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Chapter 11: Chi-Square Tests and ANOVA
Chapter 11: Chi-Square and ANOVA Tests
This chapter presents material on three more hypothesis tests. One is used to determine
significant relationship between two qualitative variables, the second is used to determine
if the sample data has a particular distribution, and the last is used to determine
significant relationships between means of 3 or more samples.
Section 11.1: Chi-Square Test for Independence
Remember, qualitative data is where you collect data on individuals that are categories or
names. Then you would count how many of the individuals had particular qualities. An
example is that there is a theory that there is a relationship between breastfeeding and
autism. To determine if there is a relationship, researchers could collect the time period
that a mother breastfed her child and if that child was diagnosed with autism. Then you
would have a table containing this information. Now you want to know if each cell is
independent of each other cell. Remember, independence says that one event does not
affect another event. Here it means that having autism is independent of being breastfed.
What you really want is to see if they are not independent. In other words, does one
affect the other? If you were to do a hypothesis test, this is your alternative hypothesis
and the null hypothesis is that they are independent. There is a hypothesis test for this
and it is called the Chi-Square Test for Independence. Technically it should be called
the Chi-Square Test for Dependence, but for historical reasons it is known as the test for
independence. Just as with previous hypothesis tests, all the steps are the same except for
the assumptions and the test statistic.
Hypothesis Test for Chi-Square Test
- State the null and alternative hypotheses and the level of significance
H
o
: the two variables are independent (this means that the one variable is not
affected by the other)
H
A
: the two variables are dependent (this means that the one variable is affected
by the other)
Also, state your α level here.
- State and check the assumptions for the hypothesis test
a. A random sample is taken.
b. Expected frequencies for each cell are greater than or equal to 5 (The expected
frequencies, E , will be calculated later, and this assumption means E ≥ 5 ).
- Find the test statistic and p-value
Finding the test statistic involves several steps. First the data is collected and
counted, and then it is organized into a table (in a table each entry is called a cell).
These values are known as the observed frequencies, which the symbol for an
observed frequency is O. Each table is made up of rows and columns. Then each
row is totaled to give a row total and each column is totaled to give a column
total.
Chapter 11: Chi-Squared Tests and ANOVA
The null hypothesis is that the variables are independent. Using the multiplication
rule for independent events you can calculate the probability of being one value of
the first variable, A , and one value of the second variable, B (the probability of a
particular cell P A and B
). Remember in a hypothesis test, you assume that H
0
is true, the two variables are assumed to be independent.
P A and B
= P A
⋅ P B
if A and B are independent
number of ways A can happen
total number of individuals
number of ways B can happen
total number of individuals
row total
n
column total
n
Now you want to find out how many individuals you expect to be in a certain cell.
To find the expected frequencies, you just need to multiply the probability of that
cell times the total number of individuals. Do not round the expected frequencies.
Expected frequency cell A and B
= E A and B
= n
row total
n
column total
n
row total ⋅column total
n
If the variables are independent the expected frequencies and the observed
frequencies should be the same. The test statistic here will involve looking at the
difference between the expected frequency and the observed frequency for each
cell. Then you want to find the “total difference” of all of these differences. The
larger the total, the smaller the chances that you could find that test statistic given
that the assumption of independence is true. That means that the assumption of
independence is not true. How do you find the test statistic? First find the
differences between the observed and expected frequencies. Because some of
these differences will be positive and some will be negative, you need to square
these differences. These squares could be large just because the frequencies are
large, you need to divide by the expected frequencies to scale them. Then finally
add up all of these fractional values. This is the test statistic.
Test Statistic:
The symbol for Chi-Square is χ
2
χ
2
O − E
2
E
where O is the observed frequency and E is the expected frequency
Chapter 11: Chi-Squared Tests and ANOVA
Table #11.1.1: Autism Versus Breastfeeding
Autism
Breast Feeding Timelines
Row
Total
None Less
than 2
months
2 to 6
months
More
than 6
months
Yes 241 198 164 215 818
No 20 25 27 44 116
Column Total 261 223 191 259 934
Solution:
- State the null and alternative hypotheses and the level of significance
H
o
: Breastfeeding and autism are independent
H
A
: Breastfeeding and autism are dependent
α = 0.
- State and check the assumptions for the hypothesis test
a. A random sample of breastfeeding time frames and autism incidence was
taken.
b. Expected frequencies for each cell are greater than or equal to 5 (ie. E ≥ 5 ).
See step 3. All expected frequencies are more than 5.
- Find the test statistic and p-value
Test statistic:
First find the expected frequencies for each cell.
E ( Autism and no breastfeeding) =
E Autism and < 2 months
E Autism and 2 to 6 months
E ( Autism and more than 6 months) =
Others are done similarly. It is easier to do the calculations for the test statistic
with a table, the others are in table #11.1.2 along with the calculation for the test
statistic. (Note: the column of O − E should add to 0 or close to 0.)
Chapter 11: Chi-Square Tests and ANOVA
Table #11.1.2: Calculations for Chi-Square Test Statistic
O E
O − E
O − E
2
O − E
2
E
Total 0.000 1
11.2166432 = χ
2
The test statistic formula is χ
2
O − E
2
E
, which is the total of the last
column in table #11.1.2.
p-value:
df = ( 2 − 1 )* ( 4 − 1 ) = 3
Using TI-83/84: χcdf (11.2166432 ,1E 99 , 3 ) ≈ 0.
Using R: 1 − pchisq (11.2166432 , 3 ) ≈ 0.
- Conclusion
Fail to reject H
o
since the p-value is more than 0.01.
- Interpretation
There is not enough evidence to show that breastfeeding and autism are
dependent. This means that you cannot say that the whether a child is breastfed or
not will indicate if that the child will be diagnosed with autism.
Example #11.1.2: Hypothesis Test with Chi-Square Test Using Technology
Is there a relationship between autism and breastfeeding? To determine if there
is, a researcher asked mothers of autistic and non-autistic children to say what
time period they breastfed their children. The data is in table #11.1.1 (Schultz,
Klonoff-Cohen, Wingard, Askhoomoff, Macera, Ji & Bacher, 2006). Do the data
provide enough evidence to show that that breastfeeding and autism are
independent? Test at the1% level.
Solution:
- State the null and alternative hypotheses and the level of significance
H
o
: Breastfeeding and autism are independent
H
A
: Breastfeeding and autism are dependent
α = 0.
Chapter 11: Chi-Square Tests and ANOVA
Now type the table in by pressing ENTER after each cell value. Figure #11.1.
contains the complete table typed in. Once you have the data in, press QUIT.
Figure #11.1.4: Data Typed into Matrix
To run the test on the calculator, go into STAT, then move over to TEST and
choose χ
2
- Test from the list. The setup for the test is in figure #11.1.5.
Figure #11.1.5: Setup for Chi-Square Test on TI-83/
Once you press ENTER on Calculate you will see the results in figure #11.1.6.
Figure #11.1.6: Results for Chi-Square Test on TI-83/
The test statistic is χ
2
≈ 11.2167 and the p-value is p ≈ 0.01061. Notice that the
calculator calculates the expected values for you and places them in matrix B. To
Chapter 11: Chi-Squared Tests and ANOVA
review the expected values, go into MATRX and choose 2:[B]. Figure #11.1.
shows the output. Press the right arrows to see the entire matrix.
Figure #11.1.7: Expected Frequency for Chi-Square Test on TI-83/
To compute the test statistic and p-value with R,
row1 = c(data from row 1 separated by commas)
row2 = c(data from row 2 separated by commas)
keep going until you have all of your rows typed in.
data.table = rbind(row1, row2, …) – makes the data into a table. You can
call it what ever you want. It does not have to be data.table.
data.table – use if you want to look at the table
chisq.test(data.table) – calculates the chi-squared test for independence
chisq.test(data.table)$expected – let’s you see the expected values
For this example, the commands would be
row1 = c(241, 198, 164, 215)
row2 = c(20, 25, 27, 44)
data.table = rbind(row1, row2)
data.table
Output:
[,1] [,2] [,3] [,4]
row1 241 198 164 215
row2 20 25 27 44
chisq.test(data.table)
Chapter 11: Chi-Squared Tests and ANOVA
Table #11.1.3: Number of Leprosy Cases
WHO Region
World Bank Income Group
Row
Total
High
Income
Upper
Middle
Income
Lower
Middle
Income
Low
Income
Americas 174 36028 615 0 36817
Eastern
Mediterranean
Europe 10 0 0 0 10
Western
Pacific
Africa 0 39 1986 15928 17953
South-East
Asia
Column Total 264 36289 158069 27923 222545
Solution:
- State the null and alternative hypotheses and the level of significance
H
o
: WHO region and Income Level when dealing with the disease of leprosy are
independent
H
A
: WHO region and Income Level when dealing with the disease of leprosy are
dependent
α = 0.
- State and check the assumptions for the hypothesis test
a. A random sample of incidence of leprosy was taken from different countries
and the income level and WHO region was taken.
b. Expected frequencies for each cell are greater than or equal to 5 (ie. E ≥ 5 ).
See step 3. There are actually 4 expected frequencies that are less than 5, and
the results of the test may not be valid. If you look at the expected
frequencies you will notice that they are all in Europe. This is because Europe
didn’t have many cases in 2011.
- Find the test statistic and p-value
Test statistic:
First find the expected frequencies for each cell.
E ( Americas and High Income) =
E ( Americas and Upper Middle Income) =
E ( Americas and Lower Middle Income) =
E ( Americas and Lower Income) =
Chapter 11: Chi-Square Tests and ANOVA
Others are done similarly. It is easier to do the calculations for the test statistic
with a table, and the others are in table #11.1.4 along with the calculation for the
test statistic.
Table #11.1.4: Calculations for Chi-Square Test Statistic
O E
O − E
O − E
2
O − E
2
E
Total 0.
328594.008 = χ
2
The test statistic formula is χ
2
O − E
2
E
, which is the total of the last
column in table #11.1.2.
p-value:
df = ( 6 − 1 )* ( 4 − 1 ) = 15
Using the TI-83/84: χcdf ( 328594.008,1E 99 , 15 ) ≈ 0
Using R: 1 − pchisq ( 328594.008, 15 ) ≈ 0
- Conclusion
Reject H
o
since the p-value is less than 0.05.
Chapter 11: Chi-Square Tests and ANOVA
Figure #11.1.8: Setup for Matrix on TI-83/
Figure #11.1.9: Results for Chi-Square Test on TI-83/
χ
2
Figure #11.1.10: Expected Frequency for Chi-Square Test on TI-83/
Press the right arrow to look at the other expected frequencies.
Chapter 11: Chi-Squared Tests and ANOVA
p-value:
p − value ≈ 0
Using R:
row1=c(174, 36028, 615, 0)
row2=c(54, 6, 1883, 604)
row3=c(10, 0, 0, 0)
row4=c(26, 216, 3689, 1155)
row5=c(0, 39, 1986, 15928)
row6=c(0, 0, 149896, 10236)
chisq.test(data.table)
Pearson's Chi-squared test
data: data.table
X-squared = 328590, df = 15, p-value < 2.2e- 16
Warning message:
In chisq.test(data.table) : Chi-squared approximation may be incorrect
chisq.test(data.table)$expected
[,1] [,2] [,3] [,4]
row1 43.67515783 6003.514404 2.615034e+04 4619.
row2 3.02144735 415.323117 1.809080e+03 319.
row3 0.01186277 1.630637 7.102788e+00 1.
row4 6.03340448 829.341724 3.612478e+03 638.
row5 21.29722977 2927.481709 1.275164e+04 2252.
row6 189.96089780 26111.708410 1.137384e+05 20091.9 62686
Warning message:
In chisq.test(data.table) : Chi-squared approximation may be incorrect
χ
2
= 328590 and p-value = 2.2 × 10
− 16
- Conclusion
Reject H
o
since the p-value is less than 0.05.
- Interpretation
There is enough evidence to show that WHO region and income level are
dependent when dealing with the disease of leprosy. WHO can decide how to
focus their efforts based on region and income level. Do remember though that
the results may not be valid due to the expected frequencies not all be more than
Chapter 11: Chi-Squared Tests and ANOVA
3.) Is there a relationship between autism and what an infant is fed? To determine if
there is, a researcher asked mothers of autistic and non-autistic children to say
what they fed their infant. The data is in table #11.1.7 (Schultz, Klonoff-Cohen,
Wingard, Askhoomoff, Macera, Ji & Bacher, 2006). Do the data provide enough
evidence to show that that what an infant is fed and autism are independent? Test
at the1% level.
Table #11.1.7: Autism Versus Breastfeeding
Autism
Feeding
Row
Total
Brest-
feeding
Formula
with
DHA/ARA
Formula
without
DHA/ARA
Yes 12 39 65 116
No 6 22 10 38
Column
Total
4.) A person’s educational attainment and age group was collected by the U.S.
Census Bureau in 1984 to see if age group and educational attainment are related.
The counts in thousands are in table #11.1.8 ("Education by age," 2013). Do the
data show that educational attainment and age are independent? Test at the 5%
level.
Table #11.1.8: Educational Attainment and Age Group
Education
Age Group Row
25 - 34 35 - 44 45 - 54 55 - 64 >64 Total
Did not
complete
HS
Competed
HS
College 1- 3
years
College 4 or
more years
Column
Total 40173 20057 22240 22034 26290 130794
Chapter 11: Chi-Square Tests and ANOVA
5.) Students at multiple grade schools were asked what their personal goal (get good
grades, be popular, be good at sports) was and how important good grades were to
them (1 very important and 4 least important). The data is in table #11.1.
("Popular kids datafile," 2013). Do the data provide enough evidence to show
that goal attainment and importance of grades are independent? Test at the 5%
level.
Table #11.1.9: Personal Goal and Importance of Grades
Goal
Grades Importance Rating
1 2 3 4 Row Total
Grades 70 66 55 56 247
Popular 14 33 45 49 141
Sports 10 24 33 23 90
Column Total 94 123 133 128 478
6.) Students at multiple grade schools were asked what their personal goal (get good
grades, be popular, be good at sports) was and how important being good at sports
were to them (1 very important and 4 least important). The data is in table
#11.1.10 ("Popular kids datafile," 2013). Do the data provide enough evidence to
show that goal attainment and importance of sports are independent? Test at the
5% level.
Table #11.1.10: Personal Goal and Importance of Sports
Goal
Sports Importance Rating
1 2 3 4 Row Total
Grades 83 81 55 28 247
Popular 32 49 43 17 141
Sports 50 24 14 2 90
Column Total 165 154 112 47 478
7.) Students at multiple grade schools were asked what their personal goal (get good
grades, be popular, be good at sports) was and how important having good looks
were to them (1 very important and 4 least important). The data is in table
#11.1.11 ("Popular kids datafile," 2013). Do the data provide enough evidence to
show that goal attainment and importance of looks are independent? Test at the
5% level.
Table #11.1.11: Personal Goal and Importance of Looks
Goal
Looks Importance Rating
Row Total 1 2 3 4
Grades 80 66 66 35 247
Popular 81 30 18 12 141
Sports 24 30 17 19 90
Column Total 185 126 101 66 478
Chapter 11: Chi-Square Tests and ANOVA
Section 11.2: Chi-Square Goodness of Fit
In probability, you calculated probabilities using both experimental and theoretical
methods. There are times when it is important to determine how well the experimental
values match the theoretical values. An example of this is if you wish to verify if a die is
fair. To determine if observed values fit the expected values, you want to see if the
difference between observed values and expected values is large enough to say that the
test statistic is unlikely to happen if you assume that the observed values fit the expected
values. The test statistic in this case is also the chi-square. The process is the same as for
the chi-square test for independence.
Hypothesis Test for Goodness of Fit Test
- State the null and alternative hypotheses and the level of significance
H
o
: The data are consistent with a specific distribution
H
A
: The data are not consistent with a specific distribution
Also, state your α level here.
- State and check the assumptions for the hypothesis test
a. A random sample is taken.
b. Expected frequencies for each cell are greater than or equal to 5 (The expected
frequencies, E , will be calculated later, and this assumption means E ≥ 5 ).
- Find the test statistic and p-value
Finding the test statistic involves several steps. First the data is collected and
counted, and then it is organized into a table (in a table each entry is called a cell).
These values are known as the observed frequencies, which the symbol for an
observed frequency is O. The table is made up of k entries. The total number of
observed frequencies is n. The expected frequencies are calculated by
multiplying the probability of each entry, p , times n.
Expected frequency entry i
= E = n * p
Test Statistic:
χ
2
O − E
2
E
where O is the observed frequency and E is the expected frequency
Again, the test statistic involves squaring the differences, so the test statistics are
all positive. Thus a chi-squared test for goodness of fit is always right tailed.
p-value:
Using the TI- 83 /84: χcdf lower limit,1E 99 , df
Using R:
1 − pchisq χ
2
, df
Where the degrees of freedom is df = k − 1
Chapter 11: Chi-Squared Tests and ANOVA
- Conclusion
This is where you write reject H
o
or fail to reject H
o
. The rule is: if the p-value
< α , then reject H
o
. If the p-value ≥ α , then fail to reject H
o
- Interpretation
This is where you interpret in real world terms the conclusion to the test. The
conclusion for a hypothesis test is that you either have enough evidence to show
H
A
is true, or you do not have enough evidence to show H
A
is true.
Example #11.2.1: Goodness of Fit Test Using the Formula
Suppose you have a die that you are curious if it is fair or not. If it is fair then the
proportion for each value should be the same. You need to find the observed
frequencies and to accomplish this you roll the die 500 times and count how often
each side comes up. The data is in table #11.2.1. Do the data show that the die is
fair? Test at the 5% level.
Table #11.2.1: Observed Frequencies of Die
Die values 1 2 3 4 5 6 Total
Observed Frequency 78 87 87 76 85 87 100
Solution:
- State the null and alternative hypotheses and the level of significance
H
o
: The observed frequencies are consistent with the distribution for fair die (the
die is fair)
H
A
: The observed frequencies are not consistent with the distribution for fair die
(the die is not fair)
α = 0.
- State and check the assumptions for the hypothesis test
a. A random sample is taken since each throw of a die is a random event.
b. Expected frequencies for each cell are greater than or equal to 5. See step 3.
- Find the test statistic and p-value
First you need to find the probability of rolling each side of the die. The sample
space for rolling a die is {1, 2, 3, 4, 5, 6}. Since you are assuming that the die is
fair, then P ( 1 ) = P ( 2 ) = P ( 3 ) = P ( 4 ) = P ( 5 ) = P ( 6 ) =
Now you can find the expected frequency for each side of the die. Since all the
probabilities are the same, then each expected frequency is the same.
Expected frequency = E = n * p = 500 *
