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Chi Square Test

The Chi-square distribution has many uses in the field of testing of hypotheses. It helps to test whether a population has given variance. It also helps to test ‘goodness of fit’ of a theoretical distribution to an observed distribution and in testing independence of attributes in a contingency table.

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What is Chi Square Test?

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Any statistical test that uses the chi square distribution can be called chi square test. Chi-square test is conducted a statistical test to investigate difference, and it is denoted by $\chi^2$. The chi-square test measures the difference between a statistically generated expected result and an actual result to see if there is a statistically significant difference between them. It measure the goodness of fit between an expected and an actual result.

Chi Square Test Formula

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The formula for Chi Square is defined as follows:

$\chi ^2 = \sum \frac{(O - E) ^2}{E}$

Where,

$\chi ^2$ - Chi Square

O - Observed sample in each category.

E - Expected frequency in corresponding category.


Chi Square Test Degrees of Freedom

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The degree of freedom for the chi square difference test is equal to the difference between degree of freedom associated with the models. Each type of two way table has its own chi-square distribution, depending on the number of rows and columns, and each chi-square distribution is identified by its degree of freedom. A two way table with r rows and c column uses a chi-square distribution with (r - 1)*(c - 1) degree of freedom.

  1. For one degree of freedom, the distribution looks like a hyperbola.
  2. For than one degree of freedom, it loos like a mound that has a long right tail.

Chi Square Test of Independence

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Chi square test is applied when we have two categorical variables from a single population. It is used to determine whether there is a significant association between the two variables. This test is applicable when the observations are independent (random). The Chi-square test for independence is also called a contingency table Chi-square test.

Chi Square Test of Independence Example


For a given population, we consider two attributes and we may find the dependence between them. We have a set of workers in a factory and we try to classify them as smokers and non-smokers. The same workers are classified again as 'men' and 'women'. Here, we may find that the number of smokers are more in men than in women. So, we say that the attributes 'smoking' and 'sex' is dependent (associated).

Chi Square Goodness of Fit Test

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This test is applicable when the observations are independent (random) and the total frequency should be large. This test is used to test association of variables in two-way tables where the assumed model of independence is evaluated against the observed data. The chi-square goodness of fit test is that it can be applied to any univariate distribution for which you can calculate the cumulative distribution function. The chi-square goodness-of-fit test can be applied to discrete distributions such as the binomial and the Poisson.

Chi-square test statistic is of the form

$\chi^2$ = $\frac{\sum (\text{Observed value - Expected value})^2}{\text{Expected value}}$

Degree of Freedom for the Chi-Square Test for Goodness of Fit

The number of degree of freedom that we calculate for the Chi-square test for goodness of fit reflects the number of categories that we are comparing minus one.

Degree of freedom (df) = c - 1

Chi Square Difference Test

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The chi square difference test is very useful both for making simpler models more complex and for making complex models simpler. A more accurate test can be obtained by performing a chi square difference test.
  • Estimating the original model.
  • Estimating the revised model in which new path has been added.
  • Calculating the difference between the two resulting chi square values.
The resulting chi square difference statistic also has a chi square distribution. The degree of freedom for the chi square difference test is equal to the difference between degree of freedom associated with the models. When the chi square difference is statisticant, the model with the smaller chi-square is considered to fit the data better than the model with the higher chi-square.

Chi Square Test of Homogeneity

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The chi square test of homogeneity is used to test the differences between two popuations that are homogeneous with respect to some characteristics. In this test categories are assumed mutually and exhaustively exclusive. The test statistics for chi square test of homogeneity is the same as that for chi square of association.

$\chi^2$ = $\sum_{i=1}^{m}\sum_{j=1}^n \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$

Where, df = (m - 1)(n - 1).

Chi Square Test of Association

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Chi-square test of association is equivalent to the Chi-square test of independence and the Chi-square test of homogeneity. The Chi-square test of association is used to determine whether there is an association between two or more categorical variables. In the Chi-square test of association the expected proportions are known a priori, for the Chi-square test of association the expected proportions are not known a priori but must be estimated from the sample data.

Chi Square Test for Trend

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The chi-square test for trend tests is a linear trend between rows and the columns of the table. It only makes sense when the rows are arranged in a natural order (such as by age or time), and are equally spaced. A large chi-square statistic indicates in the table, the observed frequencies differ markedly from the expected frequencies. When a chi-square is high, examine the table to determine which cells are responsible. In the chi-squared test for trend, we not only use the order of the categories, but attach a numerical value. The chi-squared for trend statistic is always less than the chi-squared for association statistic. The difference between the two chi-squared statistics follows a Chi-squared distribution if the null hypothesis is true, with degrees of freedom equal to the difference between the two degrees of freedom.

One Sample Chi Square Test

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The one-sample Chi-square test compares the distribution of cases across the categories of a variable with a hypothesized distribution. The Chi-square test used with one sample is described as a "goodness of fit" test. It can help you decide whether a distribution of frequencies for a variable in a sample is representative of, or "fits", a specified population distribution. The one sample Chi-square test is used to test a hypothesis such as 'suicide rate varies significant from month to month'. If the hypothesis is false, the suicide rate will be the same for one of the twelve months. The one sample Chi-square test can be used to compare observed suicide rates per month with what would be expected if the rate were equal for the all months.

Chi Square Test Interpretation

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The chi-square test measures the difference between a statistically generated expected result and an actual result to see if there is a statistically significant difference between them. After finding the Chi-square value and the degree of freedom are known, a standard table of Chi-square values can be consulted to determine the corresponding p-value. The p value indicates the probability that a Chi-square value that large would have resulted from the chance.

Chi Square Test Assumptions

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The chi-square test have some important assumption.
  • For the chi-square test to be meaningful it is imperative that each person, item or entity contributes to only one cell of the contingency table.
  • Both independent and dependent variables are categorical with two or more levels.
  • The data consist of frequencies, not scores.
  • Each randomly selected observation can be classified into only one category for the independent variable and only one category for the dependent variable.

Purpose of Chi Square Test

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Chi-square test is one of the simplest and most widely used non-parametric tests. The chi-square test is the most commonly used method for comparing frequencies or proportions. It is a statistical test used to compare observed data with data that would be expected according to a given hypothesis. It is very popularly known as test of "goodness of fit" for the reason that it enables us to ascertain how appropriately the theoretical distributions.

Chi Square Test Table

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Table for Chi square test is given below:

Chi Square Test Table

Chi Square Test Example

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Given below are some of the examples on chi square test.

Solved Examples

Question 1:

Find the chi square for the following given datas

Color

Blue

Black

Brown

Yellow

Observed frequency

5

15

10

20

Expected frequency

10

20

5

30


Solution:
 
For blue, Observed frequency - Expected frequency = 5-10 = -5

For black, Observed frequency - Expected frequency = 15-20 = -5

For brown, Observed frequency - Expected frequency = 10-5 = 5

For yellow, Observed frequency - Expected frequency = 20-30 = -10

For blue,` (O-E) ^2/E` = `(-5) ^2/10 `

= `25/10` =2.5

For black, `(O-E) ^2/E` = `(-5) ^2/20`

= `25/20` = 1.25

For brown, `(O-E) ^2/E` = `(5) ^2/5 `

= `25/5` =5

For yellow, `(O-E) ^2/E` =` (-10)^2/30`

= `100/30` = 3.3333

`chi ^2 = sum( O-E) ^2/E`

= (2.5 + 1.25 + 5 + 3.3333)

=9.58333

The above calculation is also done by tabulation method which is explained below,

Color

Observed frequency

Expected frequency

Observed frequency -Expected Frequency
(O-E)

`(O-E)^2`

`(O-E)^2/E`

Blue

5

10

-5

25

2.5

Black

15

20

-5

25

1.25

Brown

10

5

5

25

5

Yellow

20

30

-10

100

0.83333


The formula for define Chi Square test is given by,

` chi ^2 = sum( O-E) ^2/E`

= (2.5 + 1.25 + 5 + 3.3333)

=9.58333


 

Question 2:

Find the chi square for the following given datas

Color

Blue

Black

Brown

Yellow

Observed frequency

10

5

25

35

Expected frequency

15

30

30

25


Solution:
 

Color

Observed frequency

Expected frequency

Observed frequency -Expected frequency
(O-E)

`(O-E)^2`

`(O-E)^2/E`

Blue

10

15

-5

25

1.6666

Black

5

30

-25

625

20.8333

Brown

25

30

-5

25

0.8333

Yellow

35

25

10

100

4


The formula for Chi Square is given by,

`chi ^ 2 = sum (O-E) ^2/E`

= 1.6666 + 20.8333 + 0.8333 + 4.0000

=27.3332


 

Question 3:

Find the chi square for the following given datas

Color

Blue

Black

Brown

Yellow

Observed frequency

23

24

32

23

Expected frequency

12

32

25

21


Solution:
 

Color

Observed frequency

Expected frequency

Observed frequency -Expected frequency
(O-E)

`(O-E)^2`

`(O-E)^2/E`

Blue

23

12

11

121

10.0833

Black

24

32

-8

64

2.0000

Brown

32

25

7

49

1.9600

Yellow

23

21

2

4

0.1905


The formula for Chi Square is given by,

`chi ^ 2 = sum (O-E) ^2/E`

=10.0833 + 2.0000 + 1.9600 + 0.1905

=14.2338


 

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