Transcript Chi-Square Test of Independence

Parametric versus Nonparametric
Statistics-when to use them and
which is more powerful?
Dr Mahmoud Alhussami
Parametric Assumptions





The observations must be independent.
Dependent variable should be continuous (I/R)
The observations must be drawn from normally
distributed populations
These populations must have the same variances.
Equal variance (homogeneity of variance)
The groups should be randomly drawn from normally
distributed and independent populations
e.g. Male X Female
Nurse X Physician
Manager X Staff
NO OVER LAP
Parametric Assumptions

The independent variable is categorical with
two or more levels.
 Distribution for the two or more independent
variables is normal.
 large variation = less likely to have sig t or F
test = accepting null hypothesis (fail to reject)
= Type II error = a threat to power
Sending an innocent to jail for no significant
reason
Advantages of Parametric
Techniques
 They
are more powerful and more flexible
than nonparametric techniques.
 They not only allow the researcher to
study the effect of many independent
variables on the dependent variable, but
they also make possible the study of their
interaction.
Nonparametric Methods





Most of the statistical methods referred to as
parametric require the use of interval- or ratio-scaled
data.
Nonparametric methods are often the only way to
analyze nominal or ordinal data and draw statistical
conclusions.
Nonparametric methods require no assumptions
about the population probability distributions.
Nonparametric methods are often called distributionfree methods.
Nonparametric methods can be used with small
samples
Nonparametric Methods
 In
general, for a statistical method to be
classified as nonparametric, it must satisfy
at least one of the following conditions.



The method can be used with nominal data.
The method can be used with ordinal data.
The method can be used with interval or ratio
data when no assumption can be made about
the population probability distribution.
Non Parametric Tests
 Do
not make as many assumptions about
the distribution of the data as the t test.


Do not require data to be Normal
Good for data with outliers
 Non-parametric
tests based on ranks of
the data

Work well for ordinal data (data that have a
defined order, but for which averages may not
make sense).
Nonparametric Methods
 There
is at least one nonparametric test
equivalent to a parametric test
 These tests fall into several categories
1.
2.
3.
Tests of differences between groups
(independent samples)
Tests of differences between variables
(dependent samples)
Tests of relationships between variables
Advantages of Nonparametric Techniques





Sometimes there is no parametric alternative to
the use of nonparametric statistics.
Certain nonparametric test can be used to
analyze nominal data.
Certain nonparametric test can be used to
analyze ordinal data.
The computations on nonparametric statistics
are usually less complicated than those for
parametric statistics, particularly for small
samples.
Probability statements obtained from most
nonparametric tests are exact probabilities.
Advantages of Nonparametric Tests

Treat samples made up of observations from
several different populations.
 Can treat data which are inherently in ranks as
well as data whose seemingly numerical scores
have the strength in ranks
 They are available to treat data which are
classificatory
 Easier to learn and apply than parametric tests
Disadvantages of Nonparametric
Statistics
 Nonparametric
tests can be wasteful of
data if parametric tests are available for
use with the data.
 Nonparametric tests are usually not as
widely available and well known as
parametric tests.
 For large samples, the calculations for
many nonparametric statistics can be
tedious.
Criticisms of Nonparametric
Procedures
 Losing
precision/wasteful of data
 Low power
 False sense of security
 Lack of software
 Testing distributions only
 Higher-ordered interactions not dealt with
Parametric vs. Nonparametric
Statistics

Parametric Statistics are statistical techniques
based on assumptions about the population
from which the sample data are collected.
 Assumption that data being analyzed are
randomly selected from a normally
distributed population.
 Requires quantitative measurement that
yield interval or ratio level data.
 Nonparametric Statistics are based on fewer
assumptions about the population and the
parameters.


Sometimes called “distribution-free” statistics.
A variety of nonparametric statistics are available
for use with nominal or ordinal data.
Summary Table of Statistical Tests
Level of
Measurement
Sample Characteristics
1
Sample
Categorical
or Nominal
Χ2
Rank or
Ordinal
Parametric
(Interval &
Ratio)
z test
or t test
2 Sample
Correlation
K Sample (i.e., >2)
Independent
Dependent
Independent
Dependent
Χ2
Macnarmar’s
Χ2
Cochran’s Q
Mann
Whitney U
Wilcoxin
Matched
Pairs Signed
Ranks
Kruskal Wallis
H
Friendman’s
ANOVA
Spearman’s
rho
t test
between
groups
t test within
groups
1 way ANOVA
between
groups
1 way
ANOVA
(within or
repeated
measure)
Pearson’s r
Χ2
Factorial (2 way) ANOVA
Chi-Square
Types of Statistical Tests
When running a t test and ANOVA
 We compare:


We assume






Mean differences between groups
random sampling
the groups are homogeneous
distribution is normal
samples are large enough to represent population
(>30)
DV Data: represented on an interval or ratio scale
These are Parametric tests!
Types of Tests
When the assumptions are violated:
 Subjects were not randomly sampled
 DV Data:



Ordinal (ranked)
Nominal (categorized: types of car, levels of
education, learning styles, Likert Scale)
The scores are greatly skewed or we have no
knowledge of the distribution
We use tests that are equivalent to t test and
ANOVA
Non-Parametric Test!
Requirements for ChiSquare test
18
 Must
be a random sample from population
 Data must be in raw frequencies
 Variables must be independent
 A sufficiently large sample size is required
(at least 20)
 Actual count data (not percentages)
 Observations must be independent.
 Does not prove causality.
Different Scales, Different Measures
of Association
Scale of Both
Variables
Nominal Scale
Measures of
Association
Pearson ChiSquare: χ2
Ordinal Scale
Spearman’s rho
Interval or Ratio
Scale
Pearson r
Chi Square

Used when data are nominal (both IV and DV)


Comparing frequencies of distributions occurring in
different categories or groups
Tests whether group distributions are different
• Shoppers’ preference for the taste of 3 brands of candy

determines the association between IV and DV by
counting the frequencies of distribution
• Gender relative to study preference (alone or in group)
Important
 The
chi square test can only be used on
data that has the following characteristics:
The data must be in the form
of frequencies
The frequency data must have a
precise numerical value and must be
organised into categories or groups.
The expected frequency in any one cell
of the table must be greater than 5.
The total number of observations must be
greater than 20.
Formula
χ 2 = ∑ (O – E)2
E
χ2 = The value of chi square
O = The observed value
E = The expected value
∑ (O – E)2 = all the values of (O – E) squared then added
together
What is it?
 Test
of proportions
 Non parametric test
 Dichotomous variables are used
 Tests the association between two
factors
e.g. treatment and disease
gender and mortality
types of chi-square analysis
techniques

Tests of Independence is a chi-square
technique used to determine whether two
characteristics (such as food spoilage and
refrigeration temperature) are related or
independent.
 Goodness-of-fit test is a chi-square test
technique used to study similarities between
proportions or frequencies between groupings
(or classification) of categorical data (comparing
a distribution of data with another distribution of
data where the expected frequencies are
known).
Chi Square Test of Independence

Purpose




To determine if two variables of interest independent (not related)
or are related (dependent)?
When the variables are independent, we are saying that knowledge of
one gives us no information about the other variable. When they are
dependent, we are saying that knowledge of one variable is helpful in
predicting the value of the other variable.
The chi-square test of independence is a test of the influence or
impact that a subject’s value on one variable has on the same
subject’s value for a second variable.
Some examples where one might use the chi-squared test of
independence are:
• Is level of education related to level of income?
• Is the level of price related to the level of quality in production?

Hypotheses

The null hypothesis is that the two variables are independent. This will
be true if the observed counts in the sample are similar to the
expected counts.
• H0: X and Y are independent
• H1: X and Y are dependent
Chi Square Test of
Independence
 Wording



Are X and Y independent?
Are X and Y related?
The research hypothesis states that the
two variables are dependent or related.
This will be true if the observed counts for
the categories of the variables in the
sample are different from the expected
counts.
 Level

of Research questions
of Measurement
Both X and Y are categorical
Assumptions
Chi Square Test of Independence

Each subject contributes data to only one cell

Finite values


Observations must be grouped in categories. No assumption is
made about level of data. Nominal, ordinal, or interval data may be
used with chi-square tests.
A sufficiently large sample size


In general N > 20.
No one accepted cutoff – the general rules are
• No cells with observed frequency = 0
• No cells with the expected frequency < 5
• Applying chi-square to small samples exposes the researcher to an
unacceptable rate of Type II errors.
Note: chi-square must be calculated on actual count data, not
substituting percentages, which would have the effect of pretending
the sample size is 100.
Chi Square Test of Goodness of Fit
 Purpose


To determine whether an observed
frequency distribution departs significantly
from a hypothesized frequency distribution.
This test is sometimes called a One-sample
Chi Square Test.
 Hypotheses

The null hypothesis is that the two variables are
independent. This will be true if the observed
counts in the sample are similar to the expected
counts.
• H0: X follows the hypothesized distribution
• H1: X deviates from the hypothesized distribution
Chi Square Test of Goodness of Fit
 Sample



Do students buy more Coke, Gatorade or
Coffee?
Does my sample contain a
disproportionate amount of Hispanics as
compared to the population of the county
from which they were sampled?
Has the ethnic composition of the city of
Amman changed since 1990?
 Level

Research Questions
of Measurement
X is categorical
Assumptions
Chi Square Test of Goodness of Fit

The research question involves the comparison of
the observed frequency of one categorical variable
within a sample to the expected frequency of that
variable.

The observed and theoretical distributions must
contain the same divisions (i.e. ‘levels’ or
‘classes’)

The expected frequency in each division must be
>5

There must be a sufficient sample (in general
N>20)
Steps in Test of Hypothesis
1.
2.
3.
4.
5.
6.
Determine the appropriate test
Establish the level of significance:α
Formulate the statistical hypothesis
Calculate the test statistic
Determine the degree of freedom
Compare computed test statistic against a
tabled/critical value
1. Determine Appropriate Test





Chi Square is used when both variables are
measured on a nominal scale.
It can be applied to interval or ratio data that
have been categorized into a small number of
groups.
It assumes that the observations are
randomly sampled from the population.
All observations are independent (an
individual can appear only once in a table and
there are no overlapping categories).
It does not make any assumptions about the
shape of the distribution nor about the
homogeneity of variances.
2. Establish Level of
Significance
α
is a predetermined value
 The convention
• α = .05
• α = .01
• α = .001
3. Determine The Hypothesis:
Whether There is an
Association or Not
 Ho
: The two variables are independent
 Ha : The two variables are associated
4. Calculating Test Statistics

Contrasts observed frequencies in each cell of a
contingency table with expected frequencies.
 The expected frequencies represent the number
of cases that would be found in each cell if the
null hypothesis were true ( i.e. the nominal
variables are unrelated).
 Expected frequency of two unrelated events is
product of the row and column frequency divided
by number of cases.
Fe= Fr Fc / N
Expected frequency = row total x column total
Grand total
4. Calculating Test Statistics
 ( Fo  Fe ) 
  

F
e


2
2
4. Calculating Test Statistics
 ( Fo  Fe ) 
  

F
e


2
2
5. Determine Degrees of
Freedom
df = (R-1)(C-1)
6. Compare computed test statistic
against a tabled/critical value
 The
computed value of the Pearson chisquare statistic is compared with the
critical value to determine if the
computed value is improbable
 The critical tabled values are based on
sampling distributions of the Pearson
chi-square statistic
 If calculated 2 is greater than 2 table
value, reject Ho
Decision and Interpretation

If the probability of the test statistic is less than
or equal to the probability of the alpha error rate,
we reject the null hypothesis and conclude that
our data supports the research hypothesis. We
conclude that there is a relationship between the
variables.
 If the probability of the test statistic is greater
than the probability of the alpha error rate, we
fail to reject the null hypothesis. We conclude
that there is no relationship between the
variables, i.e. they are independent.
Example
 Suppose
a researcher is interested in
voting preferences on gun control issues.
 A questionnaire was developed and sent
to a random sample of 90 voters.
 The researcher also collects information
about the political party membership of the
sample of 90 respondents.
Bivariate Frequency Table or
Contingency Table
Favor
Neutral Oppose
f row
Democrat
10
10
30
50
Republican
15
15
10
40
f column
25
25
40
n = 90
Bivariate Frequency Table or
Contingency Table
Favor
Neutral Oppose
f row
Democrat
10
10
30
50
Republican
15
15
10
40
f column
25
25
40
n = 90
Row frequency
Bivariate Frequency Table or
Contingency Table
Favor
Neutral Oppose
f row
Democrat
10
10
30
50
Republican
15
15
10
40
f column
25
25
40
n = 90
Bivariate Frequency Table or
Contingency Table
Favor
Neutral Oppose
f row
Democrat
10
10
30
50
Republican
15
15
10
40
f column
25
25
40
n = 90
Column frequency
1. Determine Appropriate Test
1.
2.
Party Membership ( 2 levels) and
Nominal
Voting Preference ( 3 levels) and
Nominal
2. Establish Level of
Significance
Alpha of .05
3. Determine The Hypothesis
•
Ho : There is no difference between D & R
in their opinion on gun control issue.
•
Ha : There is an association between
responses to the gun control survey and
the party membership in the population.
4. Calculating Test Statistics
Favor
Neutral Oppose
Democrat
fo =10
fo =10
fe =13.9 fe =13.9
Republican fo =15
fo =15
fe =11.1 fe =11.1
f column
25
25
f row
fo =30
fe=22.2
fo =10
fe =17.8
50
40
n = 90
40
4. Calculating Test Statistics
Favor
Neutral Oppose
f row
= 50*25/90
Democrat
fo =10
fo =10
fe =13.9 fe =13.9
Republica fo =15
fo =15
n fe =11.1 fe =11.1
f column
25
25
fo =30
fe=22.2
fo =10
fe =17.8
50
40
n = 90
40
4. Calculating Test Statistics
Favor
Neutral Oppose
f row
fo =10
fo =10
fo =30
fe =13.9 fe =13.9 fe=22.2
= 40* 25/90
Republica fo =15
fo =15
fo =10
n fe =11.1 fe =11.1 fe =17.8
50
Democrat
f column
25
25
40
40
n = 90
4. Calculating Test Statistics
2
2
2
(
10

13
.
89
)
(
10

13
.
89
)
(
30

22
.
2
)
2 



13.89
13.89
22.2
(15  11.11) 2 (15  11.11) 2 (10  17.8) 2


11.11
11.11
17.8
= 11.03
5. Determine Degrees of
Freedom
df = (R-1)(C-1) =
(2-1)(3-1) = 2
6. Compare computed test statistic
against a tabled/critical value
α
= 0.05
 df = 2
 Critical tabled value = 5.991
 Test statistic, 11.03, exceeds critical value
 Null hypothesis is rejected
 Democrats & Republicans differ
significantly in their opinions on gun
control issues
SPSS Output for Gun Control
Example
Chi-Square Tests
Pearson Chi-Square
Likelihood Ratio
Linear-by-Linear
Association
N of Valid Cases
Value
11.025a
11.365
8.722
2
2
Asymp. Sig.
(2-sided)
.004
.003
1
.003
df
90
a. 0 cells (.0%) have expected count less than 5. The
minimum expected count is 11.11.
Additional Information in SPSS
Output
that might distort χ2
Assumptions
 Exceptions


Associations in some but not all categories
Low expected frequency per cell
 Extent
of association is not same as
statistical significance
Demonstrated
through an example
Another Example Heparin Lock
Placement
Complica tion Incidence * He parin Lock Placement Time Group Crosstabulation
Complication
Incidence
Total
Had Compilca
Count
Expected Count
% within Heparin Lock
Placement Time Group
Had NO Compilca Count
Expected Count
% within Heparin Lock
Placement Time Group
Count
Expected Count
% within Heparin Lock
Placement Time Group
from Polit Text: Table 8-1
Heparin Lock
Placement Time Group
1
2
9
11
10.0
10.0
Total
20
20.0
18.0%
22.0%
20.0%
41
40.0
39
40.0
80
80.0
82.0%
78.0%
80.0%
50
50.0
50
50.0
100
100.0
100.0%
100.0%
100.0%
Time:
1 = 72 hrs
2 = 96 hrs
Hypotheses in Heparin Lock Placement
 Ho:There
is no association between
complication incidence and length of
heparin lock placement. (The variables are
independent).
 Ha:There is an association between
complication incidence and length of
heparin lock placement. (The variables are
related).
More of SPSS Output
Pearson Chi-Square

Pearson Chi-Square =
.250, p = .617
Since the p > .05, we fail
to reject the null
hypothesis that the
complication rate is
unrelated to heparin lock
placement time.
 Continuity correction is
used in situations in
which the expected
frequency for any cell in a
2 by 2 table is less than
10.
More SPSS Output
Symmetric Measures
Nominal by
Nominal
Interval by Interval
Ordinal by Ordinal
N of Valid Cases
Phi
Cramer's V
Pearson's R
Spearman Correlation
Value
-.050
.050
-.050
-.050
100
Asymp.
a
Std. Error
Approx. T
.100
.100
-.496
-.496
a. Not assuming the null hypothesis.
b. Using the asymptotic standard error assuming the null hypothesis.
c. Based on normal approximation.
b
Approx. Sig.
.617
.617
.621c
.621c
Phi Coefficient


Pearson Chi-Square
provides information
about the existence of
relationship between 2
nominal variables, but not
about the magnitude of
the relationship
Phi coefficient is the
measure of the strength
of the association
Symmetric Measur es
Nominal by
Nominal
Interval by Interval
Ordinal by Ordinal
N of Valid Cases
Phi
Cramer's V
Pearson's R
Spearm an Correlati on
Value
-.050
.050
-.050
-.050
100
As ym
Std. E
a. Not as sum ing the null hypothesis .
b. Us ing the asym ptoti c standard error assum ing the null hypot
c. Based on norm al approximation.


N
2
Cramer’s V


When the table is larger than 2
by 2, a different index must be
used to measure the strength
of the relationship between the
variables. One such index is
Cramer’s V.
If Cramer’s V is large, it means
that there is a tendency for
particular categories of the first
variable to be associated with
particular categories of the
second variable.
Symmetric Measur es
Nominal by
Nominal
Interval by Interval
Ordinal by Ordinal
N of Valid Cases
Phi
Cramer's V
Pearson's R
Spearm an Correlati on
Value
-.050
.050
-.050
-.050
100
As ymp
Std. Er
.1
.1
a. Not as sum ing the null hypothesis .
b. Us ing the asym ptoti c standard error assum ing the null hypoth
c. Based on norm al approximation.

V
N (k  1)
2
Cramer’s V


When the table is larger than 2
by 2, a different index must be
used to measure the strength
of the relationship between the
variables. One such index is
Cramer’s V.
If Cramer’s V is large, it means
that there is a tendency for
particular categories of the first
variable to be associated with
particular categories of the
second variable.
Symmetric Measur es
Nominal by
Nominal
Interval by Interval
Ordinal by Ordinal
N of Valid Cases
Phi
Cramer's V
Pearson's R
Spearm an Correlati on
Value
-.050
.050
-.050
-.050
100
As ymp
Std. Er
.1
.1
a. Not as sum ing the null hypothesis .
b. Us ing the asym ptoti c standard error assum ing the null hypoth
c. Based on norm al approximation.

V
N (k  1)
2
Number of
cases
Smallest of number
of rows or columns
Interpreting Cell Differences in
a Chi-square Test - 1
A chi-square test of
independence of the
relationship between sex
and marital status finds a
statistically significant
relationship between the
variables.
Interpreting Cell Differences in
a Chi-square Test - 2
Researcher often try to identify try to identify which cell or cells are the
major contributors to the significant chi-square test by examining the
pattern of column percentages.
Based on the column percentages, we would identify cells on the
married row and the widowed row as the ones producing the
significant result because they show the largest differences: 8.2% on
the married row (50.9%-42.7%) and 9.0% on the widowed row
(13.1%-4.1%)
Interpreting Cell Differences in
a Chi-square Test - 3
Using a level of significance of 0.05, the critical value for a
standardized residual would be -1.96 and +1.96. Using
standardized residuals, we would find that only the cells on the
widowed row are the significant contributors to the chi-square
relationship between sex and marital status.
If we interpreted the contribution of the married marital status,
we would be mistaken. Basing the interpretation on column
percentages can be misleading.
Chi-Square Test of
Independence: post hoc test
in SPSS (1)
You can conduct a chi-square test of
independence in crosstabulation of
SPSS by selecting:
Analyze > Descriptive Statistics
> Crosstabs…
Chi-Square Test of
Independence: post hoc test
in SPSS (2)
First, select and move the
variables for the question to
“Row(s):” and “Column(s):”
list boxes.
The variable mentioned first
in the problem, sex, is used
as the independent variable
and is moved to the
“Column(s):” list box.
Second, click on
“Statistics…” button to
request the test
statistic.
The variable mentioned
second in the problem,
[fund], is used as the
dependent variable and is
moved to the “Row(s)” list
box.
Chi-Square Test of
Independence: post hoc test
in SPSS (3)
First, click on “Chi-square” to
request the chi-square test of
independence.
Second, click on “Continue”
button to close the Statistics
dialog box.
Chi-Square Test of
Independence: post hoc test
in SPSS (6)
In the table Chi-Square Tests result,
SPSS also tells us that “0 cells have
expected count less than 5 and the
minimum expected count is 70.63”.
The sample size requirement for the
chi-square test of independence is
satisfied.
Chi-Square Test of
Independence: post hoc test
in SPSS (7)
The probability of the chi-square test
statistic (chi-square=2.821) was
p=0.244, greater than the alpha level
of significance of 0.05. The null
hypothesis that differences in "degree
of religious fundamentalism" are
independent of differences in "sex" is
not rejected.
The research hypothesis that
differences in "degree of religious
fundamentalism" are related to
differences in "sex" is not supported by
this analysis.
Thus, the answer for this question is
False. We do not interpret cell
differences unless the chi-square test
statistic supports the research
hypothesis.
SPSS Analysis
Chi Square Test of Goodness
of Fit

Analyze –
Nonparametric Tests –
Chi Square
X variable
goes here
If all expected
frequencies are
the same, click
this box
If all expected
frequencies are
not the same, enter
the expected value
for each division
here
Examples
(using the Montana.sav data)
All
expected frequencies
are the same
All
expected frequencies
are not the same
Thank You for Listening
Questions?