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


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).


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!

http://edweb.sdsu.edu/courses/ED690MJ/Class11/NumberScales.html

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


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


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


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


Contingency Table




f column 25 25 40 n = 90

Row

frequen

cy


Contingency Table




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.



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



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



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


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



Count

Expected Count



Had Compilca

Had NO Compilca

Complication

Incidence

Total

1 2

Heparin Lock


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



N of Valid Cases

Value

Asymp.

Std. Errora

Approx. Tb

Approx. Sig.




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



N of Valid Cases

Value

Asymp.

Std. Errora

Approx. Tb

Approx. Sig.




)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



N of Valid Cases

Value

Asymp.

Std. Errora

Approx. Tb

Approx. Sig.




)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.



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%)



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


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


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


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


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?

Parametric versus Nonparametric Statistics-when … versus Nonparametric Statistics-when to use them...

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