Parametric versus Nonparametric Statistics-when … versus Nonparametric Statistics-when to use them...
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Transcript of Parametric versus Nonparametric Statistics-when … versus Nonparametric Statistics-when to use them...
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.
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 distribution-
free methods.
Nonparametric methods can be used with small
samples
Nonparametric Methods
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. Tests of differences between groups
(independent samples)
2. Tests of differences between variables
(dependent samples)
3. 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 Correlation
1
Sample
2 Sample K Sample (i.e., >2)
Independent Dependent Independent Dependent
Categorical
or Nominal
Χ2 Χ2 Macnarmar’s
Χ2Χ2 Cochran’s Q
Rank or
Ordinal
Mann
Whitney U
Wilcoxin
Matched
Pairs Signed
Ranks
Kruskal Wallis
H
Friendman’s
ANOVA
Spearman’s
rho
Parametric
(Interval &
Ratio)
z test
or t test
t test
between
groups
t test within
groups
1 way ANOVA
between
groups
1 way
ANOVA
(within or
repeated
measure)
Pearson’s r
Factorial (2 way) ANOVA
Chi-Square
Types of Statistical Tests
When running a t test and ANOVA
We compare: Mean differences between groups
We assume 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 Chi-
Square test18
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
Measures of
Association
Nominal Scale Pearson Chi-
Square: 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 total number of observations must be greater than 20.
The expected frequency in any one cell of the table must be greater than 5.
Formula
χ 2 = ∑ (O – E)2
E
χ2 = The value of chi squareO = The observed valueE = 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
IndependenceWording of Research questions
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 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 Research Questions
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 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. Determine the appropriate test
2. Establish the level of significance:α
3. Formulate the statistical hypothesis
4. Calculate the test statistic
5. Determine the degree of freedom
6. 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
e
eo
F
FF 22 )(
4. Calculating Test Statistics
e
eo
F
FF 22 )(
5. Determine Degrees of
Freedomdf = (R-1)(C-1)
6. Compare computed test statistic
against a tabled/critical value
The computed value of the Pearson chi-
square 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
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
frequen
cy
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 = 90Column frequency
1. Determine Appropriate Test
1. Party Membership ( 2 levels) and
Nominal
2. 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 f row
Democrat fo =10
fe =13.9
fo =10
fe =13.9
fo =30
fe=22.2
50
Republican fo =15
fe =11.1
fo =15
fe =11.1
fo =10
fe =17.8
40
f column 25 25 40 n = 90
4. Calculating Test Statistics
Favor Neutral Oppose f row
Democrat fo =10
fe =13.9
fo =10
fe =13.9
fo =30
fe=22.2
50
Republica
n
fo =15
fe =11.1
fo =15
fe =11.1
fo =10
fe =17.8
40
f column 25 25 40 n = 90
= 50*25/90
4. Calculating Test Statistics
Favor Neutral Oppose f row
Democrat fo =10
fe =13.9
fo =10
fe =13.9
fo =30
fe=22.2
50
Republica
n
fo =15
fe =11.1
fo =15
fe =11.1
fo =10
fe =17.8
40
f column 25 25 40 n = 90
= 40* 25/90
4. Calculating Test Statistics
8.17
)8.1710(
11.11
)11.1115(
11.11
)11.1115(
2.22
)2.2230(
89.13
)89.1310(
89.13
)89.1310(
222
2222
= 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
11.025a 2 .004
11.365 2 .003
8.722 1 .003
90
Pearson Chi-Square
Likelihood Ratio
Linear-by-Linear
Association
N of Valid Cases
Value df
Asymp. Sig.
(2-sided)
0 cells (.0%) have expected count less than 5. The
minimum expected count is 11.11.
a.
Additional Information in SPSS
Output
Exceptions that might distort χ2
Assumptions
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
PlacementCom plication Incidence * Heparin Lock Placem ent Time Gr oup Crosstabulation
9 11 20
10.0 10.0 20.0
18.0% 22.0% 20.0%
41 39 80
40.0 40.0 80.0
82.0% 78.0% 80.0%
50 50 100
50.0 50.0 100.0
100.0% 100.0% 100.0%
Count
Expected Count
% w ithin Heparin Lock
Placement Time Group
Count
Expected Count
% w ithin Heparin Lock
Placement Time Group
Count
Expected Count
% w ithin Heparin Lock
Placement Time Group
Had Compilca
Had NO Compilca
Complication
Incidence
Total
1 2
Heparin Lock
Placement Time Group
Total
from Polit Text: Table 8-1
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
Sym metric Measures
-.050 .617
.050 .617
-.050 .100 -.496 .621c
-.050 .100 -.496 .621c
100
Phi
Cramer's V
Nominal by
Nominal
Pearson's RInterval by Interval
Spearman CorrelationOrdinal by Ordinal
N of Valid Cases
Value
Asymp.
Std. Errora
Approx. Tb
Approx. Sig.
Not assuming the null hypothes is.a.
Using the asymptotic standard error assuming the null hypothes is .b.
Based on normal approximation.c.
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
Sym metric Measures
-.050 .617
.050 .617
-.050 .100 -.496 .621c
-.050 .100 -.496 .621c
100
Phi
Cramer's V
Nominal by
Nominal
Pearson's RInterval by Interval
Spearman CorrelationOrdinal by Ordinal
N of Valid Cases
Value
Asymp.
Std. Errora
Approx. Tb
Approx. Sig.
Not assuming the null hypothes is.a.
Using the asymptotic standard error assuming the null hypothes is .b.
Based on normal approximation.c.
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.
Sym metric Measures
-.050 .617
.050 .617
-.050 .100 -.496 .621c
-.050 .100 -.496 .621c
100
Phi
Cramer's V
Nominal by
Nominal
Pearson's RInterval by Interval
Spearman CorrelationOrdinal by Ordinal
N of Valid Cases
Value
Asymp.
Std. Errora
Approx. Tb
Approx. Sig.
Not assuming the null hypothes is.a.
Using the asymptotic standard error assuming the null hypothes is .b.
Based on normal approximation.c.
)1(
2
kNV
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.
Sym metric Measures
-.050 .617
.050 .617
-.050 .100 -.496 .621c
-.050 .100 -.496 .621c
100
Phi
Cramer's V
Nominal by
Nominal
Pearson's RInterval by Interval
Spearman CorrelationOrdinal by Ordinal
N of Valid Cases
Value
Asymp.
Std. Errora
Approx. Tb
Approx. Sig.
Not assuming the null hypothes is.a.
Using the asymptotic standard error assuming the null hypothes is .b.
Based on normal approximation.c.
)1(
2
kNV
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.
The variable mentioned second in the problem, [fund], is used as the dependent variable and is moved to the “Row(s)” list box.
Second, click on “Statistics…” button to request the test statistic.
Chi-Square Test of
Independence: post hoc test
in SPSS (3)
Second, click on “Continue” button to close the Statistics dialog box.
First, click on “Chi-square” to request the chi-square test of independence.
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?