hypothesis to see if this survey of fish has the same proportions as historically.

12.4
The R dataset UCBAdmissions contains data on admission to UC Berkeley by gender. We wish to investigate if the distribution of males admitted is similar to that of females.

To do so, we need to first do some spade work as the data set is presented in a complex contingency table. The ftable (flatten table) command is needed. To use it try