Chi square test final presentation

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χ 2 (chi - square) Test: as a test of Independence χ 2 (chi - square) Test: as a test of Independence condition of application of chi - square test condition of application of chi - square test & & steps involved steps involved ------------------------------------------------------------------------------------------------------ A

Transcript of Chi square test final presentation

Page 1: Chi square test final presentation

χ 2 (chi - square) Test: as a test of Independence χ 2 (chi - square) Test: as a test of Independence condition of application of chi - square test condition of application of chi - square test

&&steps involvedsteps involved

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DEFINITION OF CHI-SQUARE TEST INTRODUCTION TO CHI-SQUARE TEST CHI-SQUARE AS A TEST FOR COMPARING VARIANCE CHI-SQUARE AS A NON-PARAMETRIC TEST CONDITIONS FOR THE APPLICATION OF χ2 TEST CHARACTERISTICS OF χ2 TEST STEPS INVOLVED IN APPLYING CHI-SQUARE TEST AND

A NUEMERICAL TO EXPLAIN SIGNIFICANCE OF CHI SQUARE TEST ONE TAILED AND TWO TAILED TESTS CONCLUSION REFERENCES

CONTENTS:CONTENTS:

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The Chi-Square (χ2) test could be defined as a Nonparametric test that is used to test hypotheses about distributions of frequencies across categories of data. It can be used to test for comparing variance too.

DEFINATION OF CHI-SQUARE TESTDEFINATION OF CHI-SQUARE TEST

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We often have occasions to make comparisons between two characteristics of something to see if they are linked or related to each other.

One way to do this is to work out what we would expect to find if there was no relationship between them (the usual null hypothesis) and what we actually observe.

The test we use to measure the differences between what is observed and what is expected according to an assumed hypothesis is called the chi-square test.

INTRODUCTION TO CHI-SQUARE TESTINTRODUCTION TO CHI-SQUARE TEST

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The chi-square test is often used to judge the significance of population variance.

Formula of Chi-square test of comparing variance:

χ 2 = σ2s /σ2

p (n-1)

Where σ2

s = variance of the sampleσ2

p = variance of the population(n-1) = degree of freedom, n being the number of items in the sample.

CHI-SQUARE TEST AS A TEST FOR CHI-SQUARE TEST AS A TEST FOR COMPARING VARIANCECOMPARING VARIANCE

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Chi-Square Table Chi-Square Table

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Chi-square is an important test and as such no rigid assumptions are necessary in respect of the type of population. We require only the degree of freedom (implicitly of course the size of the sample) for using this test. As a non-parametric test, chi square can be used (i) as a test of goodness of fit, (ii) as a test of homogeniety and (iii) as a test of independence.

CHI-SQUARE AS A NON-PARAMETRIC TESTCHI-SQUARE AS A NON-PARAMETRIC TEST

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As a test of goodness of fit - It is used to test the single variable i.e., it is fit to the population or not. It means Sample is supporting the population or not.

As a test of homogeneity- It is used to test a single categorical variable from two different populations. It is used to determine whether frequency counts are distributed identically across different populations.

As a test of independence – To explain that variable are how much attached with each other.

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There must be two observed set of data or one observed set and one expected set [generally there are n - rows and c - columns].The number of observations or items must be reasonably large [i.e., normally equal or greater than 50]. Expected frequency should not be very small should not be less than five.Observations recorded and used are collected on random basis.All the times in the sample must be independent.

CONDITIONS FOR THE APPLICATIONS OF CONDITIONS FOR THE APPLICATIONS OF χχ2 2 TESTTEST

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Chi-square is based on the observed frequency and not on the parameters mean and standard deviation.It is a technique for testing the hypothesis and not used for the purposes of estimation.It is useful test for research and is used to complex contingency table with several Classes. In contingency table variables are arranged in different formats in a table; It is a way to present the different attributes of a population.The text involves less mathematical details and there is no need of any parameter values eg. Mean and standard deviation.It shows additive property.

CHARACTERISTICS OF CHARACTERISTICS OF χχ2 2 TESTTEST

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Step 1-First of all expected frequencies are calculated on the basis of null hypothesis. The expected frequency of any given cell is calculated on the basis of following formula:

Expected frequency of any cell = (Row total of that cell) X (Column Total of the cell) / Grand Total of the Table.Step 2-Obtained difference between observed and expected frequency and obtained square of such differences, i.e. (Oi-Ei)2

STEPS INVOLVED IN CHI-SQUARESTEPS INVOLVED IN CHI-SQUARE

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Step 3-Divide the quantity (Oi-Ei) by the corresponding expected frequencies to get (Oi-Ei) 2 / Ei

Step 4-Find the summation (∑) of there value to get ∑ (Oi-Ei) 2 / Ei

This is the required chi-square value or χ2 value as the formula of χ2 test is as follows:χ2 =∑ (Oi-Ei) 2 / Ei Step 5-Thus the calculated value of chi-square should be compared with relevant table value of χ2 and inference can be drawn as stated above.

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Data obtained during out break of small pox in town.

Test the effectiveness of vaccination in preventing the attack from smallpox. Test your result with the help of χ2 test at 5 % level of significance.Attributes:VaccinationAttacked Null Hypothesis (H0)- Vaccination (A) and Small Pox (B) are Independent and i.e., vaccination is not effective in small pox prevention.Alternative Hypothesis (Ha)- Vaccination (A) is effective in small pox prevention.

Numerical Numerical

Attacked (B) Not Attacked (b) Total

Vaccinated (A) 31 469 500

Not vaccinated (a) 185 1315 1500

Total 216 1784 2000

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Solution

Contingency Table:

Calculation of Expected Frequencies for (CT):AB = 500 X 216 / 2000 = 54aB = 1500 X 216 / 2000 = 162Ab = 500 X 1784 / 2000 = 446Ab = 1500 X 1784 / 2000= 1338

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Attacked (B) Not Attacked (b) Total

Vaccinated (A) (AB) 54 (Ab) 446 500

Not vaccinated (a)

(aB) 162 (ab) 1338 1500

Total 216 1784 2000

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Putting the value in table:

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Group Oi Ei Oi - Ei (Oi - Ei)2 (Oi - Ei)2 / Ei

AB 31 54 31-54 = -23 529 9.796

Ab 469 446 469-446 = 23 529 1.186

aB 185 162 185-162 = 23 529 3.265

ab 1315 1338 1315-1338 = -23 529 0.395

∑ (Oi - Ei)2 / Ei14.642

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χ2 = (Oi - Ei)2 / E2

χ2 = 14.642

Degrees of freedom in this case = (r-1) (C-1)Degrees of freedom in this case = (2-1) (2-1)Degrees of freedom in this case = (1) (1) Degrees of freedom in this case = 1

Table value of χ2 for 1 degree of freedom at 5% level of significance is 3.841. The calculated value of χ2 is much higher than this table value and hence the result of the experiment does not support the hypothesis. We can, thus, conclude that vaccination is effective in preventing the attack from smallpox.

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SIGNIFICANCE OF THE CHI-SQUARE TEST IS AS FOLLOW:

It is very useful to check the Interdependence of two variables on each other.Chi Square can also be used to test differences between two or more actual samples.Being a nonparametric tests chi-square test is much easier to compute, and also to apply.The Chi Square (X2) 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 and another hypothetical or previously established distribution such as that which may be expected due to chance or probability. Chi square test can be used to treat data which have been measured on nominal (classificatory) scales. Such data cannot, on any logical basis, be ordered numerically, hence there is no possibility of using parametric statistical tests which require numerical data.

SIGNIFICANCE OF CHI-SQUARE TESTSIGNIFICANCE OF CHI-SQUARE TEST

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As conclusion we can say that the chi-square test is an important test amongst the several tests of significance developed by statisticians. Being a non-parametric test It’s calculations are easier than other tests. It tests for effect of one variable over the other, hence it is very helpful to understand the effectiveness of products. It is very useful for all researchers to Test goodness of fit.Test the significance of association between two attributes.Test the homogeneity or the significance of population variance.

ConclusionConclusion

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Fundamentals of Statistics by Elhance, D.N., Elhance,

Veena and Aggarwal, B.M., Published by Kitab Mahal

Introduction to statistics by Hooda, R.P., Published

by McMillan Publishing

Research Methodology: methods and techniques by

Kothari, C.R., Published by New Age International

Class notes of Mdm. Aarti joshi

ReferencesReferences

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