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Calculator steps:
The calculator steps for this section can be
found on YouTube or by clicking: Click
here
Bluman, Chapter 11 1
While you wait: Copy the table below.
Allow room for data to be added in each cell.
Fill in the last row and column.
Bluman, Chapter 11 2
Location No College
Four-Year
Degree
Advanced
Degree Total
Urban 15 12 8
Suburban 8 15 9
Rural 6 8 7
Total `
Tests Using
Contingency
Tables Sec: 11.2
Bluman, Chapter 11 3
11.2 Tests Using Contingency
Tables When data can be tabulated in table form in terms of
frequencies, several types of hypotheses can be tested
by using the chi-square test.
The test of independence of variables is used to
determine whether two variables are independent of or
related to each other when a single sample is selected.
The test of homogeneity of proportions is used to
determine whether the proportions for a variable are
equal when several samples are selected from
different populations.
Bluman, Chapter 11 4
Test for Independence
The chi-square goodness-of-fit test can be used
to test the independence of two variables.
The hypotheses are:
H0: There is no relationship between two
variables.
H1: There is a relationship between two
variables.
If the null hypothesis is rejected, there is some
relationship between the variables.
Bluman, Chapter 11 5
Test for Independence
In order to test the null hypothesis, one must
compute the expected frequencies, assuming
the null hypothesis is true.
When data are arranged in table form for the
independence test, the table is called a
contingency table.
Bluman, Chapter 11 6
Contingency Tables
Bluman, Chapter 11 7
The dimension of the contingency table
is given: ROW by COLUMN
Contingency Tables
The degrees of freedom for any contingency
table are d.f. = (rows – 1) (columns – 1) =
(R – 1)(C – 1).
Bluman, Chapter 11 8
The table above is 2 by 3. Only count the
numerical cells.
Test for Independence
The formula for the test for independence:
where
d.f. = (R – 1)(C – 1)
O = observed frequency
E = expected frequency =
Bluman, Chapter 11 9
2
2
O E
E
row sum column sum
grand total
Example (see page 618): College
Education and Place of Residence A sociologist wishes to see whether the number of years
of college a person has completed is related to her or his
place of residence. A sample of 88 people is selected
and classified as shown. At α = 0.05, can the sociologist
conclude that a person’s location is dependent on the
number of years of college?
Bluman, Chapter 11 10
Location
No
College
Four-Year
Degree
Advanced
Degree Total
Urban 15 12 8 35
Suburban 8 15 9 32
Rural 6 8 7 21
Total 29 35 24 88
Example 11-5: College Education and
Place of Residence
Bluman, Chapter 11 11
Step 1: State the hypotheses and identify the claim.
H0: A person’s place of residence is independent
of the number of years of college completed.
H1: A person’s place of residence is dependent on
the number of years of college completed
(claim).
Step 2: Find the critical value.
The critical value is 9.488, since the degrees of
freedom are (3 – 1)(3 – 1) = 4.
Bluman, Chapter 11 12
Location
No
College
Four-Year
Degree
Advanced
Degree Total
Urban 15
12 8 35
Suburban 8
15 9 32
Rural 6
8 7 21
Total 29 35 24 88
Compute the expected values.
row sum column sum
grand totalE
1,1
35 2911.53
88 E
(11.53)
(10.55)
(6.92)
(13.92)
(12.73)
(8.35)
(9.55)
(8.73)
(5.73)
See next two slides for
calculator use.
Bluman, Chapter 11 13
[A]
Enter Observed values
as Matrix 1: 2nd x-1
[B]
Enter Expected values as Matrix
Bluman, Chapter 11 14
Test statistics :
2=3.006
College Education and Place of
Residence
Bluman, Chapter 11 15
Step 3: Compute the test value.
2
2
O E
E
2 2 2
2 2 2
2 2 2
15 11.53 12 13.92 8 9.55
11.53 13.92 9.55
8 10.55 15 12.73 9 8.73
10.55 12.73 8.73
6 6.92 8 8.35 7 5.73
6.92 8.35 5.73
3.01
College Education and Place of
Residence
Bluman, Chapter 11 16
Step 4: Make the decision.
Do not reject the null hypothesis, since 3.01<9.488.
Step 5: Summarize the results.
There is not enough evidence to support the claim
that a person’s place of residence is dependent on
the number of years of college completed.
Chapter 11 Other Chi-Square Tests
Section 11-2
Example 11-6
Page # 610
Bluman, Chapter 11 17
Example 11-6: Alcohol and Gender
A researcher wishes to determine whether there is a
relationship between the gender of an individual and the
amount of alcohol consumed. A sample of 68 people is
selected, and the following data are obtained. At α = 0.10,
can the researcher conclude that alcohol consumption is
related to gender?
Bluman, Chapter 11 18
Gender
Alcohol Consumption
Total Low Moderate High
Male 10 9 8 27
Female 13 16 12 41
Total 23 25 20 68
Example 11-6: Alcohol and Gender
Step 1: State the hypotheses and identify the claim.
H0: The amount of alcohol that a person
consumes is independent of the individual’s
gender.
H1: The amount of alcohol that a person
consumes is dependent on the individual’s
gender (claim).
Step 2: Find the critical value.
The critical value is 4.605, since the degrees of
freedom are (2 – 1 )(3 – 1) = (1)(2) = 2.
Bluman, Chapter 11 19
Example 11-6: Alcohol and Gender
Bluman, Chapter 11 20
Gender
Alcohol Consumption
Total Low Moderate High
Male 10
9 8 27
Female 13
16 12 41
Total 23 25 20 68
Compute the expected values.
row sum column sum
grand totalE
1,1
27 239.13
68 E
(9.13) (9.93) (7.94)
(13.87) (15.07) (12.06)
Example 11-6: Alcohol and Gender
Step 3: Compute the test value.
Bluman, Chapter 11 21
2
2
O E
E
2 2 2
2 2 2
10 9.13 9 9.93 8 7.94
9.13 9.93 7.94
13 13.87 16 15.07 12 12.06
13.87 15.07 12.06
χ2 = 0.281
Example 11-6: Alcohol and Gender
Step 4: Make the decision.
Do not reject the null hypothesis, since
0.281 < 4.605.
.
Step 5: Summarize the results.
There is not enough evidence to support the claim
that the amount of alcohol a person consumes is
dependent on the individual’s gender.
Bluman, Chapter 11 22
Bluman, Chapter 11 23
Please record the table below in your
notes. Allow for additional col and row Household
income
Less than
$30,000 ( 24%)
$30,000-
$74,999
(33%)
$75,000-
$99,999
(38%)
$100,000 or
more (49%)
YES 24 33 38 49
Bluman, Chapter 11 24
Please record the table below in your
notes. Allow for additional col and row
Viewing Preferences
Lone Ranger Sesame Street The Simpsons
Boys 50 30 20
Girls 50 80 70
Test for Homogeneity of
Proportions Homogeneity of proportions test is used
when samples are selected from several
different populations and the researcher is
interested in determining whether the
proportions of elements that have a common
characteristic are the same for each population.
Bluman, Chapter 11 25
Test for Homogeneity of
Proportions The hypotheses are:
H0: p1 = p2 = p3 = … = pn
H1: At least one proportion is different from
the others.
When the null hypothesis is rejected, it can be
assumed that the proportions are not all equal.
Bluman, Chapter 11 26
Assumptions for Homogeneity of
Proportions
1. The data are obtained from a random sample.
2. The expected frequency for each category
must be 5 or more.
Bluman, Chapter 11 27
Chapter 11 Other Chi-Square Tests
Section 11-2
Example 11-7
Page #611
Bluman, Chapter 11 28
Money and Happiness A psychologist selected 100 people from
each of 4 income groups and asked them if
they were “VERY HAPPY”. The percent of
each group who said yes are shown. At
a=0.05 test the claim that there is no
difference in proportions.
Bluman, Chapter 11 29
Household
income
Less than
$30,000 ( 24%)
$30,000-
$74,999
(33%)
$75,000-
$99,999
(38%)
$100,000 or
more (49%)
YES 24 33 38 49
Complete the table:
Bluman, Chapter 11 30
Household
income
Less than
$30,000 (
24%)
$30,000-
$74,999
(33%)
$75,000-
$99,999
(38%)
$100,000 or
more (49%) Total
YES 24 33 38 49
NO
Total
Household
income
Less than
$30,000 (
24%)
$30,000-
$74,999
(33%)
$75,000-
$99,999
(38%)
$100,000 or
more (49%) Total
YES 24 33 38 49 144
NO 76 67 62 51 256
Total 100 100 100 100 400
Example 11-7: Money and Happiness
Step 1: State the hypotheses.
H0: p1 = p2 = p3 = p4
H1: At least one proportion differs from the other.
Step 2: Find the critical value.
The critical value is 7.815, since the degrees of
freedom are (2 – 1 )(4 – 1) = 3.
Bluman, Chapter 11 31
Example 11-7: Money and Happiness
Step 4: Make the decision.
Reject the null hypothesis, since 14.149 > 7.815
.
Step 5: Summarize the results.
There is enough evidence to reject the claim that
there is a difference proportions. Hence it seems
that there is a difference in the proportions of the
income.
Bluman, Chapter 11 32
Example 11-7: Lost Luggage
A researcher selected 100 passengers from each of 3
airlines and asked them if the airline had lost their
luggage on their last flight. The data are shown in the
table. At α = 0.05, test the claim that the proportion of
passengers from each airline who lost luggage on the
flight is the same for each airline.
Bluman, Chapter 11 33
Airline 1 Airline 2 Airline 3 Total
Yes 10 7 4 21
No 90 93 96 279
Total 100 100 100 300
Example 11-7: Lost Luggage
Step 1: State the hypotheses.
H0: p1 = p2 = p3
H1: At least one proportion differs from the other.
Step 2: Find the critical value.
The critical value is 5.991, since the degrees of
freedom are (2 – 1 )(3 – 1) = (1)(2) = 2.
Bluman, Chapter 11 34
Example 11-7: Lost Luggage
Bluman, Chapter 11 35
Compute the expected values.
row sum column sum
grand totalE
1,1
21 1007
300 E
Airline 1 Airline 2 Airline 3 Total
Yes 10
7 4 21
No 90
93 96 279
Total 100 100 100 300
(7) (7) (7)
(93) (93) (93)
Example 11-7: Luggage
Step 3: Compute the test value.
Bluman, Chapter 11 36
2
2
O E
E
2 2 2
2 2 2
10 7 7 7 4 7
7 7 7
90 93 93 93 96 93
93 93 93
2.765
Example 11-7: Lost Luggage
Step 4: Make the decision.
Do not reject the null hypothesis, since
2.765 < 5.991.
.
Step 5: Summarize the results.
There is not enough evidence to reject the claim
that the proportions are equal. Hence it seems that
there is no difference in the proportions of the
luggage lost by each airline.
Bluman, Chapter 11 37
In a study of the television viewing habits of
children, a developmental psychologist selects a
random sample of 300 first graders: 100 boys
and 200 girls. Each child is asked which of the
following TV programs they like best: The Lone
Ranger, Sesame Street, or The Simpsons.
Results are shown in the contingency table to
follow.
Bluman, Chapter 11 38
Bluman, Chapter 11 39
Viewing Preferences
Lone Ranger Sesame Street The Simpsons
Boys 50 30 20
Girls 50 80 70
Do the boys' preferences for these TV
programs differ significantly from the
girls' preferences? Use a 0.05 level of
significance.
Step 1:
Bluman, Chapter 11 40
H0: Pboys who prefer Lone Ranger = Pgirls who prefer Lone Ranger
H0: Pboys who prefer Sesame Street = Pgirls who prefer Sesame Street
H0: Pboys who prefer The Simpsons = Pgirls who prefer The Simpsons
Null hypothesis: The null hypothesis
states that the proportion of boys who
prefer the Lone Ranger is identical to
the proportion of girls. Similarly, for the
other programs.
Alternative hypothesis: At least
one of the null hypothesis
statements is false.
Calculations:
Determine the expected values:
Bluman, Chapter 11 41
Viewing Preferences Row total
Lone Ranger Sesame Street The Simpsons
Boys 50 30 20 100
Girls 50 80 70 200
Column total 100 110 90 300
On your own
Read section 11.2
and all its examples.
Sec 11.2 page 614
#1-7 all;
And #9, 17, 23, 27, 31
Bluman, Chapter 11 42
Example
Suppose a new postoperative procedure is
administered to a number of patients in a large
hospital. Do the doctors feel differently about this
procedure from the nurses, or do they feel basically
the same?
The question is not whether the
doctors prefer the procedure but
whether there is a difference of
opinion between doctors and
nurses.
Hypothesis:
H0: the opinion about the procedure is
independent of the profession.
H1: the opinion about the procedure is
dependent of the profession.
Group Prefers
new
Prefers
old
No pref
nurses 100 80 20
doctors 50 120 30
Consider the result listed
below:
Chapter 11 Other Chi-Square Tests
Section 11-2
Example 11-5
Page #606
Bluman, Chapter 11 46
Hospitals and Infections
Bluman, Chapter 11 47
Chapter 11 Other Chi-Square Tests
Section 11-2
618
Page #606
Bluman, Chapter 11 48