Www Psych Utah Edu Stat Introstats Web Text Chi Square Assoc

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Chi Square Test ofAssociation

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Copyright 2000, Tom Malloy
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Top Test of Association

There are two different Chi Square tools. We discussed the Goodness of Fit Chi

Square in the previous lecture. Now we will discuss the Chi Square Test of

Association.

The Chi Square Test of Association was derived mathematically by Karl Pearson

early in the century, and is often known as Pearson's Chi Square Test of

Association. Pearson showed that the formulas for the Goodness of Fit statistic

we learned last lecture and the Test of Association statistic we will learn in this

lecture both have Chi Square as their sampling distribution. In this class we don't

have sufficient mathematical prerequisites to follow these proofs, but suffice it to

say they were great insights at their time, insights which have allowed substantial

scientific sophistication to be brought to bear on frequency-categorical data.

Quite often we have two categorical variables such as gender (male, female) and

job status (managers, clerks). We wonder whether there is an association (or

correlation) between the two categorical variables; that is, is there a relationship

between a person's gender and their job status? As another example, we might


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interested in the potential association of Political Party (Republican, Democratic,

Independent, Other) and Environmental Attitudes (Preservation, Development,

Other). The Chi Square Test of Association allows us to evaluate associations

(i.e., correlations) between categorical variables such as these.

We will begin our discussion of the Chi Square Test of Association with the threecriteria that are necessary for its appropriate use.


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THREE CRITERIA

1) Recall that in the Chi Square Goodness of Fit we needed a partition (a system of

mutually exclusive and exhaustive categories). Now, in the Chi Square Test of

Association, we need TWO partitions. 2) We also need some number of independent

observations. 3) Finally, we need frequency data.

CRITERION #1: 2 Partitions.In probability theory a partition is a mutuallyexclusive and an exhaustive set of categories. As you know, categories are pigeon holes,

or places where we can put things conceptually. To be a partition, a set of categories have

to be both mutually exclusive and exhaustive. Mutually exclusive means that observation


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can go into one and only one category. No idea (nor thing, nor observation) may go in

more than one category.

Exhaustive means that the set of categories

cover every possible case. It means that every

object we observe can be put into one of thecategories. More detail on these terms is

available in the Chi Square Goodness of Fit

lecture.

Often when you fill out a survey you'll find that

there is an extra category called "Other.""Other" is a great category because it ensures

that whatever set of choices is offered must be exhaustive, because if you don't fit in any

of the existing categories then you're in "Other".

Examples of Partitions

As we discussed in the Goodness of Fit lecture,

gender is a partition. It is a set of mutually

exclusive and exhaustive categories. There are

only two categories in gender--male and female.

You cannot be both at the same time so so male


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and female are mutually exclusive. Moreover,

for human beings, gender is exhaustive

because everybody goes in one of the two categories. Male and female exhaust the

possibilities.

Let's look at another example of a partition. Let's say at a particular corporation weclassify people according to their job status. At this particular corporation the job status

is either managerial or clerical and there are no other categories. (This unrealistic, but it

keeps the calculations in this example simple.) These categories are mutually exclusive.

Any employee is either going to be a manager or they're going to be a clerk; they can't

be both. And, since we've said that those are the only job types, those two job types

exhaust all the possibilities.

Example using 2 Partitions

Now let's create a running example for this

lecture. We will classify every employee of a

business both by gender and by job status.

That is, we will apply two partitions in

categorizing people. As you can see from the

graphic, that results in four possibilities: Male

clerk, Female clerk, Male manager, and Female


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manager.

Typically, in this kind of chi square you will make some kind of table. For this small

example we have a two by two (2 x 2) table; but the table could be seven by four table (7

x 4) or whatever, depending on how many pigeon holes are in each partition.

CATEGORICAL VARIABLES. Another way to speak about this example is that we have

two categorical variables: Gender and Job Status. When you classify by two categorical

variables, thus creating all the possible combinations of the two (clerical male, clerical

female, managerial male, managerial female) it is called "crossing" the variables.

To do a chi squares test of association you must classify each of your observations not

with just one but with two partitions.

In this simple example, male and female is one dimension and job status is the other

dimension. You'll notice that our requirement is that every person we observe in our

study can be classified by each of the partitions. So everybody will go into one of these

"cells" created by crossing these two partitions with each other.

CRITERION #2. N

Independent

Observations


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The second criterion for the Chi Square Test

of Association is that we have some number

of observations, each independent of the

others. In our running example, say that we

classify 200 people who work at a corporation by gender and job status. We will assumethat these people are independent of each other and that they are just picked out at

random.

In the Goodness of Fit lecture we discussed the case of a rat in a T-maze. In a T-maze a

rat has to right or left so you have a partition for each trial. However, if you observe the

same rat over many trials, these observations would not be independent because the

particular rat may have a turning bias.

The important point is that you need to think about the observations that you are

making and decide whether they're independent or not. For our example we are

assuming that for each person who works at this particular company, their gender is

independent of the next person. That fact that one person was born a man, doesn't

have anything to do with someone else who works in a different part of the corporation

being a woman. Their births are independent of each other and have no relationship to

each other.The same must be true for job status.

CRITERION #3: Frequency Data. The third criterion is frequency data.


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Frequency data means that we don't measure anything when we observe, all we do is put

the observations in categories and count the number of observation that fall into each

category.

Each time someone falls into a category you can make a little hash mark in one of the cells

of the table. We are just counting, not measuring.

Think about how simply counting is different from the measuring we did for the t tests. Our

dependent variable in t-tests has been things like someone's height or someone's weight,

or some number of puzzles solved correctly. In those examples when we observe a person

our dependent variable generated a measurement number - how many inches tall they are;

how many pounds they weigh; how many puzzles they solve; and we actually assign thatnumber to that person. That kind of data is called measurement data; and it requires

statistics like t-tests, correlation coefficients, and so on.

For Chi Square, we do not measure the participants; we just count their frequency in

various categories.


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Scientific Hypotheses. Let's say that the scientific hypothesis is that there isgender bias in hiring and promotion in this corporation. That means job status will depend

upon gender. More specifically it is more likely that the men will be managers and the

women will be clerks, which would be a classic kind of gender bias.

In contrast, the skeptic (or maybe corporation's lawyer or public relations representative)

will say that hiring and promotion are completely fair. They will note that there will be

different proportions of men and women in different categories but this is simply due to

chance and there is nothing systematic about how different genders fall into different job

status categories.

So we have our scientific hypothesis of gender bias, and our skeptical hypothesis which

is saying hiring practices are fair and if the data shows differently then it is only due to

chance.


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What is the Question?The essential question being asked by the Chi Square Test ofAssociation is: "Is one way of categorizing things related to the other way of categorizing things, or are

they independent?" Another way of putting it is, "Are the two partitions (categorical variables) correlated

(associated) with each other or are they unrelated (independent)?"

In our example we are trying to determine whether gender is related to job status. Is there a correlations

(association) between a person's gender and her/his job status?

To answer this question we collect some data. We go to the business and collect the relevant information

on 200 employees. We put the information into a table like the one we've been showing which crosses

job status with gender.

ASSOCIATION MEANS PREDICTABILITY. So let's repeat our essential question in yet another way. Are

the two classifications correlated or associated or are they independent? That is, can you predict one

f h h ? M ifi ll di ' j b f h i d If h i


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from the other? More specifically, can you predict someone's job status from their gender. If there is an

association between gender and status, it will mean that you can make a good guess about their job

status by knowing their gender.

Let's examine how the frequency data in the table would look in several cases from complete

dependence to complete independence. We examine 200 people from the business. I've made theexample so that of these 200, 110 are women and 90 are men.

Complete Dependence

We'll start with the extreme case--complete

dependence between gender and job status. If

your study showed the observed frequencies in

the table on the graphic (not a single male clerk

and not a single female manager), that data would

have to come from watching old sitcoms from the

1950's. In any event, if your data looked like this,

where all 90 of the men are managers and all 110 of the women are clerks, the data wouldindicate complete dependence or complete predictability. In this example, the data would

demonstrate a case of extreme gender bias. The association between gender and job

status is as high as it can be.

If I told you that someone is a clerk and then asked you to guess what their gender is, you

would be able to predict their gender perfectly If a person is a clerk then she must be


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would be able to predict their gender perfectly. If a person is a clerk, then she must be

female. If I said someone is a man, you could guess his job status perfectly; he must be a

manager. This is complete dependence of gender and status.

Strong Dependence

Here is an example of observed frequency data

indicating partial but strong dependence. This

example isn't quite so clear cut, there is still a

great deal of predictability between the two

categorical variables.

In the current graphic, out of 200 people, there

were 20 male clerks, 70 male managers; there

were 100 clerical females, and 10 females managers. In other words, 22% of the men are

clerks whereas 91% of the women are clerks.

Now you can't make a perfect prediction from gender to job status, but you can make apretty good guess. If I say that somebody is a manager and asked you to guess whether

they're a male or a female, you could make a pretty good guess. You'd guess a manager

would be male. Now you wouldn't be right all of the time, but you'd be right most of the

time.

In this example there's strong predictability from gender to job status and vice versa In


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In this example there s strong predictability from gender to job status and vice versa. In

other words there is a strong association between gender and job status.

Independence

What would the data look like if the partitions

were independent or had no relationship

whatsoever? In such a case, you would not

be able to predict job status from gender any

better than chance.

I've changed the example such that half theemployees are managers and half are clerks.

In line with this, you'll notice that half of the women are clerical and half are managers.

The same goes for the men, half of them are clerical and half of them are managers. In

other words, 50% of all employees are managers, AND 50% of the men are managers

and 50% of the women are managers.

I've made the example such that men and women are not equal in number. Out of every

100 people there is 45 men and 55 women. So 45% of the employees are man and 55%

of the employees are women. If we look at the 100 clerks, we find that 45% of them are

men and 55% of them are women. That is the same as the percentage of men and

women in the company. There appears to be no bias whatsoever, not even chance


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variations.

NO ASSOCIATION. In this case then, we have independence of the two partitions or the

two category systems. The category system called gender is independent of the

category system called job status. Or in the language used in the Chi Square, there is

no association between gender and job status.

So that's the question is for this kind of test statistic. It's a different kind of question than the one you

would have for a t test which uses measurement data.

The Research Data

For our running example, let's suppose that our

research yields the data shown in the current

graphic.

N = 200 total people. Males = 90 out of 200 or

45%. Females = 110 out 200 or 55%. Clerks =

120 out of 200 or 60%. Managers = 80 out of 200

or 40%.


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DATA PATTERNIn this particular case, the data pattern fits the scientific hypothesis. You can argue it anynumber of ways. For instance, 120 out of 200, or 60% of all employees are clerical, but 90

out of 110, or 82% of females are clerical, and 30 out of 90 or 30% of the males are clerical.

These are the kinds of arguments that someone who thought there was gender biaswould make. They'd say "Look, 60% of all employees are clerical, but 82% of the females

and only 33% of the males are clerical. There appears to be gender bias."

CHANCE

The skeptic would say, "Well I think that data

pattern is just happening by chance." The PCH

of chance says that these data could have


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of chance says that these data could have

come about by chance variations in hiring and

promotion.

EVALUATE THE PCH OFCHANCE

To deal with the plausible competing hypothesis

(PCH) of chance, we're going to do a Chi Square

Test of Association.

Go To Top


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Go To Top

Let's calculate the Expected Frequencies

Review

The slide to the left and the two slides below

summarize the example. Review these three

slides and then we'll start calculating the expected

frequencies.


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FORMULA FOR EXPECTEDFREQUENCIES

This formula will make most sense after you

have worked through the example.

Calculating the Expected


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Calculating the Expected

Frequencies

We already have the observed frequencies for

each cell in our table. Now we will calculate an

expected frequency for each cell. The symbol

for expected frequency is fe. We will learn how

to calculate the expected frequencies by going

through all the cells in the example.

NOTATION. We want to be able to note and communicate which of the four cells we are

talking about at any given moment. The table is 2-dimensional, so we need two

dimensions to describe a location in it. By convention, these two dimensions we will call

"j" and "k." As the blue arrow in the graphic shows, the indexjruns down rows; it tells

us which row we are in. And the index kruns across columns; it tells us which column

we are in. [It's arbitrary which dimension we call j and which we call k. We just have to

agree on which is which.] The general symbol for the expected frequency in a particular,

unspecified cell is fe(jk). This can be read as the expected frequency for the cell where

row j intersects with column k. [Generally, "jk" is a subscript but that is currently difficult

to write subscripts in web html text. So I'll use a parenthesis around jk when I need to be

clear. In obvious cases I'll just write the indices without the parentheses.]

Our notation is such that we put row (j) first and column (k) last when indexing a cell in a


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table. So fe(11) is the expected frequency for the first row in the first column. fe(12)

indicates the expected frequency in row 1, column 2. Fe(21) indicates the second row in

the first column; and Fe(22) is the cell that is in the second row of the second column.

CALCULATING THE EXPECTED FREQUENCY FOR CELL 1-1. To repeat, the symbol,

fe11, is the designation we give for the cell where j =1 and k =1. In this example cell 1-1 is

the upper left hand cell (which is where we placed clerical male employees). The

expected frequency for that cell is determined by the total number in that row (which is

120) times the total number in that column (which is 90) divided by the total number of

observations in the whole table (which is 200).

This is somewhat confusing to describe in words. But it's really simple if you look at the

examples in the graphics. All you need to do is find the row total, multiply it times the

column total, and divide by the total total.

EXPECTED FREQUENCY FOR CELL 1,1. So you calculate the expected frequency for

the first cell as 120 times 90 over 200. fe11is equal to 54.

Attempt to find the expected frequencies for the other cells on your own before you look

at the results below.

Ex ected Fre uencies for

h O h C ll


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the Other Cells

Go ahead and find the expected frequency

for cell 1-2 (row 1, column 2) or the female

clerks cell.

One convenient way to summarize the

information you will need when we

eventually get to the formula is to write the

expected frequency in the cell with the

observed frequency.

Fe(1,2).The expected frequency for row1, column 2 is 120 times 110, over 200 which

you can see equals 66.


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Fe(2,1). For row 2, column 1 the expected frequency works out to be 90 times 80 over


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( , ) , p q y200, or 36.

Fe(2,2). Then the expected frequency for the cell in row two, column two is 110 times 80over 200, or 44.

In the final graphic of the series, the table has observed frequencies, which are the black

colored numbers, and expected frequencies which are shown in blue.

The observed frequencies are the data. The expected frequencies is what the data should

have come out to be if Gender were NOT associated with Job Status.

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The Formula


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Go To Top

Let's look at the formula. It tells you to sum up

the squared differences between the expected

and the observed frequencies and divide each

squared difference by the expected frequency.

This is the same as the Goodness of Fit, exceptthat it is for a 2-dimensional case, so the

formula uses a double summation notation.

DEGREES OF FREEDOM. The degrees of

freedom are the number of rows minus one

times the number of columns minus one. Interms of notation, capital J represents the

number of rows, and capital K represents the

number of columns.


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Go To Top

Calculations

Here is what our 2 x 2 table would be like like

given our data and the calculated expected

frequencies.

Next we determine the deviation between

observed frequency and the expected

frequency for each cell. We square the

deviation for each cell. Finally we divide the

squared deviation by the expected frequency

for that cell. The four graphics below show all

the calculations.


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IN WORDS. Get a value for each cell by determining (fo - Fe) squared over Fe In the

example the values for the four cells are 10.67, 8.73, 16, and 13.09.

Then sum up the values for all the cells to get the final chi square value. In the example chi

square = 10.67 + 8.73 + 16 + 13.09 = 48.58.

Degrees of Freedom

The formula for degrees of freedom is (the

number of rows minus one) times (the

number of columns minus one). It can be

symbolized by J - 1 times K minus 1 or (J -

1)(K - 1). In this case, the degrees of freedom

l 2 i 1 ti 2 i 1 j t 1


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equal 2 minus 1 times 2 minus 1, or just 1.

[Note: Capital J is used to indicate the

number of rows. As we've already said, little j

is the index for some particular row. The

same is true of K and k.]

Now we can go on to the topic of statistical

conclusion validity.

STATISTICAL HYPOTHESES

The observed frequencies (fo's) are the data we

collect. The expected frequencies are what the

data should be if the two categorical variablesare independent (not associated).

NULL HYPOTHESIS. The skeptic thinks thatthere is no association between categorical

variables (in this case, between gender andstatus) so the observed frequencies shouldequal the expected frequencies other than

chance differences. So the corresponding null hypothesis is that we expect the

difference between observed and expected frequency in each cell to equal zero.

ALTERNATIVE HYPOTHESIS. The scientist things that there IS an association between

the two categorical variables. So the scientist thinks the data will differ from the expectedfrequencies. So the corresponding null hypothesis is that the difference between the


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frequencies. So the corresponding null hypothesis is that the difference between the

observed and expected frequencies will NOT be zero.

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Statistical Conclusion

Validity

Here we show the sampling distribution of chi

square again The number line along the bottom


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square again. The number line along the bottom

of the graph goes from zero to positive infinity

for chi square. Chi square is a squared entity -

everything in it is squared. Even if you get

negative numbers from your cell calculations,they are going to be squared and made into

positive numbers. You cannot get a chi square

below zero. If you calculate a chi square below

zero, you made a mistake.

So the range of the Chi Square test statisticgoes from zero to positive infinity.

What H0 predicts

If you picture the null hypothesis in your mind,

you'll remember that we expect the difference

between observed expected frequencies to be

zero. If H0 is actually true, then every term (for

every cell) in the Chi Square formula would be

zero. That is, there would be no differences

between the observed and the expected

frequencies in any cell Therefore each of those


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frequencies in any cell. Therefore, each of those

differences would be zero, zero squared is zero,

and the whole Chi Square would be equal to

zero. So H0 is predicting values of Chi Square

near zero. So high values of Chi Square are notwhat H0 is predicting.

The Rejection Region

To find the critical value you need to know the

degrees of freedom and your selected alpha

level. Then you just look the critical value up in

your table. The critical value of chi square, with

one degree of freedom and alpha of .05, is 3.84.

Chi Square tables are available on the course

web site.

You draw your "Reject H0" and "Do not reject

H0" regions based on this critical value of 3.84.

We found that our calculated chi square of

48.58 falls in the rejection region therefore, we

would reject H0


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would reject H0.

Why we Reject H0

H0 is predicting that there will be no differencebetween expected and observed frequencies

and so you should get a chi square in the

neighborhood of zero if Ho is true. By chance

alone, you may get a Chi Square value bigger

than zero. However, the chance of getting a

value of Chi Square beyond 3.84 by chance

alone is very small.

Therefore using the logic we have used before

with other statistical tests, we'll reject H0

because it's very improbable that you would get

a Chi Square of this magnitude by chance

alone.


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Sampling Distribution of

Chi Square

Let's review the sampling distribution of chi-

square. This slide shows the overall 4 step

process and is the same slide you saw for the

Goodness of Fit Chi Square.

The sampling distribution of the test statistic

you just calculated is called the Chi Square

probability distribution. This distribution starts

at zero and goes to positive infinity. Notice also

that it is not symmetrical. It's different than a


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that it is not symmetrical. It s different than a

bell curve, it has a big lump down by zero

where most of the probability is and it has only

one tail going off toward positive infinity.

Go To Top

Copyright 1997, 2000 Tom Malloy

Www Psych Utah Edu Stat Introstats Web Text Chi Square Assoc

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