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.
Note
Hypothesis Test for Chi-Square Test
State the null and alternative hypotheses and the level of significanceH o H_{o} H A H_{A} α \alpha State and check the assumptions for the hypothesis test
A random sample is taken. 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 E \geq 5 Find the test statistic and p-value 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.A, and one value of the second variable, B (the probability of a particular cell P ( A and B ) ) P(A \text { and } B) ) H o H_{o}
P
(
A
and
B
)
=
P
(
A
)
·
P
(
B
)
if
A
and
B
a
r
e
i
n
d
e
p
e
n
d
e
n
t
=
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
\begin{align*} P(A \text { and } B) &=P(A) \cdot P(B) \text { if } A \text { and } B are independent \\[4pt] &=\dfrac{\text { number of ways } A \text { can happen }}{\text { total number of individuals }} \cdot \dfrac{\text { number of ways } B \text { can happen }}{\text { total number of individuals }} \\[4pt] &= \dfrac{\text { row total }}{n} * \dfrac{\text { column total }}{n} \end{align*}
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 ) (\operatorname{cell} A \text { and } B)=E(A \text { and } B)
=
n
(
row total
n
·
column total
n
)
=
row total -column total
n
\begin{array}{l}{=n\left(\dfrac{\text { row total }}{n} \cdot \dfrac{\text { column total }}{n}\right)} \\ {=\dfrac{\text { row total -column total }}{n}}\end{array}
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 \chi^{2}
χ
2
=
∑
(
O
−
E
)
2
E
\chi^{2}=\sum \dfrac{(O-E)^{2}}{E}
where O is the observed frequency and E is the expected frequency
Distribution of Chi-Square
χ 2 \chi^{2} Figure 1 \PageIndex{1} ). Since the test statistic involves squaring the differences, the test statistics are all positive. A chi-squared test for independence is always right tailed.
p-value:
Using the TI-83/84: χ cdf (lower limit, 1 E 99 , d f ) \chi \text { cdf (lower limit, } 1 \mathrm{E} 99, d f )
Using R: 1 − pchisq ( x 2 , d f ) 1-\text { pchisq }\left(x^{2}, d f\right)
Where the degrees of freedom is d f = ( # of rows − 1 ) * ( # of columns − 1 ) d f=(\# \text { of rows }-1) *(\# \text { of columns }-1)
4. Conclusion
This is where you write reject H o H_{o} H o H_{o} α \alpha H o H_{o} ≥ α \geq \alpha H o H_{o}
5. 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 H_{A} H A H_{A}
Example 1 hypothesis test with chi-square test using formula
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.
Show solution
1. State the null and alternative hypotheses and the level of significance
H o H_{o}
H A H_{A}
α \alpha
2. State and check the assumptions for the hypothesis test
A random sample of breastfeeding time frames and autism incidence was taken. Expected frequencies for each cell are greater than or equal to 5 (ie. E ≥ 5 E \geq 5 3. Find the test statistic and p-value
Test statistic:
First find the expected frequencies for each cell
E
(
Autism and no breastfeeding
)
=
818
*
261
934
≈
228.585
E(\text { Autism and no breastfeeding })=\dfrac{818^{*} 261}{934} \approx 228.585
E
(
Autism and
<
2
months
)
=
818
*
223
934
≈
195.304
E(\text { Autism and }<2 \text { months })=\dfrac{818^{*} 223}{934} \approx 195.304
E
(
Autism and
2
to
6
months
)
=
818
*
191
934
≈
167.278
E(\text { Autism and } 2 \text { to } 6 \text { months })=\dfrac{818^{*} 191}{934} \approx 167.278
E
(
Autism and more than
6
months
)
=
818
*
259
934
≈
226.833
E(\text { Autism and more than } 6 \text { months })=\dfrac{818 * 259}{934} \approx 226.833
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.)
The test statistic formula is χ 2 = ∑ ( O − E ) 2 E \chi^{2}=\sum \dfrac{(O-E)^{2}}{E} 2 \PageIndex{2}
p-value:
d
f
=
(
2
−
1
)
*
(
4
−
1
)
=
3
d f=(2-1)^{*}(4-1)=3
Using TI-83/84: χ cdf ( 11.2166432 , 1 E 99 , 3 ) ≈ 0.01061 \chi \operatorname{cdf}(11.2166432,1 \mathrm{E} 99,3) \approx 0.01061
Using R: 1 − pchisq ( 11.2166432 , 3 ) ≈ 0.01061566 1-\text{pchisq}(11.2166432,3) \approx 0.01061566
pchisq(11.2166432, 3)4. Conclusion
Fail to reject H o H_{o}
5. 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 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 Example 1 \PageIndex{1}
Show solution
1. State the null and alternative hypotheses and the level of significance
H o H_{o}
H A H_{A}
α \alpha
2. State and check the assumptions for the hypothesis test
A random sample of breastfeeding time frames and autism incidence was taken. Expected frequencies for each cell are greater than or equal to 5 (ie. E ≥ 5 E \geq 5 3. Find the test statistic and p-value
Test statistic:
To use the TI-83/84 calculator to compute the test statistic, you must first put the data into the calculator. However, this process is different than for other hypothesis tests. You need to put the data in as a matrix instead of in the list. Go into the MATRX menu then move over to EDIT and choose 1:[A]. This will allow you to type the table into the calculator. Figure 2 \PageIndex{2} shows what you will see on your calculator when you choose 1:[A] from the EDIT menu.
The table has 2 rows and 4 columns (don’t include the row total column and the column total row in your count). You need to tell the calculator that you have a 2 by 4. The 1 X1 (you might have another size in your matrix, but it doesn’t matter because you will change it) on the calculator is the size of the matrix. So type 2 ENTER and 4 ENTER and the calculator will make a matrix of the correct size. See Figure 3 \PageIndex{3} .
Now type the table in by pressing ENTER after each cell value. Figure 4 \PageIndex{4} contains the complete table typed in. Once you have the data in, press QUIT.
To run the test on the calculator, go into STAT, then move over to TEST and choose χ 2 \chi^{2} Figure 5 \PageIndex{5} .
Once you press ENTER on Calculate you will see the results in Figure 6 \PageIndex{6} .
The test statistic is χ 2 ≈ 11.2167 \chi^{2} \approx 11.2167 p ≈ 0.01061 p \approx 0.01061 Figure 7 \PageIndex{7} shows the output. Press the right arrows to see the entire matrix.
To compute the test statistic and p-value with R,
For this example, the commands would be
row1 = c(241, 198, 164, 215)row2 = c(20, 25, 27, 44)data.table = rbind(row1, row2)Output:
chisq.test(data.table)
chisq.test(data.table)Output:
data: data.table
chisq.test(data.table)$expected
Output: [,1] [,2] [,3] [,4]
The test statistic is χ 2 ≈ 11.217 \chi^{2} \approx 11.217 p ≈ 0.01061 p \approx 0.01061
4. Conclusion
Fail to reject H o H_{o}
5. 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 3 hypothesis test with chi-square test using formula
The World Health Organization (WHO) keeps track of how many incidents of leprosy there are in the world. Using the WHO regions and the World Banks income groups, one can ask if an income level and a WHO region are dependent on each other in terms of predicting where the disease is. Data on leprosy cases in different countries was collected for the year 2011 and a summary is presented in Table 3 \PageIndex{3} ("Leprosy: Number of," 2013). Is there evidence to show that income level and WHO region are independent when dealing with the disease of leprosy? Test at the 5% level.
Show solution
1. State the null and alternative hypotheses and the level of significance
H o H_{o}
H A H_{A}
α \alpha
2. State and check the assumptions for the hypothesis test
A random sample of incidence of leprosy was taken from different countries and the income level and WHO region was taken. Expected frequencies for each cell are greater than or equal to 5 (ie. E ≥ 5 E \geq 5 3. Find the test statistic and p-value
Test statistic:
First find the expected frequencies for each cell.
E
(
Americas and High Income
)
=
36817
*
264
222545
≈
43.675
E(\text { Americas and High Income })=\dfrac{36817 * 264}{222545} \approx 43.675
E
(
Americas and Upper Middle Income
)
=
36817
*
36289
222545
≈
6003.514
E(\text { Americas and Upper Middle Income })=\dfrac{36817 * 36289}{222545} \approx 6003.514
E
(
Americas and Lower Middle Income)
=
36817
*
158069
222545
≈
26150.335
E (\text { Americas and Lower Middle Income) }=\dfrac{36817 * 158069}{222545} \approx 26150.335
E
(
Americas and Lower Income
)
=
36817
*
27923
222545
≈
4619.475
E(\text { Americas and Lower Income })=\dfrac{36817 * 27923}{222545} \approx 4619.475
Others are done similarly. It is easier to do the calculations for the test statistic with a table, and the others are in Example 4 \PageIndex{4}
The test statistic formula is χ 2 = ∑ ( O − E ) 2 E \chi^{2}=\sum \dfrac{(O-E)^{2}}{E} 2 \PageIndex{2}
p-value:
d
f
=
(
6
−
1
)
*
(
4
−
1
)
=
15
d f=(6-1) *(4-1)=15
Using the TI-83/84: χ cdf ( 328594.008 , 1 E 99 , 15 ) ≈ 0 \chi \operatorname{cdf}(328594.008,1 \mathrm{E} 99,15) \approx 0
Using R: 1 − pchisq ( 328594.008 , 15 ) ≈ 0 1-\text { pchisq }(328594.008,15) \approx 0
pchisq(328594.008, 15)4. Conclusion
Reject H o H_{o}
5. 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 5.
Example 4 hypothesis test with chi-square test using technology
The World Health Organization (WHO) keeps track of how many incidents of leprosy there are in the world. Using the WHO regions and the World Banks income groups, one can ask if an income level and a WHO region are dependent on each other in terms of predicting where the disease is. Data on leprosy cases in different countries was collected for the year 2011 and a summary is presented in Table 3 \PageIndex{3} ("Leprosy: Number of," 2013). Is there evidence to show that income level and WHO region are independent when dealing with the disease of leprosy? Test at the 5% level.
Show solution
1. State the null and alternative hypotheses and the level of significanceH o H_{o}
H A H_{A}
α \alpha
2. State and check the assumptions for the hypothesis test
A random sample of incidence of leprosy was taken from different countries and the income level and WHO region was taken. Expected frequencies for each cell are greater than or equal to 5 (ie. E ≥ 5 E \geq 5 3. Find the test statistic and p-value2 \PageIndex{2}
χ
2
≈
328594.0079
\chi^{2} \approx 328594.0079
Press the right arrow to look at the other expected frequencies.
p-value:
p
−
value
≈
0
p-\text {value} \approx 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)Pearson's Chi-squared test
data: data.table
Warning message:
chisq.test(data.table)$expected
[
,
1
]
[
,
2
]
[
,
3
]
[
,
4
]
row1
43.67515783
6003.514404
2.615034
e
+
04
4619.475122
row2
3.02144735
415.323117
1.809080
e
+
03
319.575281
row3
0.01186277
1.630637
7.102788
e
+
00
1.254713
row4
6.03340448
829.341724
3.612478
e
+
03
638.146793
row5
21.29722977
2927.481709
1.275164
e
+
04
2252.585405
row6
189.96089780
26111.708410
1.137384
e
+
05
20091.962686
\begin{array} {ccccc} {}&{[,1]}&{[,2]}&{[,3]}&{[,4]} \\ {\text{row1}}&{43.67515783}&{6003.514404}&{2.615034e+04}&{4619.475122}\\{\text{row2}}&{3.02144735}&{415.323117}&{1.809080e+03}&{319.575281}\\ {\text{row3}}&{0.01186277}&{1.630637}&{7.102788e+00}&{1.254713}\\{\text{row4}}&{6.03340448}&{829.341724}&{3.612478e+03}&{638.146793}\\{\text{row5}}&{21.29722977}&{2927.481709}&{1.275164e+04}&{2252.585405}\\{\text{row6}}&{189.96089780}&{26111.708410}&{1.137384e+05}&{20091.962686} \end{array}
Warning message:
χ 2 = 328590 \chi^{2}=328590 2.2 × 10 − 16 2.2 \times 10^{-16}
4. Conclusion
Reject H o H_{o}
5. 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 5.
Homework
Your Turn
In each problem show all steps of the hypothesis test. If some of the assumptions are not met, note that the results of the test may not be correct and then continue the process of the hypothesis test.
The number of people who survived the Titanic based on class and sex is in Example 5 \PageIndex{5} Researchers watched groups of dolphins off the coast of Ireland in 1998 to determine what activities the dolphins partake in at certain times of the day ("Activities of dolphin," 2013). The numbers in Example 6 \PageIndex{6} 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 Example 7 \PageIndex{7} 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 Example 8 \PageIndex{8} 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 Example 9 \PageIndex{9} 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 Example 10 \PageIndex{10} 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 ("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.
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 money were to them (1 very important and 4 least important). The data is in Table 12 ("Popular kids datafile," 2013). Do the data provide enough evidence to show that goal attainment and importance of money are independent? Test at the 5% level.
Answer
For all hypothesis tests, just the conclusion is given. See solutions for the entire answer.
1. Reject Ho
3. Reject Ho
5. Reject Ho
7. Reject Ho
Adapted from Statistics with Technology, 2nd Edition, by Kathryn Kozak (Coconino Community College), © Kathryn Kozak, licensed under CC BY-SA 4.0. Changes were made. License: CC-BY-SA-4.0 .
