MODULE 3.3
December 2 – 9, 2020
Chapter
12 Chi- Square Tests
Learning Objectives
After completing this chapter, the students will able to:
o Explain the characteristics of the chi-square distribution.
o Determines the assumptions for chi-square distribution.
o Test a hypothesis on the difference between an observed and expected set of
frequencies.
o Test a distribution for goodness-of-fit.
o Test two variables for dependence.
o Test proportions for homogeneity.
Chapter Outline
12.1 Introduction
12.2The Chi-Square Distribution
12.3Goodness-of-Fit Test
Equal Expected Frequencies
Unequal Expected Frequencies
12.4 Test of Independence and Homogeneity of Proportions
Test for Independence
Test for Homogeneity of Proportions
Statistics: the mathematical theory of ignorance.
-- Morris Kline
…….
313
12.1 Introduction
Pearson’s chi-square test, also known as the chi-square goodness-of-fit test or chi-square
test for independence. The symbol for chi-square is χ2 (Greek letter chi, pronounced
“ki”). A chi square test(also chi-squared orχ2 test) is any statistical hypothesis test in
which the sampling distribution of the test statistic is a chi-square distribution when the
null hypothesis is true, or any in which this is asymptoticallytrue, meaning that the
sampling distribution (if the null hypothesis is true) can be made to approximate a chisquare distribution as closely as desired by making the sample size large enough.
There are two types of random variables and they yield two types of data: numerical and
categorical. A chi square (χ2) statistic is used to investigate whether distributions of
categorical variables differ from one another. The table below could help you see the
differences between these two variables.
Data Type
Categorical
Numerical
Numerical
Question Type
What is your sex?
Discrete – How many credit cards do you have?
Continuous – What is your weight?
Possible Responses
Male or Female
Two or three
50 kilograms
12.2 The Chi- Square Distribution
In Chi-square significance tests, random sample data are assumed and a sufficiently
large sample size is also assumed. Applying chi-square to small samples exposes the
researcher to an unacceptable rate of Type II errors. In addition, adequate cell sizes are
also assumed. A common rule is 5 or more in all cells of a 2-by-2 table and 5 or more in
80% of cells in larger tables, but no cells with zero count.
In Chi-square significance tests observations must be independent and the same
observation can only appear in one cell. This means chi-square cannot be used to test
correlated data (e.g. before-and-after, matched pairs and panel data). Also, observations
must have the same underlying distribution. The hypothesized distribution is specified in
advance so that the number of observations that are expected to appear in each cell of the
table can be calculated without reference to observed values.
Chi-square significance tests assumed a non-directional hypothesis and test the
hypothesis at two variables are related only by chance. If a significant relationship is
found, this is not equivalent to establish the researcher’s hypothesis, this A causes B, or
that B causes A. The test also grouped the observations in categories.
Characteristics of Chi-Square Distribution
The computed chi-square is always positive.
The shape of the chi-square distribution does not depend on the size of the sample.
There is a family of chi-square distributions.
The chi-square is positively skewed.
Chi-square can be applied in nominal or ordinal level of data.
Figure 12.1 shows the several chi-square distributions with corresponding degrees of
freedom. Chi-square distribution is similar to the t and F distribution in that there is a
family of χ2 distribution, each with a different shape, depending on the number of degrees
of freedom. When the number of degrees of freedom of a chi-square distribution is small
the distribution is positively skewed, but as the number of degrees of freedom increases it
becomes symmetrical and approaches the normal distribution (at about 100 degrees of
freedom, the chi-square becomes somewhat normally distributed). The area under the
chi-square distribution is equal to 1.00 or 100%.
Figure 12.1: The Chi-Square
Family of Curves
12.3 Goodness-of-Fit Test [One Sample Test]
The chi-square goodness-of-fit test is simply a different usage of Pearsonian chisquare. Chi-square goodness-of-fit test is a non-parametric test (or distribution-free tests)
and it is used in Chi-square significance tests to test if an observed distribution conforms
to any other distribution, such as one based on theoretical expected distribution (e.g., if
the observed distribution is not significantly different from a normal distribution) or one
based on some other known distribution (e.g., if the observed distribution is not
significantly different from a known distribution).
The term goodness-of-fit is used to compare the observed sample distribution with the
expected probability distribution. This test determines how well theoretical distribution
(such as normal, binomial, or Poisson) fits the empirical distribution. In Chi-square
goodness-of-fit test, sample data is divided into intervals. Then the numbers of points that
fall into the interval are compared, with the expected numbers of points in each interval.
Assumptions in Chi-Square Goodness-of-fit test:
1. Subjects are randomly selected.
2. Categories are mutually exclusive.
Procedure for Chi-Square Goodness-of-fit test:
1. Set up the hypothesis for Chi-Square goodness-of-fit test:
H0: The null hypothesis assumes that there is no significant difference between
theobserved and expected value.
H1: The alternative hypothesis assumes that there is a significant between theobserved
and the expected value.
It can also be written as:
H0: The distribution is uniform.
H1:The distribution is not uniform.
2. Degree of freedom: It is determine by subtracting 1 to the number of categories in the
distribution or n-1.
3. Compute for the expected frequencies.
4. Compute the value of Chi-Square goodness-of-fit test using Formula 12-1:
χ2= ∑
(O−E)²
(Formula 12-1)
E
where: χ2= Chi-square value.
O = observed value.
E = expected value.
Statistical decision for hypothesis testing: The calculated value of χ2 test is
compared with the table value. If the computed value of χ2 is greater than the critical
value, we will reject the null hypothesis and conclude that there is a significant difference
between the observed and the expected frequency. On the contrary, if the computed value
of χ2 is less than the critical value, we will accept the null hypothesis and conclude that
there is no significant difference between the observed and expected value. In symbol,
If χ2computed<χ2critical, do not reject H0.
If χ2computed ≥ χ2critical, reject H0.
Region of
Rejection (α)
Do not rejectH0
Scale χ2
Critical
Value
5. State the conclusion.
It is presented in the succeeding examples how to apply chi-square test for equal and
equal expected frequencies.
A. Equal Expected Frequencies
Example1:A bank manager wanted to investigate if the percentage of loan applications
made is the same for each of the five days (Monday through Friday) of the week. He
randomly selected one week and counted the number of loan applications made on each
of the five days. This information he obtained is given in the following table.
Day
Number of loan application
Monday
17
Tuesday
20
Wednesday
16
Thursday
14
Friday
13
Using level of significance of 0.05, can we reject the claim that the proportion of people
who filed loan applications in the bank each of the five days of the week is the same?
Assume that this week is typical of all the weeks in regard to the applications of loans.
Solution:
Step1: State the hypothesis.
H0: p1=p2=p3=p4=p5 (claim)
(The proportion of people filing loan application is the same for all five days
of the week.)
H1:At least two of the five proportions are not equal to 0.20
(The proportion of people filing loan application is not the same for all five days of the
week.)
Step 2:The level of significance is α = 0.50.
Step 3:Determine the critical region. (Refer to Table D)
df = c – 1 = 5 – 1 = 4
χ2critical = 9.488
Table D: Critical values for Chi-Square Tests
Degrees of Freedom
1
2
3
4
0.10
2.706
4.605
6.251
7.779
0.05
3.841
5.991
7.815
9.488
0.025
5.024
7.378
9.348
11.143
χ2critical
Step 4: Compute for the value of χ2.
O
p
E= np
17
20
16
14
13
Total = 80
0.20
0.20
0.20
0.20
0.20
1.00
80(0.20)= 16
80(0.20) = 16
80(0.20) = 16
80(0.20) = 16
80(0.20) = 16
80
χ² = ∑
O–E
(O – E)²
1.00
4.00
0.00
-2.00
-3.00
0.00
1.00
16.00
0.00
4.00
9.00
(O − E)²
E
0.0625
1.0000
0.0000
0.2500
0.5625
∑
(O−E)²
E
= 1.8750
(O−E)²
= 1.875
E
Step 5: Decision rule.
Since the computed χ2of 1.875 is less than the χ2critical value of 9.488 at level of
significance of 0.05, the statistical decision is not to reject the null hypothesis.
Region of Rejection
(α = 0.05)
1.875
9.488
Step 6: Conclusion
Since the null hypothesis has not been rejected, we can conclude that there is
evidence that the proportion of people filing loan application is the same for all five days
of the week.
B. Unequal Expected Frequencies
Example 2: A national study of cinema admissions during three-month period revealed
these statistics concerning people watching movies and who entered the cinema during
the period:
Number of movies watch
1
2
3
4
5
Percent of total
40%
30%
15%
10%
5%
The cinema administrator is anxious to compare his SJS Cinema experience with the
national experience. He selected 200 movie goers and determined the number of times
during the three-month period that each was watched movies in their cinema. The
observation frequencies are as follows:
Number of movies watch
Number of movie goers
1
2
3
4
5
100
70
17
8
5
The 0.01 level of significance, can we reject the null hypothesis that the there is no
difference between the experience SJS Cinema and the national experience.
Solution:
Step 1: State the hypothesis
H0:There is no difference between the SJS Cinema and the national experience.
H1:There is a difference between the SJS Cinema and the national experience.
Step 2: The level of significance is α = 0.01
Step 3: Determine the critical region. (Refer to Table D)
df = c – 1 = 5 – 1 = 4
χ2critical = 13.277
Step 4: Compute for the value of χ2.
Number
of Movies
O
1
2
3
4
5
100
70
17
8
5
χ2= ∑
(O−E)2
E
p
40%
30%
15%
10%
5%
E = Op
200(0.40) = 80
200(0.30) = 60
200(0.15) = 30
200(0.10) = 20
200(0.05) = 10
O–E
(O – E)²
20
10
-13
-12
-5
400
100
169
144
25
(O − E)²
E
5.00
1.67
5.63
7.20
2.50
= 5.00 + 1.67 + 5.63 + 7.20 + 2.50 = 22
Step 5: Decision rule.
Since the computed χ2 of 22 is greater than the χ2 critical value of 13.277 at level
of significance of 0.01, the statistical decision is to reject the null hypothesis.
Region of Rejection
(α = 0.01)
Step 6: Conclusion.
Since the null hypothesis has been rejected, we can conclude that there is evidence
of significant difference between the SJS Cinema and the national experience. The
cinema administrator would conclude that the SJS Cinema situation is different in
other cinemas.
12.4 Test of Independence or Homogeneity of Proportions
In the Chi-Square test of Independence, the frequency of one nominal variable is
compared with different values of the second nominal variable. This test is used when we
have two nominal variables, data may be in the row and column form, and the test
variable may be more than two.
Chi-square test is far the most common type of chi-square significance test. It is used
to test the hypothesis of no association of columns and rows in tabular data. This statistic
can be used with nominal data. It is more likely to establish significance to the extent that
the relationship is wrong, the sample size is large and the number of values of the two
associated variables is also large.
Chi-square uses a probability of 0.05 or less, is commonly interpreted by social
scientists as the justification for rejecting the null hypothesis that the row variable is
unrelated to the column variable.
Assumptions in Chi-Square Goodness-of-fit test:
Subjects are randomly selected.
Observations have been classified simultaneously in two independent categories.
Procedure for Chi-Square Test of Independence of Homogeneity:
1. Set up the hypothesis of Chi-Square Test of Independence of Homogeneity:
For a contingency table that has r rows and c columns, the chi-square test can be
thought of as a test of independence. In a test of independence or homogeneity the null
and alternative hypotheses are:
H0: The null hypothesis assumes that there is no association between two variables.
H1: the alternative hypothesis assumes that there is an association between two variables.
It can also be written as:
H0: The two categorical variables are independent.
H1: The two categorical variables are related.
2. Degree of freedom: The degree of freedom is calculated by using the following
formula:
df = (r - 1) (c – 1)
(Formula 12-3)
where: df = degree of freedom for the Chi-square test of independence.
r = number of rows in the Chi-square test of independence.
c = number of columns in the Chi-square test of independence.
3. Compute for the expected frequencies:
Here O denotes the frequency of the observed data and E is the frequency of the expected
values. The general table would look something like the one below:
Category B
R1
R2
:
Rn
Total
C1
O1, 1
O2,1
:
On,1
CT1
Category A
…
…
…
…
…
…
C2
O1, 2
O2, 2
:
On, 2
CT2
Cm
O1, m
O2, m
:
On, m
CTm
Total
RT1
RT2
:
RTn
GT
Then we need to calculate the expected values for each cell in the table and we can do
that using the row total times the column total divided by the grand total. It is shown in
the table below the computation of the expected frequencies.
Expected Frequency =
Category B
R1
R2
:
Rn
E1,1
E2,1
En,1
(RowTable)(ColumnTotal)
(Formula 12-2)
TotalSampleSize
Category A
C2
(RT1 )(CT2 )
=
GT
(RT2 )(CT2 )
=
GT
C1
(RT1 )(CT1 )
=
E1,2
GT
(RT2 )(CT1 )
=
E2,2
GT
:
:
(RTn )(CT1 )
(RTn )(CT2 )
=
En,2 =
GT
GT
…
…
E1,m
…
E2,m
Cm
(RT1 )(CTm )
=
GT
(RT1 )(CTm )
=
GT
:
…
En,m =
(RTn )(CTm )
GT
4. Compute the value of Chi-Square Test of Independence or Homogeneity using
Formula 12-1:
χ2= ∑
(O−E)2
(Formula 12-1)
E
where: χ2 = Chi-square value.
O = observed value.
E = expected value.
5. Statistical decision for hypothesis testing:
If χ2computed<χ2critical, do not reject H0.
If χ2computed ≥ χ2critical, reject H0.
6. State the conclusion.
A. Test for Independence
Example 1: Suppose that a survey has been undertaken to determine if there a
relationship between place of residence and automobile preference. A random sample of
150 car owners from urban and 100 from rural areas were selected with the following
results:
Residence
Urban
Rural
Total
Toyota
65
40
105
Automobile
Mitsubishi
Nissan
45
15
20
30
60
35
Ford
25
10
30
Total
150
100
250
At the 0.05 level of significance, is there evidence to reject the claim of a relationship
between place of residence and automobile preference?
Solution:
Step 1: State the hypotheses.
H0: There is no relationship between place of residence and automobile
preference. (claim)
H1: There is a relationship between place of residence and automobile preference.
Step 2: The level of significance is α = 0.05
Step 3: Determine the critical region. (Refer to Table D)
df = (r - 1)(c - 1) = (2 - 1)(4 - 1) = 1(3) = 3
χ2critical = 7.815
Step 4: Compute for the value of χ2.
Observed Frequency Table
Residence
Urban
Rural
Total
Automobile
Mitsubishi
Nissan
45
15
20
30
60
35
Toyota
65
40
105
Ford
25
10
30
Total
150
100
250
Compute for the expected frequencies using Formula 12-1.
E1,1 =
E1,2 =
E1,3 =
(150)(105)
250
(150)(60)
250
(150)(35)
250
E1,4 =
E2,1 =
= 63
E2,2 =
= 39
E2,3 =
= 27
(150)(30)
250
250
(100)(60)
250
(100)(35)
250
E2,4 =
= 21
(100)(105)
= 42
= 26
= 18
(100)(30)
250
= 14
Expected Frequency Table
Residence
Urban
Rural
Toyota
63
42
Automobile
Mitsubishi
Nissan
39
27
26
18
Ford
21
14
Now we can solve for x² by applying Formula 11-12.
χ2= ∑
(O−E)2
(65 – 63)
=
63
2
E
+
(45 – 39)
(10 – 14)
39
2
+
(15 – 27)
27
2
+
(25 – 21)
21
2
+
(40 – 42)
42
2
+
(20 – 26)
26
2
+
(30 – 18)
18
2
+
2
14
= 0.063 + 0.923 + 5.333 + 0.762 + 0.095 + 1.385 + 8. 000 + 1.143
= 17.705
Step 5: Decision rule.
Since the computed χ2 of 17.705 is greater than the χ2 critical value of 7.815 at
level of significance of 0.05, the statistical decision is to reject the null hypothesis.
Step 6: Conclusion.
Since the null hypothesis has been rejected, we can conclude that there is evidence
that shows significant relationship between place of residence and automobile
preference.
B. Test for Homogeneity of Proportions
Examples 2: A researcher surveyed 80 randomly selected Certified Public Accountants
in each of the cities in Metro Manila area and asked them if they have performed work
for free for 10 or fewer clients in the last year. The results are shown in the table. At α =
0.10, is there enough evidence to reject the claim that the proportions of those who accept
free work for 10 or fewer client are the same in each city?
Free Work
Yes
No
Total
Cities
Mandaluyong
Makati
32
36
48
44
80
80
Manila
35
45
80
Marikina
30
50
80
Solution:
Step 1: State the hypotheses.
H0: p1= p2= p3= p4 (claim)
H1: At least one proportion is different from others.
Step 2: The level of significance is α = 0.10.
Step 3: Determine the critical region. (Refer to Table D)
df = (r - 1)(c - 1) = (2 - 1)(4 - 1) = 1(3) = 3
χ2critical = 6.251
Step 4: Compute for the value of χ2.
Observed Frequency Table
Free Work
Yes
No
Total
Manila
35
45
80
Mandaluyong
32
48
80
Cities
Makati
36
44
80
Marikina
30
50
80
Compute for the expected frequencies using the Formula 12-1.
E1,1 =
(133)(80)
320
= 33.25
E2,1 =
(187)(80)
320
= 46.75
Total
133
187
320
E1,2 =
(133)(80)
320
(133)(35)
E1,3 =
320
E1,4 =
E2,2 =
= 33.25
E2,3 =
= 33.25
(133)(80)
E2,4 =
= 33.25
320
(187)(80)
320
(187)(80)
320
(187)(80)
320
= 46.75
= 46.75
= 46.75
Expected Frequency Table
Free Work
Yes
No
Cities
Mandaluyong
Makati
33.25
33.25
46.75
46.75
Manila
33.25
46.75
Marikina
33.25
46.75
Now we can solve for χ2by applying Formula 12-1.
χ2= ∑
(O−E)2
=
E
(48 –46.75)
46.75
2
+
(35 – 33.25)
(44 –46.75)
46.75
33.25
2
+
2
+
(32 – 33.25)
(50 –46.75)
33.25
2
+
(36 –33.25)
33.25
2
+
(30 –33.25)
33.25
2
2
+
(45 –46.75)
46.75
+
2
46.75
= 0.092 + 0.047 + 0.227 + 0.318 + 0.066 + 0. 033 + 0.162 + 0.226
= 1.171
Step 5: Decision rule.
Since the computed χ2of 1.171 is less than the χ2critical value of 6.251 at level of
significance of 0.10 the statistical decision is not to reject the null hypothesis.
Step 6: Conclusion.
Since the null hypothesis has been rejected, we can conclude that there is not
enough evidence to reject the proportions of those who accept free work for 10 or
fewer clients are equal in each city.
Name: __________________________________ Date:___________ Score: ________
Section Exercise 12.3A
1.Suppose a market analyst wished to see whether costumers have any preference among
eight flavors of a new ice tea. A sample of 320 people provided these data:
Apple
Cherry
Grape
Lemon
Mango
Melon
Orange
Strawberry
60
30
40
35
50
25
55
25
If there were no preference, one would expect each flavor to be selected with equal
frequency. Is there enough evidence to reject the claim that there is no preference in the
selection of ice tea, use level of significance α = 0.01?
Solution:
Step 1: State the hypotheses.
H0:_______________________________________________________________
_______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2: The level of significance is _______.
Step 3: Determine the critical region.
df = _______ and
χ2critical= ________
Step 4: Compute the expected frequency table and compute for the value of χ2.
Apple Cherry Grape Lemon Mango Melon Orange Strawberry
χ2computed= ___________
Step 5: Decision rule.
__________________________________________________________________
__________________________________________________________________
Step 6: Conclusion.
__________________________________________________________________
__________________________________________________________________
Name: __________________________________ Date:___________ Score: ________
Section Exercise 12.3A
2. A distributor of refrigerator has six locations in the province of Rizal. The sales in
units for the first six months of the year were as follows. At the 0.05 significance level
do the records suggest that sales are uniformly distributed among the six locations?
Binangonan
54
Cainta
28
Taytay
36
Antipolo
31
Morong
46
Tanay
21
Solution:
Step 1: State the hypotheses.
H0:_______________________________________________________________
_______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2: The level of significance is __________.
.
Step 3:Determine the degrees of freedom and the critical region.
df = _______ and
χ2critical= ________
Step 4:Complete the expected frequency table and compute for the value of χ2.
Binangonan
Cainta
Taytay
Antipolo
Morong
Tanay
χ2computed= ___________
Step 5: Decision rule.
__________________________________________________________________
__________________________________________________________________
Step 6: Conclusion.
___________________________________________________________________
___________________________________________________________________
Name: __________________________________ Date:___________ Score: ________
Section Exercise 12.3B
1. According to recent census report, 44% of families have two parents present, 26%
have only a father, 20% have only a mother, and 10% have no parent. A random
sample of families from the municipality of Rizal revealed these results:
With Two Parents
80
Father Only
60
Mother Only
75
No Parents
65
If there sufficient evidence to conclude that the proportions of families by type of
parent(s) presence differs from those reported by census? Use level of significance α=
0.05?
Solution:
Step 1:State the hypotheses.
H0:_______________________________________________________________
_______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2: The level of significance is __________.
.
Step 3:Determine the degrees of freedom and the critical region.
df = _______ and
χ2critical= ________
Step 4:Complete the expected frequency table and compute for the value of χ2.
With Two Parents
Father Only
Mother Only
No Parents
χ2computed= ___________
Step 5: Decision rule.
__________________________________________________________________
__________________________________________________________________
Step 6: Conclusion.
__________________________________________________________________
__________________________________________________________________
Name: __________________________________ Date:___________ Score: ________
Section Exercise 12.3B
2. The credit card industry is very competitive, with about 150 billion solicitation each
year. In 2009, Visa has about 40% of the market, compared with nearly 22% for
Mastecard, 18% for American Express, 15% for Diners Club, and 5% for other credit
cards. To check if these percentage still held in 2010, a financial analyst randomly
sampled 300 households and asked which credit card they primarily used. The results
were as follows:
Visa
125
Mastercard
65
American Express
58
Diners Club
32
Others
20
Does this survey indicate that the population percentages using these credit cards
differ from the previous year? Use the level of significance α= 0.01?
Solution:
Step 1:State the hypotheses.
H0:_______________________________________________________________
_______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2: The level of significance is __________.
.
Step 3:Determine the degrees of freedom and the critical region.
χ2critical= ________
df = _______ and
Step 4:Complete the expected frequency table and compute for the value of χ2.
Visa
Mastercard
American Express
Diners Club
Others
χ2computed= ___________
Step 5: Decision rule.
__________________________________________________________________
____________________________________________________________
Step 6: Conclusion.
__________________________________________________________________
__________________________________________________________________
Name: __________________________________ Date:___________ Score: ________
Section Exercises 12.4A
1. A researcher wanted to study the relationship between gender and cellular phone
communication network. She took a sample of 500 cellular phone users and obtained
the following table.
Gender
Male
Female
Total
Communication Network
Globe
Sun
90
25
105
35
195
60
Smart
105
140
245
Total
220
280
500
At the 0.01 level of significance, is there evidence of a relationship between gender
and communication network?
Solution:
Step 1:State the hypotheses.
H0:_______________________________________________________________
_______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2: The level of significance is __________.
.
Step 3:Determine the degrees of freedom and the critical region.
df = _______ and
χ2critical= ________
Step 4:Complete the expected frequency table and compute for the value of χ2.
Gender
Male
Female
Smart
Communication Network
Globe
Sun
χ2computed= ___________
Step 5: Decision rule.
__________________________________________________________________
__________________________________________________________________
Step 6: Conclusion.
__________________________________________________________________
__________________________________________________________________
Name: __________________________________ Date:___________ Score: ________
Section Exercises 12.4A
2. A manufacturing company was interested in determining whether an association
exists between commuting time of their employees and the level of job performance.
A study of 200 workers revealed the following:
Commuting Time
Below 15 minutes
15-45 minutes
Over 45 minutes
Total
High
40
20
10
70
Job Performance
Average
Low
30
20
35
15
25
5
90
40
Total
90
70
40
200
At the 0.10 level of significance, is there evidence of a relationship between
commuting time and job performance?
Solution:
Step 1:State the hypotheses.
H0:_______________________________________________________________
_______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2: The level of significance is __________.
.
Step 3:Determine the degrees of freedom and the critical region.
df = _______ and
χ2critical= ________
Step 4:Complete the expected frequency table and compute for the value of χ2.
Commuting Time
Below 15 minutes
15-45 minutes
Over 45 minutes
High
Commuting Time
Average
Low
χ2computed= ___________
Step 5: Decision rule.
__________________________________________________________________
__________________________________________________________________
Step 6: Conclusion.
__________________________________________________________________
__________________________________________________________________
Name: __________________________________ Date:___________ Score: ________
Section Exercises 12.4A
3. The personnel director of San Sebastian College- Recoletos, is interesting in
determining whether a faculty member's educational level has an effect on his or her
job field specialization. An exam was given to a sample of 120 faculty members, and
the director would like to know whether there is an association in exam performance
among three groups (with Bachelor's degree, Master's degree and Doctorate degree).
Rather than using the actual examination scores. He rated each person's exam
performance as high, average, and low. The results of the study are:
Educational Level
Bachelor's degree
Master's degree
Doctorate degree
Total
High
10
20
10
40
Exam Performance
Average
Low
8
12
40
15
7
3
50
30
Total
30
70
20
120
Does job knowledge as measured by the exam appear to be related to the level of
faculty member's education, at this particular school? Use 0.01level of significance.
Solution:
Step 1:State the hypotheses.
H0:_______________________________________________________________
_______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2: The level of significance is __________.
.
Step 3:Determine the degrees of freedom and the critical region.
df = _______ and
χ2critical= ________
Step 4:Complete the expected frequency table and compute for the value of χ2.
Educational Level
Bachelor’s degree
Master's degree
Doctorate degree
High
Exam Performance
Average
Low
χ2computed= ___________
Step 5: Decision rule.
________________________________________________________________________
________________________________________________________________________
Step 6: Conclusion.
_______________________________________________________________________________
_______________________________________________________________________________
Name: __________________________________ Date:___________ Score: ________
Section Exercises 12.4B
1. An advertising agency has decided to ask 120 customers at each of four shopping
malls if they are willing to take part in a market research survey. According to past
studies, 40% of Filipinos refuse to take part in such surveys. The results are shown in
the table. At α= 0.05, test the claim that the proportions of those who are willing to
participate are equal.
Participation
Will participate
Will not participate
Total
A
80
40
120
Malls
C
90
30
120
B
70
50
120
D
60
60
120
Total
130
180
480
Solution:
Step 1:State the hypotheses.
H0:_______________________________________________________________
_______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2: The level of significance is __________.
.
Step 3:Determine the degrees of freedom and the critical region.
df = _______ and
χ2critical= ________
Step 4:Complete the expected frequency table and compute for the value of χ2.
Malls
Participation
A
B
C
D
Will participate
Will not participate
χ2computed= ___________
Step 5: Decision rule.
__________________________________________________________________
_________________________________________________________________
Step 6: Conclusion.
__________________________________________________________________
__________________________________________________________________
Name: __________________________________ Date:___________ Score: ________
Section Exercises 12.4B
2. The finance officer of a large supermarket chain wished to determine if their
customers made a list before going to grocery shopping. He surveyed 500 customers
in five stores. The results are shown in the table. At α= 0.01, test the claim that the
proportions of the customers in the five stores who made a list before going shopping
are equal.
Stores
Made list
No list
Total
A
20
80
100
B
30
70
100
C
50
50
100
D
40
60
100
E
55
45
100
Total
195
305
500
Solution:
Step 1:State the hypotheses.
H0:_______________________________________________________________
_______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2: The level of significance is __________.
.
Step 3:Determine the degrees of freedom and the critical region.
df = _______ and
χ2critical= ________
Step 4:Complete the expected frequency table and compute for the value of χ2.
Stores
A
B
C
D
E
Made list
No list
χ2computed= ___________
Step 5: Decision rule.
__________________________________________________________________
______________________________________________________________
Step 6: Conclusion.
__________________________________________________________________
__________________________________________________________________
Name: __________________________________ Date:___________ Score: ________
Section Exercises 12.4B
3. A record company wanted to survey its customers regarding music preferences. A
random sample of 500 frequent customers of the record company was selected, and
information was gathered on their music preference and job classification. From the
following data, can the null hypothesis of independence between type of music and
working status be rejected at the 5% significance level?
Job
Classification
Worker
Clerical
Managerial
Executive
Total
Pop
12
30
23
35
100
Classical
13
25
26
36
100
Type of Music
Rock
Jazz
15
17
33
40
20
13
32
30
100
100
R and B
16
38
28
18
100
Total
73
166
110
151
500
Solution:
Step 1: State the hypotheses.
H0:_______________________________________________________________
_______________________________________________________________
H1:_______________________________________________________________
_______________________________________________________________
Step 2: The level of significance is __________.
.
Step 3:Determine the degrees of freedom and the critical region.
df = _______ and
χ2critical= ________
Step 4:Complete the expected frequency table and compute for the value of χ2.
Job
Classification
Worker
Clerical
Managerial
Executive
Pop
Type of Music
Classical
Rock
Jazz
R and B
χ2computed= ___________
Step 5: Decision rule.
__________________________________________________________________
_________________________________________________________________
Step 6: Conclusion.
__________________________________________________________________
__________________________________________________________________
Name: __________________________________ Date:___________ Score: ________
BIBLIOGRAPHY

http://images.books24x7.com/bookimages/id_5642/fig11-19.jpg

http://blog.tutorvista.com/wp-content/uploads/2010/12/d7.bmp