Chi square using excel

Statistics for Managers
Using Microsoft® Excel
4th Edition
Chapter 11
Chi-Square Tests and
Nonparametric Tests
Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 11-1

Chapter Goals
After completing this chapter, you should be
able to:
 Perform a c2 test for the difference between two
proportions
 Use a c2 test for differences in more than two proportions
 Perform a c2 test of independence
 Apply and interpret the Wilcoxon rank sum test for the
difference between two medians
 Perform nonparametric analysis of variance using the
Kruskal-Wallis rank test for one-way ANOVA
Statistics for Managers Using
Microsoft Excel, 4e © 2004
Prentice-Hall, Inc. Chap 11-2

Contingency Tables
Contingency Tables
 Useful in situations involving multiple population
proportions
 Used to classify sample observations according
to two or more characteristics
 Also called a cross-classification table.

Contingency Table Example
Left-Handed vs. Gender
Dominant Hand: Left vs. Right
Gender: Male vs. Female
 2 categories for each variable, so
called a 2 x 2 table
 Suppose we examine a sample of
size 300

Contingency Table Example
(continued)
Sample results organized in a contingency table:
Gender
Hand Preference
Left Right
Female 12 108 120
Male 24 156 180
36 264 300
sample size = n = 300:
120 Females, 12
were left handed
180 Males, 24 were
left handed

c2 Test for the Difference
Between Two Proportions
H0: p1 = p2 (Proportion of females who are left
handed is equal to the proportion of
males who are left handed)
H1: p1 ≠ p2 (The two proportions are not the same –
Hand preference is not independent
of gender)
 If H0 is true, then the proportion of left-handed females should be
the same as the proportion of left-handed males
 The two proportions above should be the same as the proportion of
left-handed people overall

The Chi-Square Test Statistic
The Chi-square test statistic is:
(f f )
c = å -
2 o e
f
all cells e
2
 where:
fo = observed frequency in a particular cell
fe = expected frequency in a particular cell if H0 is true
c2 for the 2 x 2 case has 1 degree of freedom
(Assumed: each cell in the contingency table has expected
frequency of at least 5)

Decision Rule
c2
The c2 test statistic approximately follows a chi-squared
distribution with one degree of freedom
c2
U
Decision Rule:
If c2 > c2
U, reject H0,
otherwise, do not
reject H0
0
a
Reject H0 Do not
reject H0

Computing the
Average Proportion
p = X +
X
1 2 =
+
n n
The average
proportion is:
120 Females, 12 Here:
were left handed
180 Males, 24 were
left handed
X
n
1 2
0.12
p 12 24 = 36
=
300
= +
120 +
180
i.e., the proportion of left handers overall is 12%

Finding Expected Frequencies
 To obtain the expected frequency for left handed
females, multiply the average proportion left handed (p)
by the total number of females
 To obtain the expected frequency for left handed males,
multiply the average proportion left handed (p) by the
total number of males
If the two proportions are equal, then
P(Left Handed | Female) = P(Left Handed | Male) = .12
i.e., we would expect (.12)(120) = 14.4 females to be left handed
Statistics for Managers (.Using
12)(180) = 21.6 males to be left handed

Observed v. Expected
Frequencies
Observed frequencies vs. expected frequencies:
Gender
Hand Preference
Left Right
Female
Observed = 12
Expected = 14.4
Observed = 108
Expected = 105.6
120
Male
Observed = 24
Expected = 21.6
Observed = 156
Expected = 158.4
180
36 264 300

Gender
Hand Preference
Left Right
Female
Observed = 12
Expected = 14.4
Observed = 108
Expected = 105.6
120
Male
Observed = 24
Expected = 21.6
Observed = 156
Expected = 158.4
180
36 264 300
The test statistic is:
(f f )
c = å -
f
Statistics (12 for 14.4)
Managers 2 (108 Using
105.6)
2 (24 21.6)
2 (156 158.4)
2
Microsoft Excel, 14.4
4e © 2004
0.6848
105.6
21.6
158.4
all cells e
2
2 o e
= - + - + - + - =

Decision Rule Example
The test statistic is c2 = 0 . 6848 , c with 1 d.f. =
3.841 2U
Decision Rule:
If c2 > 3.841, reject H0,
otherwise, do not reject H0
Here,
c2 = 0.6848 < c2
U = 3.841,
so we do not reject H0
and conclude that there is
not sufficient evidence
that the two proportions
are different at a = .05
c2
a
Reject H0 Do not
reject H0
U=3.841
c2
0

c2 Test for the Difference Between
More Than Two Proportions
 Extend the c2 test to the case with more than
two independent populations:
H0: p1 = p2 = … = pc
H1: Not all of the pj are equal (j = 1, 2, …, c)

(f f )
c = å -
2 o e
f
all cells e
2
 where:
fo = observed frequency in a particular cell of the 2 x c table
c2 for the 2 x c case has (2-1)(c-1) = c - 1 degrees of freedom
Statistics Microsoft (Assumed: for Managers each cell Using
Excel, 4e © 2004
in the contingency table has expected
frequency of at least 1)

Computing the
Overall Proportion
X
n
p = X + X + +
X
The overall 
proportion is:
1 2 c =
+ + +
n n 
n
1 2 c
 Expected cell frequencies for the c categories
are calculated as in the 2 x 2 case, and the
decision rule is the same:
Decision Rule:
If c2 > c2
U, reject H0,
otherwise, do not
reject H0
Where c2
U is from the
chi-squared distribution
with c – 1 degrees of
freedom

c2 Test of Independence
 Similar to the c2 test for equality of more than
two proportions, but extends the concept to
contingency tables with r rows and c columns
H0: The two categorical variables are independent
(i.e., there is no relationship between them)
H1: The two categorical variables are dependent
(i.e., there is a relationship between them)

c2 Test of Independence
(f f )
c = å -
2 o e
f
all cells e
2
 where:
fo = observed frequency in a particular cell of the r x c table
c2 for the r x c case has (r-1)(c-1) degrees of freedom
Statistics (Assumed: for Managers each cell in Using
the contingency table has expected
Microsoft frequency Excel, 4e of at © least 2004
1)
(continued)

Expected Cell Frequencies
 Expected cell frequencies:
f row total column total e
n
= ´
Where:
row total = sum of all frequencies in the row
column total = sum of all frequencies in the column
n = overall sample size

Decision Rule
 The decision rule is
If c2 > c2
U, reject H0,
Where c2
U is from the chi-squared distribution
with (r – 1)(c – 1) degrees of freedom

Example
 The meal plan selected by 200 students is shown below:
Class
Standing
Number of meals per week
20/week 10/week none Total
Fresh. 24 32 14 70
Soph. 22 26 12 60
Junior 10 14 6 30
Senior 14 16 10 40
Total 70 88 42 200

Example
 The hypothesis to be tested is:
(continued)
H0: Meal plan and class standing are independent
(i.e., there is no relationship between them)
H1: Meal plan and class standing are dependent
(i.e., there is a relationship between them)

Class
Standing
Observed:
Number of meals
per week
Total
20/wk 10/wk none
Fresh. 24 32 14 70
Soph. 22 26 12 60
Junior 10 14 6 30
Senior 14 16 10 40
Total 70 88 42 200
Expected cell
frequencies if H0 is true:
Class
Standing
Number of meals
per week
Total
20/wk 10/wk none
Fresh. 24.5 30.8 14.7 70
Soph. 21.0 26.4 12.6 60
Junior 10.5 13.2 6.3 30
Senior 14.0 17.6 8.4 40
Total 70 88 42 200
Example for one cell:
f = row total ´
column total e
Statistics 30 for 70
Managers Using
Microsoft 10.5
200
Excel, 4e © 2004
n
= ´ =
Example:
Expected Cell Frequencies
(continued)

Example: The Test Statistic
 The test statistic value is:
(continued)
0.709
2
2 2 2
(10 8.4)
c = å -
= - + - + + - =
8.4
(32 30.8)
30.8
(f f )
2 o e
f
all cells e
(24 24.5)
24.5

c2
= 12.592 for α = .05 from the chi-squared
U distribution with (4 – 1)(3 – 1) = 6 degrees of
freedom

Example:
Decision and Interpretation
(continued)
The test statistic is c2 = 0 . 709 , c with 6 d.f. =
12.592 2U
Decision Rule:
If c2 > 12.592, reject H0,
Here,
c2 = 0.709 < c2
U = 12.592,
so do not reject H0
Conclusion: there is not
sufficient evidence that meal
plan and class standing are
related at a = .05
c2
a
Reject H0 Do not
reject H0
U=12.592
c2
0

Wilcoxon Rank-Sum Test for
Differences in 2 Medians
 Test two independent population medians
 Populations need not be normally distributed
 Distribution free procedure
 Used when only rank data is available
 Must use normal approximation if either of the
sample sizes is larger than 10

Wilcoxon Rank-Sum Test:
Small Samples
 Can use when both n1 , n2 ≤ 10
 Assign ranks to the combined n1 + n2 sample
observations
 If unequal sample sizes, let n1 refer to smaller-sized
sample
 Smallest value rank = 1, largest value rank = n1 + n2
 Assign average rank for ties
 Sum the ranks for each sample: T1 and T2
Microsoft  Obtain Excel, test 4e © statistic, 2004
T(from smaller sample)
1 Prentice-Hall, Inc. Chap 11-27

Checking the Rankings
 The sum of the rankings must satisfy the
formula below
 Can use this to verify the sums T1 and T2
T + T = n(n +
1) 1 2
2
where n = n1 + n2

Hypothesis and Decision Rule
M1 = median of population 1; M2 = median of population 2
Test statistic = T1 (Sum of ranks from smaller sample)
Two-Tail Test Left-Tail Test Right-Tail Test
H0: M1 = M2
H1: M1 ¹ M2
H0: M1 £ M2
H1: M1 > M2
H0: M1 ³ M2
H1: M1 < M2
Reject
Do Not Reject
Reject
T1L T1U
Reject
T1L
Do Not Reject
Do Not Reject Reject
T1U
Statistics Reject Hif for T< Managers TUsing
0 1 1L
Microsoft or if Excel, T> T4e © 2004
1 1U
Reject H0 if T1 < T1L Reject H0 if T1 > T1U

Small Sample Example
Sample data is collected on the capacity rates
(% of capacity) for two factories.
Are the median operating rates for two factories
the same?
 For factory A, the rates are 71, 82, 77, 94, 88
 For factory B, the rates are 85, 82, 92, 97
Test for equality of the sample medians
at the 0.05 significance level

Capacity Rank
Factory A Factory B Factory A Factory B
71 1
77 2
82 3.5
82 3.5
85 5
88 6
92 7
94 8
Ranked
Capacity
values:
97 9
Rank Sums: 20.5 24.5
Tie in 3rd and
4th places
(continued)

(continued)
Factory B has the smaller sample size, so
the test statistic is the sum of the
Factory B ranks:
T1 = 24.5
The sample sizes are:
n1 = 4 (factory B)
n2 = 5 (factory A)
Microsoft Excel, 4e © The 2004
level of significance is α = .05

n2
a n1
One-
Tailed
(continued)
Two-
Tailed 4 5
4
5
.05 .10 12, 28 19, 36
.025 .05 11, 29 17, 38
.01 .02 10, 30 16, 39
.005 .01 --, -- 15, 40
6
 Lower and
Upper
Critical
Values for
T1 from
Appendix
table E.8:
Prentice-Hall, Inc. T1L = 11 and T1U Chap = 29
11-33

Small Sample Solution
 a = .05
 n1 = 4 , n2 = 5
Two-Tail Test
H0: M1 = M2
H1: M1 ¹ M2
Do Not Reject
Reject
Reject
T1L=11 T1U=29
(continued)
Do not reject at a = 0.05
Statistics Reject Hif for T< Managers T=11
Using
0 1 1LMicrosoft or if Excel, T> T=4e 29
© 2004
1 1UPrentice-Hall, Inc. Chap 11-34
Test Statistic (Sum of
ranks from smaller sample):
T1 = 24.5
Decision:
Conclusion:
There is not enough evidence
that medians are not equal.

Wilcoxon Rank-Sum Test
(Large Sample)
 For large samples, the test statistic Tis
1 approximately normal with mean μ
T1 and
standard deviation σ
T1 :
μ = n1(n +
1)
T1
2
σ = n1n2 (n +
1)
T1
12
 Must use the normal approximation if either n1
or n2 > 10
 Can use the normal approximation for small samples
 Assign n1 to be the smaller of the two sample sizes

Wilcoxon Rank-Sum Test
(Large Sample)
 The Z test statistic is
T -
μ
1 T
σ
T
1
1
Z
=
 Where Z approximately follows a
standardized normal distribution
(continued)

Normal Approximation Example
Use the setting of the prior example:
The sample sizes were:
n1 = 4 (factory B)
n2 = 5 (factory A)
The level of significance was α = .05
The test statistic was T1 = 24.5

Normal Approximation Example
20
μ = n1(n + 1)
= 4(9 + 1)
=
T1 2
2
σ = n1n2 (n + 1)
= 4(5)(9 + 1)
=
T1 12
12
 The test statistic is
2.739
(continued)
1.64
24.5 20
-
1 T = - =
2.739
T μ
σ
Z
T
1
1
=
 Z = 1.64 is not greater than the critical Z value of 1.96
(for α = .05) so we do not reject H0 – there is not
sufficient evidence that the medians are not equal

Kruskal-Wallis Rank Test
 Tests the equality of more than 2 population
medians
 Assumptions:
 The samples are random and independent
 variables have a continuous distribution
 the data can be ranked
 populations have the same variability
 populations have the same shape

Kruskal-Wallis Test Procedure
 Obtain relative rankings for each value
 In event of tie, each of the tied values gets the
average rank
 Sum the rankings for data from each of the c
groups
 Compute the H test statistic

 The Kruskal-Wallis H test statistic: (with c – 1 degrees of freedom)
ù
2
j - +
3(n 1)
T
n
é
H 12
= å=
n(n 1)
c
j 1 j
ú úû
ê êë
+
where:
n = sum of sample sizes in all samples
c = Number of samples
Tj = Sum of ranks in the jth sample
nj = Size of the jth sample
(continued)

 Complete the test by comparing the
calculated H value to a critical c2 value from
the chi-square distribution with c – 1
degrees of freedom
 Decision rule
(continued)
 Reject H0 if test statistic H > c2
U
 Otherwise do not reject H0
c2
c2
U
0
a
Reject H0 Do not
reject H0

Kruskal-Wallis Example
 Do different departments have different class
sizes?
Class size
(Math, M)
Class size
(English, E)
Class size
(History, H)
23
45
54
78
66
55
60
72
45
70
30
40
18
34
44

 Do different departments have different class
sizes?
Class size
(Math, M) Ranking Class size
(English, E) Ranking Class size
(History, H) Ranking
23
41
54
78
66
269
15
12
55
60
72
45
70
10
11
14
8
13
30
40
18
34
44
35147
S = 44 S = 56 S = 20

H : Median Median Median
0 M E H = =
H : Not all population Medians are equal
A
 The H statistic is
(continued)
3(15 1) 6.72
- +
ù
ú úû
2 2 2
20
5
3(n 1)
56
+ +
5
R
n
2j
= å=
j 1 j
44
5
é
H 12
n(n +
1)
12
c
15(15 1)
ù
= + - úû
ê êë
é
êë
ö
÷ ÷ø
æ
ç çè
+
=

 Compare H = 6.72 to the critical value from the
chi-square distribution for 5 – 1 = 4 degrees of
freedom and a = .05:
4877 . 9 2U
c =
Since H = 6.72 <
4877 . 9 2U
c =
do not reject H0
(continued)
There is not sufficient evidence to reject
that the population medians are all equal

Chapter Summary
 Developed and applied the c2 test for the
difference between two proportions
 Developed and applied the c2 test for
differences in more than two proportions
 Examined the c2 test for independence
 Used the Wilcoxon rank sum test for two
population medians
 Small Samples
 Large sample Z approximation
 Applied the Kruskal-Wallis H-test for multiple
population medians

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