Chi Square Test for independence Lecture 38

Chi-Square Test for independence  

Lecture 38

Contingency Table

A table in which two variables, each of which has several levels, are classified in “r” rows and “c” columns is a powerful tool for analysing the two variables. This technique can be used to analyse categorical data or nominal data. The data presented in a contingency table is used to test the null hypothesis that the two variables are independent.

Let's have two variables, say A with r levels that’s A1, A2, …, Ai, …, Ar, and B with c levels that’s B1, B2, …, Bj, …, Bc. Then it can be arranged as:

$\mathrm{The\; r\times c\; contingency\; table\; is\; the\; name\; given\; to\; the\; above\; tabular\; arrangement.\; The\; observed\; data\; is\; represented\; by\; Oij.<span> The\; expected\; value\; can\; be\; obtained\; as:</span>}$

$\mathrm{<span>e<span\; style="font-size:\; x-small;">ij</span>=\; (Ai)(Bj)\; /\; n</span>}$

vi. Remarks: The computed chi-square value falls in the rejection area; the sample data does not provide sufficient evidence to accept the null hypothesis at the 10% significance level. Thus, it is concluded that the shifts and the quality of tiers are associated.

$$

Example 9.18: The following outcomes came from an experiment on how growth regulators affected muskmelon fruit setting. At the 5% significance level, determine if muskmelon fruit setting and growth regulator treatment are associated.

G / S	fruit set	Fruit not set
Treated	20	14
Control	8	25

Solution:

i. The null & alternative hypotheses may be stated as:

H0: The growth regulator and muskmelon fruit setting not associated 

vs. H1: The growth regulator and muskmelon fruit setting are associated.

ii. The significance level: α = 0.05

iii. The test statistic:

iv. Reject the null hypothesis if χ² ≥ χ²₀.₀₅(1) = 3.841

v. Computation:

$$

G / S	fruit set	Fruit not set	Total
Treated	20	14	34
Control	8	25	33
Total	28	39	67

vi. Remarks: The computed chi-square value falls in the rejection area; the sample data does not provide sufficient evidence to accept the null hypothesis at the 5% significance level. Thus, it is concluded that the growth regulator and muskmelon fruit setting are associated.

Co efficient of Contingency Table

The chi-square statistic simply indicates the independence or association of two variables in classification. It provides no information regarding the degree of association, which is occasionally required. Karl Pearson has defined a coefficient "C," also referred to as the coefficient of contingency table, which measures the strength of association between two variables. This coefficient is given by:

k is the smaller one in r and c; e.g., in table 2x3, k = 2

The coefficient measures the strength of the association between two variables in classification.

When C = 0. There is complete independence.

And when C > 0. There will be association

Cramer’s Co efficient of Contingency Table

Cramer’s proposed another measure for the strength of association in a contingency table, given by:

Where n is the sample size and k is the smaller r or c.

The value of Q lies between 0 and 1.

When Q = 0, there will be no association.

           Q < 0.2, there will be weak association

           Q = 0.5; there will be moderate association

Q > 0.7, there is a strong association.

Example 9.19: Using the data of Example 9.17. Find the strength of the association.

Shifts	Excellent	Satisfactory	Defective
1	106	124	2
2	68	86	1
3	38	72	3

Solution:

Co efficient of Contingency Table

As the number of rows and number of columns are identical, so k = 3.

$$

There is a weak association between shifts and quality of tiers.

Cramer’s Coefficient of Contingency Table

As Q < 0.2, there is a weak association between shifts and quality of tiers.

• Read More: Yates Correction & Fisher Exact Test