Non Parametric Tests ICSSR RMC DIB&C Dr.S.selvaRani

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Good Afternoon



Non-Parametric Test-( 11-6-2013)Presentation By

Dr.S.SelvaRani, Principal,

Sri Sarada Niketan College for Women,

Amaravathipudur, Karaikudi



தட நணறகண நத ற ம அ.

எண ஏ னழத இவபங ண வழம உன

ற சட றவ ற ற அதற த.

அபங வட னறக பமன க ல

.

சலவ சலவஞ சவசலவம அசலவஞ சலவ ல த.

எரள னனவய கம அரள நயரள த

.

எர வ சசல தவய ரள த.



Tests o Hypot eses Statistical tests arm the researcher to objectively interpret

the data, without intuitive, biased or unconcerned

generalization or particularization. Medical test information from the diagnostic labs tells like:

Your BloodTotal Sugar

Borderlinehigh

High Interpretation

172mg/dL 200-239mg/dL

>239mg/dL

Fine; you are not a Sweetperson!

220mg/dL 200-239mg/dL

>239mg/dL

Pre-caution; you are

becoming a sweet person!

thethe



Parametric and Non-parametric tests Statistical tests are of two broad types- Parametric and

Non-parametric.

Parametric and nonparametric statistical proceduresthat test hypotheses involving different assumptions.

There are assumptions about:

- Shape of Data distribution- normal or any other or Nodefinite shape per se.

- Nature of Data – Measurement or Counting based

- Variance- Randomization of sampling, etc.

All these decide to use or not a particular class of test-

Parametric and Non-parametric.



Parametric statistics test hypotheses based onthe assumption that the samples come frompopulations that are normally distributed or

conform to some other distribution. Also, parametric statistical tests assume thatthere is homogeneity of variance (variances within

groups are the same).The level of measurement for parametric tests isassumed to be ratio or interval.

Parametric Tests



Nonparametric Tests

Nonparametric statistical procedurestest hypotheses that do not requirenormal distribution or variance

assumptions about the populations from which the samples were drawn.

Nonparametric statistical proceduresare designed for ordinal or nominal data.Non-Parametric Tests are versatile.



Nominal data are data that consist of names, labels, or

categories only. The data cannot be arranged in an orderscheme (such as low to high).

For example the number 24, 28, 18, … on the shirts of the ateam of football players are substitutes for names. They don’t

count or measure anything, so they are categorical data.If you record a number (width, height, speed, errors, etc.,) it’s ameasurement.

If you record a label it’s Nominal (sex, popularity, beauty,etc., )

If the data capable of being ranked. These are then Ordinal

. You know there are Interval and Ratio data.



Choosing A Statistical Procedure - Guidelines

Two Independent Variables

Interval or RatioIndependent

t-test

Dependent

t-test

One-Way

ANOVA

Repeated

Measures

ANOVA

Two -Factor

ANOVA

Two-Factor

ANOVA

Repeated

Measures

OrdinalMann-

Whitney UWilcoxon

Kruskal-

WallisFriedman

Nominal Chi-Square Chi-Square Chi-Square

Multiple

Independent

Groups

Multiple

Dependent

Groups

Factorial Designs

Independent

Groups

Dependent

Groups

MeasurementScale of the

Dependent

Variable

One Independent Variable

Two Levels More than 2 Levels

Two

Independent

Groups

Two

Dependent

Groups



Easy to Compute. Really easy, if weknow!!!

Without making assumptions about

population values or parameters.

Distribution-free Tests.

They compare medians rather than means.



Nonparametric Tests: Features

Many nonparametric methods do notuse the raw data and instead use

the rank order of data for analysis

Nonparametric methods can be used

with small samples

Not requiring the assumption of

Normality or the assumption of

Homogeneity of variance.



Many Non-Parametric Tests Exist

Tests Description

Chi-Square Test Tests for significance of differencebetween expected and actualfrequency distributions

Anderson- DarlingTest

Tests whether a sample is drawn froma given distribution

Friedman Test Two

Way ANOVA Row-

wise Ranks

Tests whether k treatments in

randomized block designs have

identical effects

Kendall’s Tau Measures statistical dependence

between two variables



Many Non-Parametric TestsTests Description

Mann- Whitney Uor Wilcoxon Rank

Sum Test

Tests whether two samples are drawn fromthe same distribution, as compared to a

given alternative hypothesis

Median Test Tests whether two samples are drawn from

distributions with equal medians

Kolmogorov-

Smirnov Test

Tests whether a sample is drawn from a

given distribution, or whether two samples

are drawn from the same distribution

Kruskal-Wallis One-

way ANOVA Rank

data

Tests whether >2 independent samples are

drawn from the same distribution



Many Non-Parametric Tests

Tests DescriptionKuiper’s Test Tests whether a sample is drawn from a

given distribution, sensitive to cyclic

variations such as day of the week

Sign Test Tests whether matched pair samples aredrawn from distributions with equal medians

Spearman’s Rank

Correlation

Coefficient Test

Measures statistical dependence between two

variables using a monotonic function

Wilcoxon Signed

Rank Test

Tests whether matched pair samples are

drawn from populations with different mean

ranks



Caution About Using Non-Parametric tests

The main weakness of nonparametric tests is

that they are less powerful than parametrictests.

They are less likely to reject the null hypothesis

when it is false. We loose some information when data are

ordinal changed.

So, when the assumptions of parametric testscan be met, parametric tests should be usedbecause they are the most powerful testsavailable.



Type I Error – Null hypothesis is rejected when it isactually true. Probability of Type I error set by , α ( Alpha).

Type II Error– Null hypothesis is accepted when itis actually false

Probability of making a Type II error is called, β,Beta.

Increasing alpha decreases beta and vice versa

Setting alpha and beta depends upon the cost of making either type of error.



Chi-Square Test



Chi-Square TestThe Chi Square (X 2) test is undoubtedly the

most important and most used member of the nonparametric family of statistical tests.

Chi Square is employed to test the

difference between an actual sample andanother hypothetical or previously established distribution such as that whichmay be expected due to chance orprobability.



Chi-SquareTestChi Square can also be used to test

differences between two or more actualsamples.

We don’t have scores, we don’t have means. We just have numbers, or frequencies. Inother words, we have nominal data or head-count data.

Counting the Heads, nor their weight!!!

It is versatile with many applications.



Chi-Square Distribution



Chi-Square Distribution CDF



Determining the Cutoff for a C- hi-. It is allChi-Square Distribution Starts with

Value Zero and Extends upto Infinity



Degreesof

Freedom(df)

Probability ( p)

Chi-Square vales for different Alpha and Degrees of Freedom

-- 0.95 0.90 0.80 0.70 0.50 0.30 0.20 0.10 0.05 0.01 0.001

1 0.004 0.02 0.06 0.15 0.46 1.07 1.64 2.71 3.84 6.64 10.83

2 0.10 0.21 0.45 0.71 1.39 2.41 3.22 4.60 5.99 9.21 13.82

3 0.35 0.58 1.01 1.42 2.37 3.66 4.64 6.25 7.82 11.34 16.27

4 0.71 1.06 1.65 2.20 3.36 4.88 5.99 7.78 9.49 13.28 18.47

5 1.14 1.61 2.34 3.00 4.35 6.06 7.29 9.24 11.07 15.09 20.52

6 1.63 2.20 3.07 3.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46

7 2.17 2.83 3.82 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32

8 2.73 3.49 4.59 5.53 7.34 9.52 11.03 13.36 15.51 20.09 26.12

9 3.32 4.17 5.38 6.39 8.34 10.66 12.24 14.68 16.92 21.67 27.88

10 3.94 4.86 6.18 7.27 9.34 11.78 13.44 15.99 18.31 23.21 29.59Nonsignificant Significant

A i Wh ?



A great question now…. What?Isn’t Chi-Square Parametric!!!!

Chi-square is a statistic that is related to the centrallimit theorem in the sense that proportions are infact means, and that proportions are normally distributed (with a mean of p [not 3.141592653...]

and a variance of [(p) (1-p)].Therefore, we can perform a normal curve test forexamining the difference between proportions suchthat Z squared = chi square on one degree of freedom. Since Z is indubitably a parametric test,and chi square can be related to Z, we can infer thatit is, in fact, parametric.

Chi Sq are Test for Goodness of Fit



Chi-Square Test for Goodness of Fit Is any theoretical distribution like, Beta, Binomial,

Poisson, Normal or any other a good fit in a given case?

For instance we may assume, gender of new born isbinomial distributed, with q=p=0.5. In families each with four children, we test this. Binomial distribution’sacceptance needs testing here. We may use Chi-Square.

In project management the activity time estimates(pessimistic, most-likely and optimistic) is said to be afour parameter ‘beta’ distributed. We may test thisusing Chi-Square.

In an empirical distribution like, 2 : 3 : 5 as the ratio of number of new-born babies with underweight, normal weight and overweight in a rich country, to its goodnessof fit, we may use Chi-Square.



D bi d f d l t



Do birds forage randomly on any tree or are

they choosing particular kind of trees? A study on bird foraging behavior in a forest in Oregon

revealed the following. In a managed forest, 54% of thecanopy volume was Douglas fir, 40% was ponderosapine, 5% was grand fir, and 1% was western larch.

On 156 observations of foraging by nuthatches; 70

observations (45% of the total) in Douglas fir, 79 (51%)in ponderosa pine, 3 (2%) in grand fir, and 4 (3%) in

western larch.

The biological null hypothesis is that the birds forage

randomly, without regard to what species of treethey're in; The statistical null hypothesis is that theproportions of foraging events are equal to theproportions of canopy volume.



Problem Expressed in FrequencyTypes of

Trees

Douglas

Fir

Ponderosa

Pine

Grand

Fir

Western

Larch

Total

Birds Actual TreeForage

Distribution = O

70 79 3 4 156

Birds ExpectedTree PreferenceDistribution =Tree Distributionin the ManagedForest = E .

84 62 8 2 156

The difference in proportions is significant ( =11.296,

table Value= 7.82 for 3 d.f. without merging) P=0.0035.



* The formula for calculatingchi-square

Degrees of freedom (df) = n-1 where n is the number of classes



Contingency Table based Chi-square Contingency table is also referred to a cross

tabulation or cross tab. T is a type of matrix tabledisplaying multiple variables in frequency. The term was originally used by great statistician Karl Pearsonin the context of his study. But now it stands

generalized. Are variables in questions are independent or

dependent?

Are the attributes Related or Un-related, Associatedor Un-associated?

We use Chi-square. Most of you know this very well.

Yet have a cursory glance over an example.

Suppose you have the following categorical data



Asia Africa South

America Totals

Malaria A 31 14 45 90

Malaria B 2 5 53 60

Malaria C 53 45 2 100

Suppose you have the following categorical data

set. Incidence of three types of malaria in three

tropical regions.



Observed Expected |O -E| (O— E)2 (O— E)2 / E

31 30.96 0.04 0.0016 0.0000516

14 23.04 9.04 81.72 3.546

45 36.00 9.00 81.00 2.25

2 20.64 18.64 347.45 16.83

5 15.36 10.36 107.33 6.99

53 24.00 29.00 841.00 35.04 53 34.40 18.60 345.96 10.06

45 25.60 19.40 376.36 14.70

2 40.00 38.00 1444.00 36.10

We could now set up the following table:

Chi Square = 125.516; DF= (c - 1)(r - 1) = 2(2) = 4

H R j d S B l



H0 Rejected. See Below.

Df 0.5 0.10 0.05 0.02 0.01

1 0.455 2.706 3.841 5.412 6.635

2 1.386 4.605 5.991 7.824 9.210

3 2.366 6.251 7.815 9.837 11.345

4 3.357 7.779 9.488 11.668 13.277

5 4.351 9.236 11.070 13.388 15.086

Reject Ho because 125.516 is greater than 9.488 (for alpha = 0.05) Thus, we would reject the null hypothesis that there is no

relationship between location and type of malaria.

Inference : There is a relationship between type of malaria and

location.





Mann Whitney U Test Nonparametric equivalent of the

independent t test

Two independent groups

Ordinal measurement of the

Decision Variable The sampling distribution of U is

known and is used to testhypotheses in the same way as the t distribution.

Other Names: Mann– Whitney – Wilcoxon (MWW ), Wilcoxonrank-sum test, or Wilcoxon–

Mann– Whitney test)

Income Rank Income Rank



Mann Whitney U Test

To compute the Mann Whitney U:

Rank the scores inboth groups together

from highest tolowest.

Sum the ranks of thescores for each group.

The sum of ranks foreach group are used tomake the statisticalcomparison.

Rank Total

Income Rank Income Rank

25 12 27 10

32 5 19 17

36 3 16 20

40 1 33 4

22 14 30 7

37 2 17 19

20 16 21 15

18 18 23 1331 6 26 11

29 8 28 9

85 125



Non-Directional Hypotheses

Null Hypothesis: There is nodifference in scores of the twogroups (i.e. the sum of ranks for

group 1 is no different from thesum of ranks for group 2).

Alternative Hypothesis: There is adifference between the scores of

the two groups (i.e. the sum of ranks for group 1 is significantly different from the sum of ranks forgroup 2).



Computing the Mann Whitney U Using SPSS

Enter data into SPSS spreadsheet; two columns 1st column: groups; 2nd column: scores (ratings)

Analyze Nonparametric 2 Independent Samples

Select the independent variable and move it to theGrouping Variable box Click Define Groups Enter 1 for group 1 and 2 for group 2

Select the dependent variable and move it to the Test Variable boxMake sure Mann Whitney is selected

Click OK



Interpreting the OutputRanks

10 12.50 125.00

10 8.50 85.00

20

Income Status

Income Producing

No Income

Total

Equal Rights Atti tudes

N Mean Rank Sum of Ranks

Test Statisticsb

30.000

85.000

-1.512.131

.143a

Mann-Whitney U

Wilcoxon W

Z

Asymp. Sig. (2-tai led)

Exact Sig. [2*(1-tailed

Sig.)]

Equal Rights

Attitudes

Not corrected for ties.a.

Grouping Variable: Income Statusb.

The output provides a zscore equivalent of theMann Whitney U statistic.

It also gives significancelevels for both a one-tailedand a two-tailedhypothesis.



where n1

is the sample size for sample 1, and R 1

is the

sum of the ranks in sample 1. Note that it doesn't matter

which of the two samples is considered a sample. Take

either U1 or U2 as U and compute z. Same ǁ result gets.

where, mU andσ

U are the mean andstandard deviation of U . Z ≈ A standardnormal deviate whose significance can bechecked in tables of the normal distbn.



MW ‘U’ Test

U=R1- [(n1)(n1+1)]/2 =85-[(10)(11)]/2

=85-55=30.

σ= Sq.Rt.

[(n1.n2(n1+n2+1)/12]

= Sq.Rt. (2100/12)=13.23

Mu = n1*n2/2= 50

Z= [U- Mu]/ σ = [30-50]/13.23 =-1.512

U=R2- [(n2)(n2+1)]/2 =125-[(10)(11)]/2

=125-55=70

σ= Sq.Rt.[(n1.n2(n1+n2+1)/12]

= Sq.Rt. (2100/12)=13.23

Mu = n1*n2/2= 50 Z= [70- Mu]/ σ

= [70-50]/13.23 = 1.512

hi



Mann-Whitney U test II

Pair No Unmaried Married

thorax thorax

width width

1 4 2.8

2 3 2.7

3 2.6 2.6

4 3.85 2.7

5 2.65 2.6

6 2.7 2.6

7 2.85 2.7

8 2.85 2.8

9 3.2 2.9

10 2.9 2.6



Unmarried Married

thorax thorax

Rank width Rank width

3 2.6 3 2.6

6 2.65 3 2.6

8.5 2.7 3 2.6

13.5 2.85 3 2.6

13.5 2.85 8.5 2.7

15.5 2.9 8.5 2.7

17 3 8.5 2.7

18 3.2 11.5 2.8

19 3.85 11.5 2.8

20 4 15.5 2.9

Mann-Whitney U test

•Rank both lists as one combined list

•I found this a time consuming task



Mann-Whitney U test

•Sum the ranks for each sample

•N1= # obs in 1 N2= # obs in 2

Unmarried Married

thorax thoraxRank width Rank width

3 2.6 3 2.6

6 2.65 3 2.6

8.5 2.7 3 2.6

13.5 2.85 3 2.613.5 2.85 8.5 2.7

15.5 2.9 8.5 2.7

17 3 8.5 2.7

18 3.2 11.5 2.8

19 3.85 11.5 2.820 4 15.5 2.9

R1=134 R2=76



Wilcoxon Tests

Wilcoxon Dependent Sample



Wilcoxon Dependent Sample

Signed-Rank Test

Nonparametric equivalent of thedependent (paired-samples) t test

Two dependent groups (withindesign)

Ordinal level measurement of theDV.

The test statistic is T, and thesampling distribution is the T distribution.

Let N be the sample size, thenumber of pairs. Thus, there are atotal of 2N data points.



Procedure for Testing Hypotheses AboutDifferences in Related Samples

Data : The Data

are computed from pairs of measurementson each of the n elements in the sample.

ssumptions : The random variables

are independent and identically distributed, and theirdistribution is symmetric.

Test Statistic



Test Statistic

Let R = and

The test statistics

The reading scores in this section represent



The reading scores in this section represent

differences for 10 randomly selected individuals

who were measured before (Y) and after (X)

taking a speed reading course. These differenceswere determined by the Lilliefors test to be not

normally distributed. Therefore, the Wilcoxon

signed ranks test is used to test with = .01.

Alternate Hypothesis (The mean readingscores are higher after the course).

Null Hypothesis (The mean reading scores are notchanged by the course)

Person Score After Score Before Difference in Rank of Singed Rank



Course

X1

Course

Y1

Scores

D1 = X1 - Y1

[ D1] R 1

1 261 251 10 7 7

2 292 247 45 8 8

3 317 308 9 6 6

4 253 258 -5 3 -3

5 271 267 4 2 2

6 305 256 49 9 9

7 238 230 8 5 5 8 320 268 52 10 10

9 267 269 -2 1 -1

10 281 275 6 4 4



= 4.27

TR =Mean of R/(SR /√n) =4.7/(4.27/√10) = 4.7/1.35= 3.48



From t distribution with 9 degrees

of freedom the critical value =

2.8214. Since, test statisticT= 3.48 is greater than 2.8214, Ho

is Rejected.Inference : The mean reading

scores are changed by the course.

Computing the Wilcoxon Test Using SPSS



Computing the Wilcoxon Test Using SPSS

Enter data into SPSS spreadsheet; two columns

1st

column: pretest scores; 2nd

column: post-testscores

Analyze Nonparametric 2 Related Samples

Highlight both variablesmove to the TestPair(s) List Click OK

To Generate Descriptives:

Analyze Descriptive Statistics Explore

Both variables go in the Dependent box

Click StatisticsMake sure Descriptives arechecked Click OK

Wilcoxon Test



Wilcoxon TestTo compute the Wilcoxon

T: Determine the

differences betweenscores.

Rank the absolute valuesof the differences.

Place the appropriatesign with the rank (eachrank retains the positiveor negative value of itscorresponding

difference) T = the sum of the ranks

with the less frequentsign

Pretest Posttest DifferenceSgn.

Rank

36 21 15 1123 24 -1 -1

48 36 12 10

54 30 24 12

40 32 8 732 35 -3 -3

50 43 7 6

44 40 4 4

36 30 6 529 27 2 2

33 22 11 9

45 36 9 8



Non-Directional Hypotheses

Null Hypothesis: There is nodifference in scores beforeand after an intervention (i.e.the sums of the positive and

negative ranks will besimilar). Non-Directional Research

Hypothesis: There is a

difference in scores beforeand after an intervention (i.e.the sums of the positive andnegative ranks will bedifferent).

Interpreting the Output



Interpreting the OutputRanks

10a 7.40 74.00

2b 2.00 4.00

0c

12

Negative Ranks

Positive Ranks

Ties

Total

POSTTEST - PRETES

N Mean Rank Sum of Ranks

POSTTEST < PRETESTa.

POSTTEST > PRETESTb.

POSTTEST = PRETESTc.

Test Statisticsb

-2.746a

.006

Z

Asymp. Sig. (2-tail ed)

POSTTEST -

PRETEST

Based on positive ranks.a.

Wilcoxon Signed Ranks Testb.

The T test statistic is the sumof the ranks with the lessfrequent sign.

The output provides theequivalent z score for the teststatistic.

Two-Tailed significance is

given.





The Kruskal-Wallis TestThe Kruskal-Wallis test is a nonparametric test that can be used to

determine whether three or more independent samples were selected

from populations having the same distribution.

H0: There is no difference in the population distributions.

Ha: There is a difference in the population distributions.

Combine the data and rank the values. Then separate

the data according to sample and find the sum of theranks for each sample.

R i= the sum of the ranks for sample i.

Kruskal Wallis Test



Kruskal-Wallis Test

) )13

1

122

2

2

2

1

2

1

= N

n

R

n

R

n

R

N N H

k

k

Given three or more independent samples, the test

statistic H for the Kruskal-Wallis test is

where k represents the number of samples, niis the size

of the ith sample, N is the sum of the sample sizes, and

Riis the sum of the ranks of the ith sample.

Reject the null hypothesis when H is

greater than the critical number.

(always use a right tail test.)



You know the one-way ANOVA is an extension of thetwo independent groups t-test to a 3 or morepopulation problem.

The Kruskal-Wallis test is an extension of theMann-Whitney U test to a 3 or more populationproblem

The Kruskal-Wallis test handles k-independent

groups of samples, based on chi-square. Here it isNon-parametric.

Like the Mann-Whiteny U test, this test uses ranks.

Note: Kruskal-Wallis test in the ranktransformed pattern is One-way ANOVA, basedon F test. Here it is Parametric.



Procedure

1. Combine the observations of the variousgroups

2. Arrange them in order of magnitudefrom lowest to highest

3. Assign ranks to each of the observationsand replace them in each of the groups

4. Original ratio data have thereforebeen converted into ordinal orrankeddata



5. Ranks are summed in eachgroup and the test statistic, H iscomputed

6. R anks assigned to observationsin each of the k groups are

added separately to give k ranksums

You want to compare the hourly pay rates of accountants



You want to compare the hourly pay rates of accountants

who work in Michigan, New York and Virginia. To do so,

you randomly select 10 accountants in each state and record

their hourly pay rate as shown below. At the .01 level, can

you conclude that the distributions of accountants’ hourly

pay rates in these three states are different?

MI(1) NY(2) VA(3)

14.24 21.18 17.02

14.06 20.94 20.6314.85 16.26 17.47

17.47 21.03 15.54

14.83 19.95 15.38

19.01 17.54 14.9

13.08 14.89 20.48

15.94 18.88 18.5

13.48 20.06 12.8

16.94 21.81 15.57



H0 : There is no difference in the hourly pay rate in the 3 states.Ha : There is a difference in the hourly pay in the 3 states.

0.01 =

1. Write the null and alternative hypothesis

2. State the level of significance

3. Determine the sampling distribution

The sampling distribution is chi-square with d.f. = 3-1 = 2

From Table ,the critical value is 9.210.

2 5. Find the rejection region

4. Find the critical value

Data State Rank

12 8 VA 1



Test Statistic12.8 VA 1

13.08 MI 2

13.48 MI 3

14.06 MI 4

14.24 MI 5

14.83 MI 6

14.85 MI 714.89 NY 8

14.9 VA 9

15.38 VA 10

15.54 VA 11

15.57 VA 12

15.94 MI 13

16.26 NY 14

16.94 MI 1517.02 VA 16

17.47 MI 17.5

17.47 VA 17.5

17.54 NY 19

18.5 VA 20

18.88 NY 21

19.01 MI 22

19.95 NY 2320.06 NY 24

20.48 VA 25

20.63 VA 26

20.94 NY 27

21.03 NY 28

21.18 NY 29

21.81 NY 30

Michigan salaries are in ranks:

2, 3, 4, 5, 6, 7, 13, 15, 17.5, 22The sum = 94.5

New York salaries are in ranks:

8, 14, 19, 21, 23, 24, 27, 28, 29, 30The sum is 223

Virginia salaries are in ranks:

1, 9, 10, 11, 12, 16, 17.5, 20, 25, 26The sum is 147.5

R 94 5 R 223 R 147 5



R 1= 94.5, R 2 = 223, R 3 =147.5

n1 = 10, n2=10 and n3

=10, so N = 30

H =2 2 212 94.5 223 147.5 3(30 1) 10.76

30(30 1) 10 10 10 =

10.769.210

Make Your Decision

Interpret your decision

The test statistic, 10.76 falls in the rejection region, soReject the null hypothesis

There is a difference in the salaries of the 3 states.

Find the test statistic



Purpose –

Allows a scientist to test the influence of theindependent variable upon the dependent variable

Controls for the influence of other variables

Conclusions

Primary question – “How reasonable are these results if chance alone were responsible?”

If the results are not due to chance, then the results areattributed to the experimental manipulation

Example. A study is being conducted on whether entering college students gain weightduring the freshman year. Below are the "Before" and "After" weights for a random



g y gsample of 30 students. Test to see whether there is a significant "gain" in weights afterthe freshman year in college.

Before After Before fter

133 135 + 121 125 +

152 160 + 144 140 _

169 180 + 106 108 +

156 154 _ 182 175 +

178 185 + 122 120 _

220 226 + 110 114 +

145 150 + 130134 +

138 140 + 165 165 0

218 225 + 158 160 +

140 140 0 106 105 _

148 143 _ 160 166 +

98 102 + 122 125 +

142 138 _ 146 155 +

170 182 + 112 115 +

108 112 + 145 144 _

Sign test Calculation



Sign test Calculation No. of positive Ranks =20; Negative Ranks =8; Ties = 2

Null hypothesis: No difference in weight in pre-poststages

H0: Assumed Population ‘P’ = ‘Q’=0.5

Sample p =20\28 =0.71 & III ly, Sample: q =8\28 =0.29.

SE of Proportion σ= Sq.Rt [(PQ)/n]= Sq.Rt [(.5 * .5)/28]=

0.0298

Z= [p-P]/ SE = [0.71-0.5]/ 0.0298 = 0.21/ 0.0298= 7.05

H0 Rejected at 5% Significance level.

SPSS



SPSS

After entering the data in the appropriate lists and executing the SIGNTEST

program by entering 2 for the alternative X < Y, we see that there are 20

persons who increased in weight (pos. changes) out of 28 persons who

actually changed weight (changes). If there were "no difference" (i.e., if p =

P(After > Before) = 0.5) ), then there would be only a 0.01785 probability

(from the right-tail P-value) of there being as many as 20 people out of 28

who gained weight. This low p-value gives evidence to reject the claim that

there is no difference in favor of the alternative that there is tendency to gain

weight.



Thank you

Date post:	03-Apr-2018
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Non Parametric Tests ICSSR RMC DIB&C Dr.S.selvaRani

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