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Chapter 9

Chi-Square Distribution

Abstract

This chapter discusses about the chi-square distribution. Basically chi-square distribution is the measure of which enables us to find out the degree of degree of discrepancy between the observed and expected frequencies. In this chapter we discuss the uses of chi-square test as a test for goodness of fit, as a test for independence of attributes and homogeneity chi-square.

Keywords

Calculation of expected frequencies; chi-square distribution; mean of chi-square distribution; variance of chi-square distribution; conditions for using chi-square; test for goodness of fit; test for independence of attributes and homogeneity chi-square

9.1 Introduction

In this chapter we introduce chi-square distribution, the measure of which enables us to find out the degree of discrepancy between the observer and expected frequencies and then to determine whether the discrepancy between the observed and expected frequencies is due to error of sampling or due to chance.

The chi-square is denoted by the symbol χ². It was discovered by Helmert in 1875 and was rediscovered by Karl Pearson in 1900. Chi-square is always positive. The value of χ² lies between 0 and ∞.

Since χ² is not derived from the observations in a population, it is not a parameter.

Chi-square test is not a parametric test.

The χ² distribution is used for testing the goodness of fit. It is used for finding association and relation between attributes. It is also used to test the homogeneity of independent estimates of population. Chi-square is computed on the basis of frequencies in a sample and the value of χ² so obtained is a statistic.

Chi-Square Test (χ²):

χ² test is defined as

$χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$

where O_i=Observed frequency; E_i=Expected frequency.

9.2 Contingency Table

A classification table containing r rows and c columns figuring observed frequencies is called contingency table. A 2×2 contingency table is of the form:

a	b
c	d

If the data is divided into m classes A₁, A₂, …, A_m according to an attribute A and n classes B₁, B₂, …, B_n according to another attribute B, the m×n contingency table can be formed as follows:

Attributes	B₁	B₂	…	B_j	…	B_n
A₁	A₁₁	A₁₂	…	A_1j	…	O_1n
A₂	A₂₁	A₂₂	…	A_2j	…	O_2n

A_i	A_i1	A_i2	…	A_ij	…	O_in

A_m	A_m1	A_m2	…	A_mj	…	O_mn

where O_ij denote the ith row jth column frequency of the cell belonging to both the classes A_i and B_j.

9.3 Calculation of Expected Frequencies

Consider the 2×2 contingency table (Table 9.1).

Table 9.1

2×2 Contingency table

		Total
a	b	a+b
c	d	c+d
Total a+c	b+d	N=a+b+c+d

where a, b, c, and d are observed frequencies.

The expected frequencies corresponding the cell frequencies a, b, c, and d can be calculated and expressed in the form of the given Table 9.2.

Table 9.2

Expected frequency table for 2×2 contingency table

$\frac{(a + c) (a + b)}{N}$ $\frac{(a + c) (a + b)}{N}$	$\frac{(b + d) (a + b)}{N}$ $\frac{(b + d) (a + b)}{N}$
$\frac{(a + c) (c + d)}{N}$ $\frac{(a + c) (c + d)}{N}$	$\frac{(b + d) (c + d)}{N}$ $\frac{(b + d) (c + d)}{N}$

This method can be extended to compute the expected frequencies of a m×n contingency table.

9.4 Chi-Square Distribution

Let samples of size n be drawn from a normal population with standard deviation σ and if for each sample we calculate: a sampling distribution of χ² can be obtained. It is given by

$\begin{array}{l} y = f (χ^{2}) & = y_{0} e^{- \frac{1}{2} χ^{2}} \cdot {(χ^{2})}^{\frac{1}{2} (ν - 2)} \\ = y_{0} e^{- \frac{1}{2} χ^{2}} \cdot {(χ)}^{(ν - 2)} \end{array}$ $\begin{array}{l} y = f (χ^{2}) & = y_{0} e^{- \frac{1}{2} χ^{2}} \cdot {(χ^{2})}^{\frac{1}{2} (ν - 2)} \\ = y_{0} e^{- \frac{1}{2} χ^{2}} \cdot {(χ)}^{(ν - 2)} \end{array}$

where $ν = n - 1$ $ν = n - 1$ is the number of degrees of freedom and y₀ is a constant depending on such that the total area under the curve one. The χ² distribution corresponding to various values of are shown in the Fig. 9.1.

Maximum value of y:

We have

$y = y_{0} e^{- 1 / 2 χ^{2}} {(χ^{2})}^{1 / 2 (ν - 2)}$ $y = y_{0} e^{- 1 / 2 χ^{2}} {(χ^{2})}^{1 / 2 (ν - 2)}$ (9.1)

(9.1)

Differentiating Eq. (9.1) with respect to χ² we get

$\frac{d y}{d χ^{2}} = y_{0} [\frac{1}{2} (ν - 2) {(χ^{2})}^{1 / 2 (ν - 4)} e^{- 1 / 2 χ^{2}} - \frac{1}{2} e^{- χ^{2} / 2} {(χ^{2})}^{1 / 2 (ν - 2)}]$ $\frac{d y}{d χ^{2}} = y_{0} [\frac{1}{2} (ν - 2) {(χ^{2})}^{1 / 2 (ν - 4)} e^{- 1 / 2 χ^{2}} - \frac{1}{2} e^{- χ^{2} / 2} {(χ^{2})}^{1 / 2 (ν - 2)}]$

Taking $\frac{d y}{d x} = 0$ $\frac{d y}{d x} = 0$ for maximum value of y we get

$χ^{2} = ν - 2$ $χ^{2} = ν - 2$ for $ν \geq 2$ $ν \geq 2$ (since χ² cannot be negative)

The constant y₀ is related in such a way that the area χ²=0 to ∞ is unity.

In this case

$y_{0} = \frac{1}{2^{ν / 2} Γ (ν / 2)}$ $y_{0} = \frac{1}{2^{ν / 2} Γ (ν / 2)}$

9.4.1 Characteristic Function of χ² Distribution

We have

$\begin{array}{l} φ (χ^{2} (t)) & = E (e^{i t χ^{2}}) \\ = \int_{0}^{\infty} e^{i t χ^{2}} f (χ^{2}) d χ^{2} \\ = \frac{1}{2^{ν / 2} Γ (ν / 2)} \int_{0}^{\infty} e^{- 1 / 2 (1 - 2 i t) χ^{2}} {(χ^{2})}^{1 / 2 (n - 1)} d χ^{2} \\ = {(1 - 2 i t)}^{- n / 2} \end{array}$ $\begin{array}{l} φ (χ^{2} (t)) & = E (e^{i t χ^{2}}) \\ = \int_{0}^{\infty} e^{i t χ^{2}} f (χ^{2}) d χ^{2} \\ = \frac{1}{2^{ν / 2} Γ (ν / 2)} \int_{0}^{\infty} e^{- 1 / 2 (1 - 2 i t) χ^{2}} {(χ^{2})}^{1 / 2 (n - 1)} d χ^{2} \\ = {(1 - 2 i t)}^{- n / 2} \end{array}$

9.5 Mean and Variance of Chi-Square

The moment generating function of χ² with respect to the origin is

$\begin{array}{l} M_{0} (t) & = \int_{0}^{\infty} e^{t χ^{2}} \frac{1}{2^{ν / 2} Γ (ν / 2)} e^{- \frac{1}{2} (χ^{2})} {(\frac{1}{2} χ^{2})}^{\frac{1}{2} (ν - 1)} d (χ^{2}) \\ = \frac{1}{2^{\frac{ν}{2}} Γ (ν / 2)} \int_{0}^{\infty} \exp {- (\frac{1 - 2 t}{2}) χ^{2}} {(χ^{2})}^{\frac{ν}{2} - 1} d (χ^{2}) \\ = \frac{1}{2^{\frac{ν}{2}} Γ (ν / 2)} . \frac{Γ (\frac{1}{2} ν)}{{\frac{1}{2} (1 - 2 t)}^{\frac{ν}{2}}} \\ = {(1 - 2 t)}^{- \frac{ν}{2}}, (| 2 t | < 1) \\ = 1 + \frac{ν}{2} (2 t) + \frac{1}{2} \frac{ν (\frac{1}{2} ν + 1)}{2!} {(2 t)}^{2} + \dots \\ + \frac{\frac{1}{2} ν (\frac{1}{2} ν + 1) \dots (\frac{1}{2} ν + r - 1)}{r!} {(2 t)}^{r} + \dots \end{array}$ $\begin{array}{l} M_{0} (t) & = \int_{0}^{\infty} e^{t χ^{2}} \frac{1}{2^{ν / 2} Γ (ν / 2)} e^{- \frac{1}{2} (χ^{2})} {(\frac{1}{2} χ^{2})}^{\frac{1}{2} (ν - 1)} d (χ^{2}) \\ = \frac{1}{2^{\frac{ν}{2}} Γ (ν / 2)} \int_{0}^{\infty} \exp {- (\frac{1 - 2 t}{2}) χ^{2}} {(χ^{2})}^{\frac{ν}{2} - 1} d (χ^{2}) \\ = \frac{1}{2^{\frac{ν}{2}} Γ (ν / 2)} . \frac{Γ (\frac{1}{2} ν)}{{\frac{1}{2} (1 - 2 t)}^{\frac{ν}{2}}} \\ = {(1 - 2 t)}^{- \frac{ν}{2}}, (| 2 t | < 1) \\ = 1 + \frac{ν}{2} (2 t) + \frac{1}{2} \frac{ν (\frac{1}{2} ν + 1)}{2!} {(2 t)}^{2} + \dots \\ + \frac{\frac{1}{2} ν (\frac{1}{2} ν + 1) \dots (\frac{1}{2} ν + r - 1)}{r!} {(2 t)}^{r} + \dots \end{array}$

$\begin{array}{l} {μ'}_{r} & = coefficient \frac{t^{r}}{r!} \\ = 2^{r} \frac{1}{2} ν (\frac{1}{2} ν + 1) (\frac{1}{2} ν + 2) \dots (\frac{1}{2} ν + r - 1) \\ = ν (ν + 2) (ν + 4) \dots (ν + 2 r - 2) \end{array}$ $\begin{array}{l} {μ'}_{r} & = coefficient \frac{t^{r}}{r!} \\ = 2^{r} \frac{1}{2} ν (\frac{1}{2} ν + 1) (\frac{1}{2} ν + 2) \dots (\frac{1}{2} ν + r - 1) \\ = ν (ν + 2) (ν + 4) \dots (ν + 2 r - 2) \end{array}$

Hence the mean value of χ² is v and variance is v (v+2)−v²=2v.

Thus $χ^{2} - n / \sqrt{2 n}$ $χ^{2} - n / \sqrt{2 n}$ is a standard variate.

9.6 Additive Property of Independent Chi-Square Variate

Theorem 1

If $χ_{1}^{2}$ $χ_{1}^{2}$ and $χ_{2}^{2}$ $χ_{2}^{2}$ are independent χ² variates with n₁ and n₂ degrees of freedom, then $χ_{1}^{2} + χ_{2}^{2}$ $χ_{1}^{2} + χ_{2}^{2}$ is a χ² variate with n₁+n₂ degrees of freedom.

Proof

The moment generating function of

$\begin{array}{l} (χ_{1}^{2} + χ_{2}^{2}) & = (m .g .f of χ_{1}^{2}) (m .g .f of χ_{2}^{2}) \\ = {(1 - 2 t)}^{- \frac{n_{1}}{2}} {(1 - 2 t)}^{- \frac{n_{2}}{2}} \\ = {(1 - 2 t)}^{- \frac{(n_{1} + n_{2})}{2}} \end{array}$ $\begin{array}{l} (χ_{1}^{2} + χ_{2}^{2}) & = (m .g .f of χ_{1}^{2}) (m .g .f of χ_{2}^{2}) \\ = {(1 - 2 t)}^{- \frac{n_{1}}{2}} {(1 - 2 t)}^{- \frac{n_{2}}{2}} \\ = {(1 - 2 t)}^{- \frac{(n_{1} + n_{2})}{2}} \end{array}$

which is the moment generating function of χ² variate with n₁+n₂ degrees of freedom.

Hence proved.

Theorem 2

The chi-square distribution tends to normal distribution as n tends to infinity.

Proof

We have

$\begin{array}{l} M (t) & = e^{- \frac{n t}{\sqrt{2 n}}} M_{o} (\frac{t}{\sqrt{2 n}}) \\ = e^{- \frac{n t}{\sqrt{2 n}}} {(1 - \frac{2 t}{\sqrt{2 n}})}^{\frac{- n}{2}} \end{array}$ $\begin{array}{l} M (t) & = e^{- \frac{n t}{\sqrt{2 n}}} M_{o} (\frac{t}{\sqrt{2 n}}) \\ = e^{- \frac{n t}{\sqrt{2 n}}} {(1 - \frac{2 t}{\sqrt{2 n}})}^{\frac{- n}{2}} \end{array}$

Therefore

$\log_{e} M (t) = \frac{- n t}{\sqrt{2 n}} - \frac{n}{2} \log_{e} (1 - \frac{2 t}{\sqrt{2 n}})$ $\log_{e} M (t) = \frac{- n t}{\sqrt{2 n}} - \frac{n}{2} \log_{e} (1 - \frac{2 t}{\sqrt{2 n}})$

$\log_{e} M (t) = \frac{- n t}{\sqrt{2 n}} + \frac{n}{2} (\frac{2 t}{\sqrt{2 n}} + \frac{1}{2} {(\frac{2 t}{\sqrt{2 n}})}^{2} + \dots)$ $\log_{e} M (t) = \frac{- n t}{\sqrt{2 n}} + \frac{n}{2} (\frac{2 t}{\sqrt{2 n}} + \frac{1}{2} {(\frac{2 t}{\sqrt{2 n}})}^{2} + \dots)$

$\log_{e} M (t) = \frac{1}{2} t^{2} + O (\frac{1}{n})$ $\log_{e} M (t) = \frac{1}{2} t^{2} + O (\frac{1}{n})$

Further, as

$n \to \infty \log_{e} M (t) \to \frac{1}{2} t^{2}$ $n \to \infty \log_{e} M (t) \to \frac{1}{2} t^{2}$

$M (t) \to e^{\frac{1}{2} t^{2}}$ $M (t) \to e^{\frac{1}{2} t^{2}}$

Hence the result.

Example 9.1: Show that the value of χ² for the contingency table.

Classes	A	A¹	Total
B	a	b	a+b
B¹	c	d	c+d
Total	a+c	b+d	N=a+b+c+d

a, b, c, and d are cell frequencies

Calculated from the independent frequencies is

$χ^{2} = \frac{N {(a d - b c)}^{2}}{(a + b) (c + d) (b + d) (a + c)}, (N = a + b + c + d)$ $χ^{2} = \frac{N {(a d - b c)}^{2}}{(a + b) (c + d) (b + d) (a + c)}, (N = a + b + c + d)$

Solution: Since the marginal total is fixed the probability for a member belonging to class A is $\frac{a + c}{N}$ $\frac{a + c}{N}$ and is a constant.

Further the attributes being independent, the probability for it to belong to both classes A and B is $(\frac{a + b}{N}) (\frac{a + c}{N})$ $(\frac{a + b}{N}) (\frac{a + c}{N})$ .

The expected frequency of the class A, denoted by E(A) is given by

$E (A) = N \cdot (\frac{a + b}{N}) (\frac{a + c}{N}) = \frac{(a + b) (a + c)}{N}$ $E (A) = N \cdot (\frac{a + b}{N}) (\frac{a + c}{N}) = \frac{(a + b) (a + c)}{N}$

The expected frequency in each cell=product of column total and row total/whole total

Similarly we get

$\begin{array}{l} E (B) & = \frac{(a + b) (b + d)}{N} \\ E (C) & = \frac{(c + d) (a + c)}{N} \\ E (D) & = \frac{(c + d) (b + d)}{N} \end{array}$ $\begin{array}{l} E (B) & = \frac{(a + b) (b + d)}{N} \\ E (C) & = \frac{(c + d) (a + c)}{N} \\ E (D) & = \frac{(c + d) (b + d)}{N} \end{array}$

By the definition of we have

$\begin{array}{l} χ^{2} & = \frac{{[a - E (A)]}^{2}}{E (A)} + \frac{{[b - E (B)]}^{2}}{E (B)} + \frac{{[c - E (C)]}^{2}}{E (C)} + \frac{{[d - E (D)]}^{2}}{E (D)} \\ - \frac{{[a - \frac{(a + b) (a + c)}{a + b + c + d}]}^{2}}{\frac{(a + b) (a + c)}{a + b + c + d}} + \frac{{[b - \frac{(a + b) (b + d)}{a + b + c + d}]}^{2}}{\frac{(a + b) (b + d)}{a + b + c + d}} \\ + \frac{{[c - \frac{(c + d) (a + c)}{a + b + c + d}]}^{2}}{\frac{(c + d) (a + c)}{a + b + c + d}} + \frac{{[d - \frac{(c + d) (b + d)}{a + b + c + d}]}^{2}}{\frac{(c + d) (b + d)}{a + b + c + d}} \\ = \frac{{(a d - b c)}^{2}}{a + b + c + d} [\frac{1}{(a + b) (a + c)} + \frac{1}{(a + b) (b + d)} + \frac{1}{(c + d) (a + c)} + \frac{1}{(c + d) (b + d)}] \\ = \frac{{(a d - b c)}^{2}}{a + b + c + d} [\frac{a + b + c + d}{(a + b) (a + c) (b + d)} + \frac{a + b + c + d}{(a + c) (c + d) (b + d)}] \\ = {(a d - b c)}^{2} [\frac{1}{(a + b) (a + c) (b + d)} + \frac{1}{(a + c) (c + d) (b + d)}] \\ = {(a d - b c)}^{2} [\frac{c + d + a + b}{(a + b) (a + c) (b + d) (c + d)}] \\ = [\frac{(c + d + a + b) {(a d - b c)}^{2}}{(a + b) (a + c) (b + d) (c + d)}] \\ = \frac{N {(a d - b c)}^{2}}{(a + b) (c + d) (b + d) (a + c)} \end{array}$ $\begin{array}{l} χ^{2} & = \frac{{[a - E (A)]}^{2}}{E (A)} + \frac{{[b - E (B)]}^{2}}{E (B)} + \frac{{[c - E (C)]}^{2}}{E (C)} + \frac{{[d - E (D)]}^{2}}{E (D)} \\ - \frac{{[a - \frac{(a + b) (a + c)}{a + b + c + d}]}^{2}}{\frac{(a + b) (a + c)}{a + b + c + d}} + \frac{{[b - \frac{(a + b) (b + d)}{a + b + c + d}]}^{2}}{\frac{(a + b) (b + d)}{a + b + c + d}} \\ + \frac{{[c - \frac{(c + d) (a + c)}{a + b + c + d}]}^{2}}{\frac{(c + d) (a + c)}{a + b + c + d}} + \frac{{[d - \frac{(c + d) (b + d)}{a + b + c + d}]}^{2}}{\frac{(c + d) (b + d)}{a + b + c + d}} \\ = \frac{{(a d - b c)}^{2}}{a + b + c + d} [\frac{1}{(a + b) (a + c)} + \frac{1}{(a + b) (b + d)} + \frac{1}{(c + d) (a + c)} + \frac{1}{(c + d) (b + d)}] \\ = \frac{{(a d - b c)}^{2}}{a + b + c + d} [\frac{a + b + c + d}{(a + b) (a + c) (b + d)} + \frac{a + b + c + d}{(a + c) (c + d) (b + d)}] \\ = {(a d - b c)}^{2} [\frac{1}{(a + b) (a + c) (b + d)} + \frac{1}{(a + c) (c + d) (b + d)}] \\ = {(a d - b c)}^{2} [\frac{c + d + a + b}{(a + b) (a + c) (b + d) (c + d)}] \\ = [\frac{(c + d + a + b) {(a d - b c)}^{2}}{(a + b) (a + c) (b + d) (c + d)}] \\ = \frac{N {(a d - b c)}^{2}}{(a + b) (c + d) (b + d) (a + c)} \end{array}$

Example 9.2: Show that for 2 degrees of freedom the probability p of a value of χ² greater than $χ_{0}^{2}$ $χ_{0}^{2}$ is $e^{- \frac{1}{2} χ_{0}^{2}}$ $e^{- \frac{1}{2} χ_{0}^{2}}$ and hence that $χ_{0}^{2} = 2 \log_{e} (\frac{1}{p})$ $χ_{0}^{2} = 2 \log_{e} (\frac{1}{p})$ . Deduce the value of when p=0.05.

Solution: We know that

$p (χ^{2}) = \frac{1}{2^{ν - 2 / 2} Γ (ν / 2)} \int_{χ_{0}}^{\infty} e^{- \frac{χ^{2}}{2}} χ^{ν - 1} d (χ^{2})$ $p (χ^{2}) = \frac{1}{2^{ν - 2 / 2} Γ (ν / 2)} \int_{χ_{0}}^{\infty} e^{- \frac{χ^{2}}{2}} χ^{ν - 1} d (χ^{2})$

when $ν = 2$ $ν = 2$

$\begin{array}{l} p (χ^{2}) & = \frac{1}{2^{0} Γ (2 / 2)} \int_{χ_{0}}^{\infty} e^{- \frac{χ^{2}}{2}} χ d (χ^{2}) \\ = \int_{χ_{0}}^{\infty} e^{- \frac{χ^{2}}{2}} χ d (χ^{2}) \\ = {[- e^{- \frac{χ^{2}}{2}}]}_{χ 0}^{\infty} = e^{- \frac{1}{2} χ_{0}^{2}} \end{array}$ $\begin{array}{l} p (χ^{2}) & = \frac{1}{2^{0} Γ (2 / 2)} \int_{χ_{0}}^{\infty} e^{- \frac{χ^{2}}{2}} χ d (χ^{2}) \\ = \int_{χ_{0}}^{\infty} e^{- \frac{χ^{2}}{2}} χ d (χ^{2}) \\ = {[- e^{- \frac{χ^{2}}{2}}]}_{χ 0}^{\infty} = e^{- \frac{1}{2} χ_{0}^{2}} \end{array}$

$∴ e^{- \frac{1}{2} χ_{0}^{2}} = p or e^{\frac{1}{2} χ_{o}^{2}} = (\frac{1}{p})$ $∴ e^{- \frac{1}{2} χ_{0}^{2}} = p or e^{\frac{1}{2} χ_{o}^{2}} = (\frac{1}{p})$

$\frac{χ_{0}^{2}}{2} = \log_{e} (\frac{1}{p})$ $\frac{χ_{0}^{2}}{2} = \log_{e} (\frac{1}{p})$

$χ_{0}^{2} = 2 \log_{e} (\frac{1}{p}) .$ $χ_{0}^{2} = 2 \log_{e} (\frac{1}{p}) .$

When p=0.05 we have

$\begin{array}{l} χ_{0}^{2} & = 2 \log_{e} (\frac{1}{0.05}) \\ χ_{0}^{2} & = 2 \log_{e} (\frac{1}{\frac{1}{20}}) \\ χ_{0}^{2} & = 2 \log_{e} (20) = 3.012 \end{array}$ $\begin{array}{l} χ_{0}^{2} & = 2 \log_{e} (\frac{1}{0.05}) \\ χ_{0}^{2} & = 2 \log_{e} (\frac{1}{\frac{1}{20}}) \\ χ_{0}^{2} & = 2 \log_{e} (20) = 3.012 \end{array}$

Example 9.3: Prove that for a χ² distribution with n degrees of freedom

$μ_{r + 1} = 2 r (μ_{r} + n μ_{r - 1}), (ν > 0)$ $μ_{r + 1} = 2 r (μ_{r} + n μ_{r - 1}), (ν > 0)$

Solution: The moment generating function of χ² distribution with n degrees of freedom about the mean is

$μ (t) = e^{- n t} {(1 - 2 t)}^{n / 2}$ $μ (t) = e^{- n t} {(1 - 2 t)}^{n / 2}$

Applying logarithms on both sides and differentiating we get

$\frac{μ' (t)}{μ (t)} = - n + \frac{n}{2} (\frac{2}{1 - 2 t})$ $\frac{μ' (t)}{μ (t)} = - n + \frac{n}{2} (\frac{2}{1 - 2 t})$

i.e.,

$(1 - 2 t) μ' (t) = 2 n t μ (t)$ $(1 - 2 t) μ' (t) = 2 n t μ (t)$

Using Leibnitz theorem and differentiating with respect to t we get

$(1 - 2 t) μ^{r + 1} (t) + 2 (- 2) μ^{r} (t) = 2 n t μ^{r} (t) + 2 n r μ^{r - 1} (t)$ $(1 - 2 t) μ^{r + 1} (t) + 2 (- 2) μ^{r} (t) = 2 n t μ^{r} (t) + 2 n r μ^{r - 1} (t)$

Pitting t=0 and using the relation

$μ_{r} = {[\frac{d}{d t} μ^{r} (t)]}_{t = 0} = μ^{r} (ν)$ $μ_{r} = {[\frac{d}{d t} μ^{r} (t)]}_{t = 0} = μ^{r} (ν)$

We get

$μ_{r + 1} = 2 r (μ_{r} + n μ_{r - 1})$ $μ_{r + 1} = 2 r (μ_{r} + n μ_{r - 1})$

9.7 Degrees of Freedom

It is the number of values in a set, which may be arbitrarily assigned. The number of independent variables usually called the degrees of freedom. It is denoted by the symbol v. If these are normalized variables subject to k linear constraints then the degrees of freedom is v=n−k.

If the data is given in the form of a row containing n observations, then the degrees of freedom is n−1. Similarly if the data is given in the form of a column, the degrees of freedom is n−1. Where n is the number of observation in the column. If there are r rows and c columns. The degree of freedom is given by (r−1) (c−1).

9.8 Conditions for Using Chi-Square Test

1. The total number observations used in this test must be large (i.e., n≥50).

2. Each of the observations making up sample for χ² test should be independent of each other.

3. The test is wholly dependent on the degree of freedom.

4. The frequencies used in χ² test should be absolute and not relative in terms.

5. The expected frequency of any item or cell should not be less than 5. If it is less than 5, then the frequencies from the adjacent items or cells should be pooled together in order to make it 5 or more than 5 (preferably not less than 10).

6. The observations collected for χ² should be based on the method of random sampling.

The constraints on the cell frequencies if any should be linear.

9.9 Uses of Chi-Square Test

Chi-square test an important test. If we require only the degrees of freedom for using this test. It is a powerful test. It is used

1. as a test of goodness of fit.

2. as a test of independence of attributes, and

3. as a test of homogeneity.

9.9.1 Chi-Square Test as a Test of Goodness of Fit

Chi-square test is applied as a test of goodness of fit to determine whether the actual (i.e., observed) the expected (i.e., theoretical) frequencies. The degrees of freedom in this case are $ν = n - 1$ $ν = n - 1$ where n is the number of observations.

Example 9.4: In 90 throws of a die, face 1 turned 9 times, face 2 or 3 turned 24 times, face 4 or 5 turned 36 times, and face 6 turned 18 times. Test at 10% level, if the die is honest, it being given that χ² for 3 df=6.25 at 10% level of significance.

Solution:

H₀: The die is honest.

H₁: The die is not honest.

Expected frequencies for $each face = 90 \times \frac{1}{6} = 15$ $each face = 90 \times \frac{1}{6} = 15$

Level of significance=10% (i.e., 0.01)

Degrees of freedom=4−1=3

Chi-square value for 3 df at 10% level of significance=6.25

We have the following table:

Face turned	Observed O_i	Expected E_i	$(O_{i} - E_{i})$ $(O_{i} - E_{i})$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
1	9	15	−6	2.4
2 or 3	27	30	−3	0.3
4 or 5	36	30	6	1.2
6	18	15	3	0.6
Total	90	90		4.5

$χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 4.5$ $χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 4.5$

Since the calculated value of χ² is less than the table value at 10% level of significance and for 3 df, we accept the null hypothesis and conclude that the die is honest.

Example 9.5: Find whether or not the following observed distribution of phenotypes in a sample of 384 Drosophila flies have a significance goodness of fit with proposed median 9:3:3:1 distribution (test at 5% level of significance)?

Phenotypes	AB	Ab	aB	ab	Total
Number of files	232	76	58	18	384

Solution: Degrees of freedom=4−1=3

Null hypothesis H₀: 9:3:3:1

i.e., Variation is not significant.

Alternative hypothesis H₁: 1:1:1:1

Level of significance=α=0.05

Table value χ²=7.82 for 3 df

Observed O_i	Expected E_i	$(O_{i} - E_{i})$ $(O_{i} - E_{i})$	${(O_{i} - E_{i})}^{2}$ ${(O_{i} - E_{i})}^{2}$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
232	$\frac{384}{16} \times 9 = 216$ $\frac{384}{16} \times 9 = 216$	16	256	1.185
76	$\frac{384}{16} \times 3 = 72$ $\frac{384}{16} \times 3 = 72$	4	16	0.223
58	$\frac{384}{16} \times 3 = 72$ $\frac{384}{16} \times 3 = 72$	−14	196	2.723
18	$\frac{384}{16} \times 124$ $\frac{384}{16} \times 124$	−6	36	2.25
Total=384				6.381

The calculated value of χ² for 3 df at 5% level of significance is less than the table value. Hence we accept H₀, i.e., variation is not significant.

Example 9.6: Among 64 offspring’s of a certain cross between guinea pigs 32 were red, 10 were black, and 22 were white. According to the genetic model these numbers should be in the ratio 9:3:4. Are the data consistent with the model at 5% level?

Solution:

H₀: Data are consistent with the model.

H₁: Data are not consistent with the model.

Level of significance=5%

Degrees of freedom=n−1=3−1=2

Table value of χ² for 2 df at 5% level=5.991

Expected frequencies are in the ratio 9:3:4

i.e.,

$64 (\frac{9}{16}) = 36, 64 (\frac{3}{16}) = 12, 64 (\frac{4}{16}) = 16$ $64 (\frac{9}{16}) = 36, 64 (\frac{3}{16}) = 12, 64 (\frac{4}{16}) = 16$

Observed O_i	Expected E_i	${(O_{i} - E_{i})}^{2}$ ${(O_{i} - E_{i})}^{2}$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
32	36	16	0.4444
10	12	4	0.3333
22	16	36	2.25
Total	64	56	3.0277

$χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 3.0277$ $χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 3.0277$

The calculated value of χ² for 2 df at 5% level of significance is less than the table value. Hence we accept H₀, i.e., data are consistent with the model.

Example 9.7: A sample analysis of examination results of 500 students was made. It was found that 230 students had failed. one hundred sixty had secured third class, 80 were placed in second class, and 30 got first class. Do these figures commensurate with the general examination result, which is in the ratio 4:3:2:1 for various categories respectively?

Solution: Null Hypothesis H₀: The observed results commensurate with general examination result

Alternative Hypothesis H₁: It is not true that the observed results commensurate with general examination result.

Level of significance=5%

Degrees of freedom=n−1=4−1=3

The total frequency=N=500

Table value χ² for 3 df at 5% level of significance=7.81 for 3 df

Dividing 500 in the ratio 4:3:2:1

We get 200, 150, 100, and 50

Therefore the expected frequencies are 200, 150, 100, and 50 corresponding to the observed frequencies 230, 160, 80, and 30.

Class/division	Observed O_i	Expected E_i	$(O_{i} - E_{i})$ $(O_{i} - E_{i})$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
Failed	230	200	30	4.500
Third	160	150	10	0.6666
Second	80	100	–20	4.0000
First	30	50	–20	8.00

$χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 17.1666$ $χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 17.1666$

Since the calculated value χ² is greater than the table value of χ²—the null hypothesis is rejected.

Example 9.8: The table below gives the number of aircraft accidents that occurred during the various days of the week. Test whether the accidents are uniformly distributed over the week?

Days	Mon	Tue	Wed	Thu	Fri	Sat
No. of accidents	14	18	12	11	15	14

Solution: Null Hypothesis H₀: The accidents are uniformly distributed over the week.

Alternative Hypothesis H₁: The accidents are not uniformly distributed over the week.

Total number of accidents=14+18+12+11+15+14=84

Level of significance=5%

Degrees of freedom=6−1=5

Table value of χ²=11.07

The expected frequency of each day $accidents is = \frac{84}{6} = 14$ $accidents is = \frac{84}{6} = 14$

Day	Observed O_i	Expected E_i	$(O_{i} - E_{i})$ $(O_{i} - E_{i})$	${(O_{i} - E_{i})}^{2}$ ${(O_{i} - E_{i})}^{2}$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
Mon	14	14	0	0	0
Tue	18	14	4	16	1.1428
Wed	12	14	−2	4	0.2857
Thu	11	14	−3	9	0.6428
Fri	15	14	1	1	0.0714
Sat	14	14	0	0	0.000
Total	84	84			2.1427

Since 2.1427<11.07, the calculated value of χ² at 5% level, for 5 df is less than the table value of χ².

The null hypothesis is accepted. That is the accidents are uniformly distributed over the week.

Example 9.9: Four coins are tossed 160 times and the following results were obtained:

Number of heads	0	1	2	3	4
Observed frequencies	17	52	54	31	6

Under the assumption that coins are balanced, find the expected frequencies of getting 0, 1, 2, 3, or 4 heads and test the goodness of fit.

Solution:

Null Hypothesis H₀: The coins are balanced.

Alternative hypothesis H₁: The coins are not balanced.

Level of significance=5%

Degrees of freedom=5−1=4

Table value of χ²=9.488

We have N=160, $p = \frac{1}{2}$ $p = \frac{1}{2}$ , $q = \frac{1}{2}$ $q = \frac{1}{2}$

The expected frequencies of 0, 1, 2, 3, or 4 successes are

$160 \times {}^{4}C_{0} {(\frac{1}{2})}^{0} {(\frac{1}{2})}^{4} = 10$ $160 \times {}^{4}C_{0} {(\frac{1}{2})}^{0} {(\frac{1}{2})}^{4} = 10$	$160 \times {}^{4}C_{1} {(\frac{1}{2})}^{1} {(\frac{1}{2})}^{3} = 40$ $160 \times {}^{4}C_{1} {(\frac{1}{2})}^{1} {(\frac{1}{2})}^{3} = 40$
$160 \times {}^{4}C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{2} = 60$ $160 \times {}^{4}C_{2} {(\frac{1}{2})}^{2} {(\frac{1}{2})}^{2} = 60$	$160 \times {}^{4}C_{3} {(\frac{1}{2})}^{3} {(\frac{1}{2})}^{1} = 40$ $160 \times {}^{4}C_{3} {(\frac{1}{2})}^{3} {(\frac{1}{2})}^{1} = 40$
$160 \times {}^{4}C_{4} {(\frac{1}{2})}^{4} {(\frac{1}{2})}^{0} = 10$ $160 \times {}^{4}C_{4} {(\frac{1}{2})}^{4} {(\frac{1}{2})}^{0} = 10$

No. of heads	Observed O_i	Expected E_i	$(O_{i} - E_{i})$ $(O_{i} - E_{i})$	${(O_{i} - E_{i})}^{2}$ ${(O_{i} - E_{i})}^{2}$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
0	17	10	7	49	4.900
1	52	40	12	144	3.600
2	54	60	−6	36	0.600
3	31	40	−9	81	2.025
4	6	10	−4	16	1.600
Total	160	160			12.725

$χ^{2} = \frac{\sum {(O_{i} - E_{i})}^{2}}{E_{i}} = 12.725$ $χ^{2} = \frac{\sum {(O_{i} - E_{i})}^{2}}{E_{i}} = 12.725$

The calculated value of χ² is greater than the table value of χ² at 5% level of significance and 4 df. Therefore null hypothesis is rejected. The coins are not balanced. Hence the fit is poor.

Example 9.10: Fit a Poisson distribution to the following data and test the goodness of fit:

x	0	1	2	3	4	5	6
F	275	72	30	7	5	2	1

Solution: Null Hypothesis H₀: The Poisson fit is good to the given data.

Alternative Hypothesis H₁: The Poisson fit is not a good fit to the given data.

Level of significance=5%

$\begin{array}{l} Mean of the distribution & = \frac{\sum f_{i} x_{i}}{\sum f_{i}} \\ = \frac{0 + 72 + 60 + 21 + 20 + 10 + 6}{275 + 72 + 30 + 7 + 5 + 2 + 1} \\ = \frac{189}{392} = 0.482 \end{array}$ $\begin{array}{l} Mean of the distribution & = \frac{\sum f_{i} x_{i}}{\sum f_{i}} \\ = \frac{0 + 72 + 60 + 21 + 20 + 10 + 6}{275 + 72 + 30 + 7 + 5 + 2 + 1} \\ = \frac{189}{392} = 0.482 \end{array}$

The frequencies of 0, 1, 2, 3, 4, 5, and 6 successes by using recurrence formula for Poisson distribution are the expected frequencies.

Therefore the expected frequencies are

$\begin{array}{l} N (0) = 392 e^{- 0.482} = 242.1 \\ N (1) = 392 e^{- 0.482} (0.482) = 116.7 \\ N (2) = 116.7 (\frac{0.482}{2}) = 28.12 \\ N (3) = 28.12 (\frac{0.482}{3}) = 4.52 \\ N (4) = 4.52 (\frac{0.482}{4}) = 0.54 \\ N (5) = 0.54 (\frac{0.482}{5}) = 0.052 \\ N (6) = 0.052 (\frac{0.482}{6}) = 0.004 \end{array}$ $\begin{array}{l} N (0) = 392 e^{- 0.482} = 242.1 \\ N (1) = 392 e^{- 0.482} (0.482) = 116.7 \\ N (2) = 116.7 (\frac{0.482}{2}) = 28.12 \\ N (3) = 28.12 (\frac{0.482}{3}) = 4.52 \\ N (4) = 4.52 (\frac{0.482}{4}) = 0.54 \\ N (5) = 0.54 (\frac{0.482}{5}) = 0.052 \\ N (6) = 0.052 (\frac{0.482}{6}) = 0.004 \end{array}$

The frequency table is:

x	0	1	2	3	4	5	6
Observed frequency	275	72	30	7	5	2	1
Expected frequency	242.1	116.7	28.1	4.5	0.5	0.1	0

Since the last four frequencies are small, we regroup the last four frequencies and obtain the following table:

Calculation of χ²

Observed O_i	Expected E_i	$(O_{i} - E_{i})$ $(O_{i} - E_{i})$	${(O_{i} - E_{i})}^{2}$ ${(O_{i} - E_{i})}^{2}$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
275	242.1	32.9	1082.41	4.4709
72	116.7	44.7	1998.09	17.1216
30	28.1	1.9	3.61	0.1285
15	5.1	9.9	98.01	19.217
Total				40.938

$χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 40.938$ $χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 40.938$

Degrees of freedom for the Poisson fit=n−2=4−2=2

Table value of χ² at 5% for 2 df=5.991

The calculated value of χ² is 40.938 which is greater than the table value.

We reject null hypothesis and conclude that the Poisson fit is not a good fit to the given data.

9.9.2 Test for Independence of Attributes

The χ² test can also be applied to test the association between the attributes such as honesty, smoking, drinking, etc. when the sample data is presented in the form of contingency table with any number of rows and columns.

Example 9.11: The following table gives classification of 150 workers according to sex and nature of work. Test whether the nature of work is independent of the sex of the work?

	Stable	Unstable	Total
Males	60	30	90
Females	15	45	60
Total	75	75	150

Solution: Null Hypothesis H₀: The nature of the work is independent of the sex of the worker.

Alternative Hypothesis H₁: The nature of the work is not independent of the sex of the worker.

Degrees of freedom=(r−1) (c−1)=(2−1) (2−1)=1

Level of significance=5%

Table value of χ²=3.84

Expected frequencies are given in the following table:

	Stable	Unstable
Males	$\frac{75 \times 90}{150} = 45$ $\frac{75 \times 90}{150} = 45$	$\frac{75 \times 90}{150} = 45$ $\frac{75 \times 90}{150} = 45$
Females	$\frac{75 \times 90}{150} = 30$ $\frac{75 \times 90}{150} = 30$	$\frac{75 \times 90}{150} = 30$ $\frac{75 \times 90}{150} = 30$

Calculation of χ²

Observed O_i	Expected E_i	$(O_{i} - E_{i})$ $(O_{i} - E_{i})$	${(O_{i} - E_{i})}^{2}$ ${(O_{i} - E_{i})}^{2}$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
60	45	15	225	5.00
15	30	−15	225	7.50
30	45	−15	225	5.00
45	30	15	225	7.50
Total				25

$χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 25$ $χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 25$

The calculated value of χ² is greater than the table value at 5% and df. Hence we reject the null hypothesis.

We conclude that the nature of work is independent of the sex of the worker.

Example 9.12: In a certain sample of 2000 families, 1400 families are consumers of tea, out of 1800 Hindu families 1236 families consume tea. Use χ² test and state whether there is any significant difference between consumption of tea among Hindu and nonhindu families?

Solution: From the given data, the 2×2 contingency table that can be formed is given below.

	Hindu	Nonhindu	Total
Consuming tea	1236	164	1400
Not consuming tea	564	36	600
Total	1800	200	200

Null Hypothesis H₀: The attributes are independent, i.e., there is no significant difference between the communities as far as consuming of tea is concerned.

Alternative Hypothesis H₁: The attributes are not independent.

Level of significance: 5%

df=1, Table value of χ² at 5% for 1 df=3.841

The expected frequencies corresponding to the given observed frequencies are given in the table below:

$\frac{1800 \times 1400}{2000} = 1264$ $\frac{1800 \times 1400}{2000} = 1264$	$\frac{200 \times 1400}{2000} = 140$ $\frac{200 \times 1400}{2000} = 140$
$\frac{1800 \times 600}{2000} = 540$ $\frac{1800 \times 600}{2000} = 540$	$\frac{200 \times 600}{2000} = 60$ $\frac{200 \times 600}{2000} = 60$

Computation of χ²

Observed O_i	Expected E_i	$(O_{i} - E_{i})$ $(O_{i} - E_{i})$	${(O_{i} - E_{i})}^{2}$ ${(O_{i} - E_{i})}^{2}$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
1236	1260	−24	576	0.457
564	540	24	576	1.068
164	140	24	576	4.114
36	60	−24	576	9.600
Total				15.239

$χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 15.239$ $χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 15.239$

The calculated value of χ² is higher than the table value of χ² at 5% and 1 df. Therefore the null hypothesis is rejected. The two communities differ significantly as far as consumption of tea is concerned.

Example 9.13: A tobacco company claims that there is no relationship between smoking and lung ailments. To investigate the claims, a random sample of 300 males in the age group of 40 and 50 years is given a medical test. The observed sample results are tabulated below:

	Lung ailment	Nonlung ailment	Total
Smokers	75	105	180
Nonsmokers	25	95	120
Total	100	200	300

On the basis of this information, can it be concluded that smoking and long ailments are independent (given $χ_{0.05}^{2} = 3.841$ $χ_{0.05}^{2} = 3.841$ for 1 df).

Solution: Null hypothesis H₀: The smoking and lung ailments are not associated.

Alternative Hypothesis H₁: The smoking and lung ailments are associated.

Level of significance=5%

Table value of χ² for 1 df=3.841

The expected frequencies are:

$\frac{100 \times 180}{300} = 60$ $\frac{100 \times 180}{300} = 60$	$\frac{200 \times 180}{300} = 120$ $\frac{200 \times 180}{300} = 120$
$\frac{100 \times 120}{300} = 40$ $\frac{100 \times 120}{300} = 40$	$\frac{200 \times 120}{300} = 80$ $\frac{200 \times 120}{300} = 80$

Computation of χ²

Observed O_i	Expected E_i	$(O_{i} - E_{i})$ $(O_{i} - E_{i})$	${(O_{i} - E_{i})}^{2}$ ${(O_{i} - E_{i})}^{2}$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
75	60	15	225	3.750
25	40	−15	225	5.625
105	120	−15	225	1.875
95	80	15	225	2.813
Total				14.063

$χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 14.063$ $χ^{2} = \sum \frac{{(O_{i} - E_{i})}^{2}}{E_{i}} = 14.063$

For 1 df at 5% level of significance the calculated value of χ² is more than the table value. Hence we reject the null hypothesis.

Therefore smoking and lung ailments are not independent.

Example 9.14: Given the following contingency table for hair color and eye color, find the value of χ²? Is there good association between the two?

		Hair color			Total
		Fair	Brown	Black
Eye color	Blue	15	5	20	40
	Gray	20	10	20	50
	Brown	25	15	20	60
Total		60	30	60	150

Solution: Null Hypothesis H₀: The two attributes hair color and eye color are independent.

Alternative Hypothesis H₁: The two attributes hair color and eye color are not independent.

Level of significance=9.488 for 4 df at 5% level of significance.

Table of expected frequencies are:

$\frac{60 \times 40}{150} = 16$ $\frac{60 \times 40}{150} = 16$	$\frac{30 \times 40}{150} = 8$ $\frac{30 \times 40}{150} = 8$	$\frac{60 \times 40}{150} = 16$ $\frac{60 \times 40}{150} = 16$
$\frac{60 \times 50}{150} = 20$ $\frac{60 \times 50}{150} = 20$	$\frac{30 \times 50}{150} = 10$ $\frac{30 \times 50}{150} = 10$	$\frac{60 \times 50}{150} = 20$ $\frac{60 \times 50}{150} = 20$
$\frac{60 \times 60}{150} = 24$ $\frac{60 \times 60}{150} = 24$	$\frac{30 \times 60}{150} = 12$ $\frac{30 \times 60}{150} = 12$	$\frac{60 \times 60}{150} = 24$ $\frac{60 \times 60}{150} = 24$

Computation of χ²

Observed O_i	Expected E_i	$(O_{i} - E_{i})$ $(O_{i} - E_{i})$	${(O_{i} - E_{i})}^{2}$ ${(O_{i} - E_{i})}^{2}$	$\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$ $\frac{{(O_{i} - E_{i})}^{2}}{E_{i}}$
15	16	−1	1	0.0625
5	8	−3	9	1.125
20	16	4	16	1
20	20	0	0	0
10	10	0	0	0
20	20	0	0	0
25	24	1	1	0.042
15	12	3	9	0.75
20	24	−4	16	0.666
Total				3.6458

$χ^{2} = \frac{\sum {(O_{i} - E_{i})}^{2}}{E_{i}} = 14.063$ $χ^{2} = \frac{\sum {(O_{i} - E_{i})}^{2}}{E_{i}} = 14.063$

Since the calculated value of χ² for (3−1) (3−1)=2×2=4 df at 5% level, is less than the table value 9.488. Hence we accept null hypothesis, i.e., the hair color and eye color are independent.

Example 9.15: The following table shows the result of an experiment to investigate the effect of vaccination induced on the animals against a particular disease. Use the χ² test to test the hypothesis that the vaccinated and unvaccinated groups, i.e., vaccination and the disease are independent.

	Got disease	Did not get disease
Vaccinated	9	42
Not vaccinated	17	28

(Value of χ² for 1 df at 5% level is equal to 3.841).

Solution: We have a=9, b=42, c=17, d=28

N=a+b+c+d=9+42+17+28=96

Null Hypothesis H₀: The vaccination and disease are independent.

Alternative Hypothesis H₁: The vaccination and disease are not independent.

Table value of χ² for 1 df at 5% level=3.841

$\begin{array}{l} χ^{2} & = \frac{N {(a d - b c)}^{2}}{(a + b) (c + d) (b + d) (a + c)} = \frac{96 {(9 \times 28 - 17 \times 42)}^{2}}{(51) (26) (70) (45)} \\ = \frac{96 {(- 462)}^{2}}{4176900} = 4.906 \end{array}$ $\begin{array}{l} χ^{2} & = \frac{N {(a d - b c)}^{2}}{(a + b) (c + d) (b + d) (a + c)} = \frac{96 {(9 \times 28 - 17 \times 42)}^{2}}{(51) (26) (70) (45)} \\ = \frac{96 {(- 462)}^{2}}{4176900} = 4.906 \end{array}$

The calculated value of χ² is more than the table value of χ² at 5% level and 1 df. Therefore we reject H₀ and conclude that the disease and vaccination are not independent.

9.9.2.1 Yate’s Correction

In a 2×2 contingency table, if the cell frequency is small, Yate’s correction is necessary. If the expected value or frequency in any observation less than 5 in 1 df we apply Yate’s correction. We subtract 0.5 from the absolute difference between observed and expected frequencies. By making Yate’s correction becomes continuous and the formula after making Yate’s correction is

$χ^{2} = \sum \frac{{(| O_{i} - E_{i} | - 0.5)}^{2}}{E_{i}} (0.5 is Yate ’ s correction)$ $χ^{2} = \sum \frac{{(| O_{i} - E_{i} | - 0.5)}^{2}}{E_{i}} (0.5 is Yate ’ s correction)$

Example 9.16: Two batches each of 12 animals are taken for test of inoculation. One batch was inoculated and the others batch was not inoculated. The frequencies of the dead and surviving animals are given below in both cases. Can the inoculation be regarded as effective against the disease?

	Dead	Survived	Total
Inoculated	2	10	12
Not inoculated	8	4	12
Total	10	14

( $χ_{0.05}^{2}$ $χ_{0.05}^{2}$ for 1 df=3.841)

Solution: Null Hypothesis H₀: Inoculation and the disease are independent.

Alternative Hypothesis H₁: Inoculation and the disease are not independent.

$D f = (r - 1) (c - 1) = (2 - 1) (2 - 1) = 1$ $D f = (r - 1) (c - 1) = (2 - 1) (2 - 1) = 1$

Level of significance=5%

Table value of χ² for 1 df at 5% level=3.841

Frequencies in the cells are small (a and d are small) so, we make Yate’s correction. The expected frequencies are given in the table below:

$\frac{10 \times 12}{24} = 5$ $\frac{10 \times 12}{24} = 5$	$\frac{14 \times 12}{24} = 7$ $\frac{14 \times 12}{24} = 7$
$\frac{10 \times 12}{24} = 5$ $\frac{10 \times 12}{24} = 5$	$\frac{14 \times 12}{24} = 7$ $\frac{14 \times 12}{24} = 7$

Computation of χ²

Observed O_i	Expected E_i	$\| O_{i} - E_{i} \|$ $\| O_{i} - E_{i} \|$	$\| O_{i} - E_{i} \| - 0.5$ $\| O_{i} - E_{i} \| - 0.5$	${(\| O_{i} - E_{i} \| - 0.5)}^{2}$ ${(\| O_{i} - E_{i} \| - 0.5)}^{2}$	$\frac{{(\| O_{i} - E_{i} \| - 0.5)}^{2}}{E_{i}}$ $\frac{{(\| O_{i} - E_{i} \| - 0.5)}^{2}}{E_{i}}$
2	5	−3	2.5	6.25	1.25
10	7	3	2.5	6.25	0.89285
8	5	3	2.5	6.25	1.25
4	7	−3	2.5	6.25	0.89285
Total					4.2857

$χ^{2} = \sum \frac{{(| O_{i} - E_{i} | - 0.5)}^{2}}{E_{i}} = 4.2857$ $χ^{2} = \sum \frac{{(| O_{i} - E_{i} | - 0.5)}^{2}}{E_{i}} = 4.2857$

For 1 df at 5% level of significance, the calculated value of χ² is greater than the table value. Hence H₀ is rejected. The inoculation and disease and independent we conclude that inoculation is effective against the disease.

9.9.3 Homogeneity Chi-Square

Chi-square test may be used to test the homogeneity of the attributes in respect of particular characteristics. It is performed to decide whether separate samples are sufficiently uniform to be added together. Chi-square test may also be used to test the population variance.

9.9.4 Chi-Square Distribution of Sample Variance

Let σ² denote the population variance and S² denote the sample variance. The sampling distribution $(n - 1) \frac{S^{2}}{σ^{2}}$ $(n - 1) \frac{S^{2}}{σ^{2}}$ has χ² distribution with n−1 degrees of freedom. It is very useful in making inference about the population variance σ² by using sample variance S². Also I is used in making interval estimate of the population variance which is given by

$\frac{(n - 1) S^{2}}{{χ^{2}}_{α / 2}} \leq α \leq \frac{(n - 1) S}{{χ^{2}}_{1 - α / 2}}$ $\frac{(n - 1) S^{2}}{{χ^{2}}_{α / 2}} \leq α \leq \frac{(n - 1) S}{{χ^{2}}_{1 - α / 2}}$

where α is the level of significance, and $χ_{α}^{2}$ $χ_{α}^{2}$ is the value of χ² distribution giving an area to the right of $χ_{α}^{2}$ $χ_{α}^{2}$ .

The test may also be used as hypothesis test for the value of the population variance.

9.9.5 Testing a Hypothesis About the Variance of Normally Distributed Population—Decision Rule

Let $σ_{0}^{2}$ $σ_{0}^{2}$ denote the hypothesized value of the population variance. Then the decision rule for accepting or rejecting H₀ is as follows:

1. Two-tailed test	Decision rule
$H_{0} : σ^{2} = σ_{0}^{2}$ $H_{0} : σ^{2} = σ_{0}^{2}$	Accept H₀ if computed value of $χ^{2} > χ_{α / 2}^{2}$ $χ^{2} > χ_{α / 2}^{2}$ (table value)
$H_{1} : σ^{2} \neq σ_{0}^{2}$ $H_{1} : σ^{2} \neq σ_{0}^{2}$	Reject H₀ if computed value of $χ^{2} > χ_{α / 2}^{2}$ $χ^{2} > χ_{α / 2}^{2}$ (table value)
2. One-tailed test
$H_{0} : σ^{2} \geq σ_{0}^{2}$ $H_{0} : σ^{2} \geq σ_{0}^{2}$	Accept H₀ if computed value of χ²< table value of $χ_{α}^{2}$ $χ_{α}^{2}$
$H_{1} : σ^{2} < σ_{0}^{2}$ $H_{1} : σ^{2} < σ_{0}^{2}$	Reject H₀ if computed value of χ²> table value of $χ_{α}^{2}$ $χ_{α}^{2}$

9.9.5.1 Solved Examples

Example 9.17: According to the census report of 2004, the numbers of women to every 1000 men in the following 5 states/union territories are as follows:

In Andaman 846, in Delhi 821, in Chandigarh 777, in Daman and Diu 710, in Dadar and Nagar Haveli 812. By an appropriate statistical method determine whether 1:1 sex ratio among human population can be attributed to 5 regions ( $χ_{0.05}^{2}$ $χ_{0.05}^{2}$ for 4 df=7.88).

Solution: Null hypothesis H₀: Sex ratio is 1:1 among human population of 5 states, Alternative hypothesis H₁: Sex ratio is not 1:1 among the human population of 5 states. Table value of χ² for 4 df at 5% level=7.88

Computation of χ² for 5 states

	Observed O_i	Expected E_i	$\| O_{i} - E_{i} \|$ $\| O_{i} - E_{i} \|$	$\frac{{(\| O_{i} - E_{i} \|)}^{2}}{E_{i}}$ $\frac{{(\| O_{i} - E_{i} \|)}^{2}}{E_{i}}$	χ²	df
Andaman	Men=1000	923	77	6.42	12.84	1
Andaman	Women=846	923	77	6.42	12.84	1
Delhi	Men=1000	910.5	89.5	8.79	17.58	1
Delhi	Women=821	910.5	89.5	8.79	17.58	1
Chandigarh	Men=1000	888.5	111.5	13.99	27.98	1
Chandigarh	Women=777	888.5	111.5	19.99	27.98	1
Damn and Diu	Men=1000	855	145	24.59	49.18	1
Damn and Diu	Women=710	855	145	24.59	49.18	1
Dadar and Nagar Haveli	Men=1000	906	94	9.75	19.5	1
Dadar and Nagar Haveli	Women=812	906	94	9.75	19.5	1
Men	5000	Total			127.08	5
Women	3966	Total			127.08	5

Chi-square for summed data

Observed O_i	Expected E_i	$\| O_{i} - E_{i} \|$ $\| O_{i} - E_{i} \|$	$\frac{{(\| O_{i} - E_{i} \|)}^{2}}{E_{i}}$ $\frac{{(\| O_{i} - E_{i} \|)}^{2}}{E_{i}}$	df
Men=500	4483	517	59.62	2−1=1
Women=3966	4483	517	59.62	2−1=1

	Chi-square	df
Total	127.08	5
Summed	119.2	1
Homogeneity	7.88	4

The calculated value of χ² at 5% level of significance for 4 df is less than the table value of χ². We accept H₀.

Example 9.18: Heights in cm of 10 students are given below:

61, 65, 67, 66, 68, 70, 64, 65, 68, 71

Can we say that variance of the distribution heights of all students from which the above sample of 10 students was drawn is equal to 50?

Solution: We have

$\begin{array}{l} Mean of the sample & = \frac{61 + 65 + 62 + 66 + 68 + 70 + 64 + 65 + 68 + 71}{10} \\ = \frac{660}{10} = 66 \end{array}$ $\begin{array}{l} Mean of the sample & = \frac{61 + 65 + 62 + 66 + 68 + 70 + 64 + 65 + 68 + 71}{10} \\ = \frac{660}{10} = 66 \end{array}$

Null Hypothesis $H_{0} : σ^{2} = σ_{0}^{2} = 50$ $H_{0} : σ^{2} = σ_{0}^{2} = 50$

Alternative Hypothesis $H_{1} : σ^{2} \neq σ_{0}^{2}$ $H_{1} : σ^{2} \neq σ_{0}^{2}$ , i.e., $σ^{2} \neq 50$ $σ^{2} \neq 50$ (Two-tailed test)

Level of significance: 5% (α=0.05) $χ_{1 - α / 2}^{2} = χ_{0.975}^{2} = 2.70$ $χ_{1 - α / 2}^{2} = χ_{0.975}^{2} = 2.70$

Computation of χ²

x_i	$x_{i} - \bar{x}$ $x_{i} - \bar{x}$	${(x_{i} - \bar{x})}^{2}$ ${(x_{i} - \bar{x})}^{2}$
61	−5	25
65	−1	1
62	−4	16
66	0	0
68	2	4
70	4	16
64	−2	4
65	−1	1
68	2	4
71	5	25

Test statistic is

$\begin{array}{l} χ^{2} & = (n - 1) \frac{S^{2}}{σ_{0}^{2}} \\ = \frac{(n - 1)}{σ_{0}^{2}} \cdot \frac{{\sum (x_{i} - \bar{x})}^{2}}{n - 1} \\ = \frac{{\sum (x_{i} - \bar{x})}^{2}}{σ_{0}^{2}} = \frac{96}{50} = 1.92 \end{array}$ $\begin{array}{l} χ^{2} & = (n - 1) \frac{S^{2}}{σ_{0}^{2}} \\ = \frac{(n - 1)}{σ_{0}^{2}} \cdot \frac{{\sum (x_{i} - \bar{x})}^{2}}{n - 1} \\ = \frac{{\sum (x_{i} - \bar{x})}^{2}}{σ_{0}^{2}} = \frac{96}{50} = 1.92 \end{array}$

Since calculated value of χ² for 9 df at 5% level of significance is less than the value of $χ_{1 - α / 2}^{2}$ $χ_{1 - α / 2}^{2}$ at 5% level, we reject null hypothesis and conclude that population variance is not 50.

Exercise 9.1

1. The following figures show the distribution of digits in numbers chosen at random from a telephone directory:

Digits	0	1	2	3	4	5	6	7	8	9
Frequency	1026	1107	997	996	1075	993	1107	972	964	853

Test whether the digits may be taken to occur equally frequently in directory ( $χ_{0.05}^{2} = 16.919$ $χ_{0.05}^{2} = 16.919$ for 9 df)?

$(Hint : Expected frequency of each observation = 10, 000 / 10 = 1000)$ $(Hint : Expected frequency of each observation = 10, 000 / 10 = 1000)$

Ans: Null hypothesis is rejected

2. A die is thrown 264 times with the following results show that the die is biased:

No. appeared on the die	1	2	3	4	5	6
Frequency	40	32	28	58	54	60

(Given χ² for 5 df at 5% level=11.07)

3. On the basis of information given below about the treatment of 200 patients suffering from a disease, state whether the new treatment is comparatively superior to the conventional treatment?

	Favorable	Not favorable	Total
New treatment	60	30	90
Conventional treatment	40	70	110

( $χ_{0.05}^{2} = 3.841$ $χ_{0.05}^{2} = 3.841$ for 1 df)

Ans: Null hypothesis is rejected. The new treatment is superior to conventional statement

4. Two hundred digit were chosen at random from a set of tables the frequencies of the digits were

Digit	0	1	2	3	4	5	6	7	8	9
Frequency	18	19	23	21	16	25	22	20	21	15

Use χ² test to assess the correctness of the hypothesis that the digits were distributed in equal numbers in the tables from which these were chosen (For df at 5% level of significance χ² value is 16.919).

5. In a sample of owls, it is found that red male are 35, red female are 70, gray male are 50, gray female are 45. Coloration is due to the plumage. Is the coloration independent of sex of the sample?

Ans: The coloration of the individuals are not independent of sex

6. The following table given the frequencies of occurrence of the digits 0, 1, 2, …, 9, in the last place in the four-figure logarithm of numbers. Examine if there is any peculiarity?

Digit	0	1	2	3	4	5	6	7	8	9
Frequency	6	16	15	10	12	12	3	2	9	5

(χ² Value for 5 df at 5% level of significance is 11.07)

Ans: H₀ accepted. There is peculiarity

7. A die was thrown 498 times. Denoting x to be the number appearing on the top face of it, the observed frequency of x is given below:

X	1	2	3	4	5	6
t	69	78	85	82	86	98

What opinion you would form for the accuracy of the die (χ² value for 5 df at 5% level of significance is 11.07)?

Ans: Die is unbiased

8. Among 64 offsprings of a certain cross between guinea pigs 34 were red, 10 were black, and 30 were white. According to the genetic model these numbers should be in the ratio 9:3:4. Are the data consistent with the model at 5% level?

Ans: The data are consistent with the model

9. In an investigation into the health and nutrition of the two groups of children of different social states, the following results are obtained:

Health	Social status		Total
	Poor	Rich
Below normal	130	20	150
Normal	102	108	210
Above normal	24	96	120

Discuss the relation between the health and their social status.

Ans: Health and social status are associated, i.e., H₀ is rejected

10. The following table gives a classification of a sample of 160 plants of their leaf color and flatness:

	Flat leaves	Curled leaves	Total
White flower	99	36	135
Red flower	20	5	25
Total	119	41	160

Test whether the flower color is independent of the flatness of leaf?

Ans: H₀ is rejected, flower color is independent of the flatness of leaf

11. From the following information, state whether the two attributes viz., condition of house and condition of child are independent:

Condition of the child	Condition of the house		Total
	Clean	Dirty
Clean	69	51	120
Fairly clean	81	20	101
Dirty	35	44	79
Total	185	115	300

(χ2 at 5% level for 2 df=5.991)

Ans: H₀ rejected. There is an association between the condition of the child and condition of the house

12. A certain drug was administered to 500 people out of a total of 800 included in the sample to test its efficiency against typhoid. The results are given below:

	Typhoid	No typhoid	Total
Drug	200	300	500
No drug	280	20	300
Total	480	320	800

On the basis of the data can we say that the drug is effective in preventing typhoid?

Ans: The drug is effective

13. In an experiment on the immunization of goats from anthrax the following results were obtained. Derive your inference on the vaccine.

	Died of anthrax	Survived	Total
Inoculated with vaccine	2	10	12
Not inoculated	6	6	12
Total	8	16

Ans: H₀ is accepted. The vaccine is ineffective in controlling the disease

14. Fifty students were selected at random from 500 students enrolled in a computer program were classified according to age and grade points giving the following data:

	20 and under	Age in years 21–30	Age above 30
	3	5	2
5.1 to 7.5	8	7	5
7.6 to 10	4	Grade parts	8
		up to 5.0

Test at 5% level of significance the hypothesis that the age and grade points are independent (χ² at 5% level for 2 df=5.991).

Ans: H₀ accepted. Grade points and age are independent of each other

15. The following data related to the sales in a time of a trade depression of a certain commodity demand:

District where sales	District not hit by depression	District hit by depression	Total
Satisfactory	350	80	330
Not satisfactory	140	30	170
Total	390	110	500

Do these data suggest that the sales are significantly affected by depression ( $χ_{0.05}^{2} = 3.841$ $χ_{0.05}^{2} = 3.841$ for 1 df)?

Ans: H₀ is accepted. The sales are not significantly affected by depression

16. A survey of 200 families having three children selected at random gave the following results:

Male births	0	1	2	3
No. of families	40	58	62	40

Test the hypothesis that male and female births are equally likely at 5 % level of significance ( $χ_{0.05}^{2} = 7.82$ $χ_{0.05}^{2} = 7.82$ for 3 df)?

17. Two researchers adopted different sampling techniques. While investigating same group of students to find the number of students falling into different intelligence level. The results are as follows:

Researchers	Below average	Average	Above average	Genius	Total
X	86	60	44	10	200
Y	40	33	25	2	100
Total	126	93	69	12	300

Would you say that the sampling techniques adopted by the two researchers are significantly different ( $χ_{0.05}^{2}$ $χ_{0.05}^{2}$ for 2 df and 3 df are 5.991 and 7.82, respectively)?

18. In the following data find whether there is any significant liking in the habit of soft designs among categories of employees?

Soft drinks	Clerks	Teachers	Officers
Pepsi	10	25	65
Thumbs-up	15	30	65
Fanta	50	60	30

(χ² for 4 df at 5% level of significance=9.4888)

Ans: H₀ rejected. Habit of drinking soft drinks depends on the category

19. A firm manufacturing rivets wants to limit variations to their length as much as possible. The lengths (in cm) of 10 rivets manufactured by a new process are

21.5	1.99	2.05	2.12	2.17
2.01	1.98	2.03	2.25	1.92

Examine whether the new process can be considered superior to the old if the population has standard deviation 0.145 cm ( $χ_{0.05}^{2} = 16.919$ $χ_{0.05}^{2} = 16.919$ for 9 df)?

Ans: H₀ accepted. The new process cannot be considered superior to the old process

20. Four coins were tossed 160 times and the following results were obtained:

No. of heads	0	1	2	3	4
Observed frequencies	17	52	54	31	6

Under the assumption that wins are balanced, find the expected frequencies of getting 0, 1, 2, 3, or 4 heads and test the goodness of fit ( $χ_{0.05}^{2}$ $χ_{0.05}^{2}$ value is 9.488)?

Ans: The fit is poor

21. One of 8000 graduates in a city, 800 are males; out of 1600 graduate employees 120 are females. Use χ² test to determine if any destruction is made in appointment on the basis of sex ( $χ_{0.05}^{2}$ $χ_{0.05}^{2}$ for 1 df=3.841).

Ans: H₀ rejected. There is distribution

22. A survey of 200 families having three children selected at random gave the following results:

Male birth	0	1	2	3
No. of families	40	58	62	40

Test the hypothesis that male and female births are equally likely at 5% level of significance ( $χ_{0.05}^{2}$ $χ_{0.05}^{2}$ value is 7.82).

Ans: H₀ rejected

23. In the accounting department of bank 100 accounts are selected at random and examined for errors. The following results have been obtained:

No. of errors	0	1	2	3	4	5	6
No. of accounts	36	40	19	2	0	2	1

Does this information verify that errors are distributed according poison probability ( $χ_{0.05}^{2} = 7.815$ $χ_{0.05}^{2} = 7.815$ for 3 df)?

Ans: H₀ accepted

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Table of Contents for Chapter 9. Chi-Square Distribution

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Chi-Square Distribution

Abstract

Keywords

9.1 Introduction

9.2 Contingency Table

9.3 Calculation of Expected Frequencies

9.4 Chi-Square Distribution

9.4.1 Characteristic Function of χ2 Distribution

9.5 Mean and Variance of Chi-Square

9.6 Additive Property of Independent Chi-Square Variate

9.7 Degrees of Freedom

9.8 Conditions for Using Chi-Square Test

9.9 Uses of Chi-Square Test

9.9.1 Chi-Square Test as a Test of Goodness of Fit

9.9.2 Test for Independence of Attributes

9.9.2.1 Yate’s Correction

9.9.3 Homogeneity Chi-Square

9.9.4 Chi-Square Distribution of Sample Variance

9.9.5 Testing a Hypothesis About the Variance of Normally Distributed Population—Decision Rule

9.9.5.1 Solved Examples

Table of Contents for
Chapter 9. Chi-Square Distribution

9.4.1 Characteristic Function of χ² Distribution