Chi Square & Anova

DIFFERENCE BETWEEN CHI SQUARE & ANOVA
BirinderSingh,AssistantProfessor,PCTE
Ludhiana
2
 It enables us to test
whether more than two
population proportions
can be considered
equal
 Analysis of Variance
(Anova) enables us to
test whether more than
two population means
can be considered
equal.
Chi Square (χ2 Test) Anova (F Test)

CHARACTERISTICS OF CHI SQUARE
 Every Chi square distribution extends indefinitely to right from
zero.
 It is skewed to right
 As df increases, Chi square curve become more bell shaped and
approaches normal distribution.
 Its mean is degree of freedom
 Its variance is twice degree of freedom
3
Ludhiana

CHI SQUARE (Χ2 TEST)
 Chi Square Test deals with analysis of categorical data in terms
of frequencies / proportions / percentages.
 It is primarily of three types:
 Test of Homogeneity: To determine whether different population are
similar w.r.t some characteristics.
 Test of Independence: Tests whether the characteristics of the
elements of the same population are related or independent.
 Test of Goodness of Fit: To determine whether there is a significant
difference between an observed frequency distribution and theoretical
probability distribution.
4
Ludhiana

COMPUTATIONAL PROCEDURE – CHI SQUARE
TEST
 Formulate Null & Alternative Hypothesis
 State type of test
 Select LOS
 Compute expected frequencies assuming H0 to be true.
 Compute χ2 calculated value using
 𝜒2
cal =
(𝑓𝑜 −𝑓𝑒)2
𝑓𝑒
 Extract 𝜒2
crit value from table
 Compare 𝜒2
cal & 𝜒2
crit and make decision.
5
Ludhiana

CHI SQUARE – TEST OF HOMOGENEITY
 Based on a study, it is expected that 50% of the students opt for
marketing, 30% for finance and 20% for HR. In a sample for 100,
it was observed that 61, 24 and 15 opt for these subjects
respectively. Do you agree with study findings at 10% LOS?
(𝜒2
cal = 4.87)
(𝜒2
crit = 4.605)
(Rejected)
6
Ludhiana

CHI SQUARE – TEST OF HOMOGENEITY
 A shoe seller has received the consignment of the order that he
had placed for 10000 pair of different sizes. Without physically
segregating the sizes and counting no. of pair of shoes of each
size, he wants to ascertain that consignment received is as per
order . Check at 5% LOS.
(𝜒2
cal = 6.87)
(𝜒2
crit = 12.592)
(Accepted)
7
Ludhiana
Size 4 5 6 7 8 9 10
Order qty 500 1500 2000 2000 2000 1500 500
Rec. qty 700 1800 2200 2000 2000 1300 0

CHI SQUARE – TEST OF INDEPENDENCE
FORMULAE TO BE USED
 Computation of expected frequency
 Fe = (RT x CT) / GT where RT = Row Total
CT = Column Total
GT = Grand Total
 Computation of degree of freedom
 Df = (r – 1) (c – 1)
8
Ludhiana

 Following data was collected when a survey was carried out on
preference for formal wear in work place:
Is there difference in preference due to sex? Check at 20% LOS.
(𝜒2
cal = 0.2522)
(𝜒2
crit = 1.642)
(Accepted)
9
Ludhiana
Gents Ladies
Yes 520 60
No 80 11

 Sample data in respect of viewership of a TV Program for various
age groups was collected and is as follows:
Is
viewership of program independent of age ? Check at 5% LOS.
(𝜒2
cal = 26.01)
(𝜒2
crit = 9.488)
(Rejected)
10
Ludhiana
15-25 26-40 41-50
Always 75 180 105
Sometimes 50 60 40
Never 25 20 5

CHI SQUARE – TEST OF GOODNESS OF FIT
 Gordon Company requires that college seniors who are seeking
positions will be interviewed. For staffing purposes, the director of
recruitment thinks that the interview process can be approximated
by a binomial distribution with p = 0.40 i.e. Can he conclude that
BD at p = 0.4 provides a good description of observed
frequencies. Check at 20% LOS.
(𝜒2
cal = 5.041)
(𝜒2
crit = 4.642)
(Rejected)
11
Ludhiana

ANALYSIS OF VARIANCE (ANOVA)
 It enables us to test for the significance of the differences among
more than two sample means.
 Using Anova, we will be able to make inferences about whether
our samples are drawn from population having the same mean.
 Examples:
 Comparing the mileage of five different brands of cars
 Testing which of the four different training methods produces the fastest
learning record
 Comparing the average salary of three different companies
 In each of these cases, we would compare the means of more
than two sample means.
 F-Distribution is used to analyze certain situations
12
Ludhiana

ASSUMPTIONS
 Populations are normally distributed
 Samples are random and independent
 Population Variances are equal.
13
Ludhiana

COMPUTATIONAL PROCEDURE IN ANOVA
(ONE WAY)
 Define Null & Alternative Hypothesis
 Select estimator & determine its distribution
 Select Significance Level
 Calculate Sum of all observations: T = Ʃxi
 Calculate correction factor: CF = T2 / nT where nT = sample size
 Calculate Sum of squares total, SST = Σ(Σ𝑥𝑖
2) − CF
 Calculate Sum of squares between columns, SSB = Σ((Σ𝑥𝑖)2
/𝑛𝑖) − CF
 Calculate Sum of squares within columns, SSW = SST – SSB
 Calculate Mean of squares between groups, MSB = SSB / (k – 1) where k
= no. of samples
 Calculate Mean of squares within groups, MSW = SSW / (nT – k)
 Calculate Fcal = MSB / MSW
 Calculate Fcrit = F(dfnum, dfden, α) where dfnum = k – 1, dfden = nT – k
 Compare Fcal & Fcrit and make your statistical & managerial decisions
14
Ludhiana

PRACTICE PROBLEM – ONE WAY ANOVA
 Three group of students are taught a statistical technique by three
different methods. When tested on one problem, sample scores of
3 students selected at random from each of the group as under:
At 0.05 LOS, do the means
of populations taught by
three methods differ?
(Fcal = 1.5, Fcrit = 5.14)
(Accepted)
15
Ludhiana
Group 1 Group 2 Group 3
3 5 3
4 7 7
5 6 5

16
Ludhiana

PRACTICE PROBLEM – ONE WAY ANOVA
 IAA wanted to find out if average sale of small cars namely Swift,
Jazz & Figo is same in Tier II cities. It obtained quarterly sales
data from 5 such cities A,B,C,D,E as shown. What conclusion can
be drawn at 0.05 LOS?
(Fcal = 1.5, Fcrit = 5.14)
(Accepted)
17
Ludhiana
City Swift Jazz Figo
A 32 26 30
B 28 34 -
C 25 - 28
D 34 33 32
E 31 31 26

COMPUTATIONAL PROCEDURE IN ANOVA
(TWO WAY)
 Define Null & Alternative Hypothesis
 Select estimator & determine its distribution
 Select Significance Level
 Calculate Sum of all observations: T = Ʃxi
 Calculate correction factor: CF = T2 / nT where nT = sample size
 Calculate Sum of squares columns, SSC = Σ((Σ𝑥𝑗)2
/𝑛𝑖) − CF
 Calculate Sum of squares rows, SSR = Σ((Σ𝑥𝑖)2/𝑛𝑗) − CF
 Calculate Sum of squares total, SST = Σ(Σ𝑥𝑖
2
) − CF
 Calculate Sum of square error, SSE = SST – (SSC + SSR)
 Calculate Mean of squares column, MSC = SSC / (c – 1)
 Calculate Mean of squares rows, MSR = SSR / (r – 1)
 Calculate Mean of squares error, MSE = SSE / (c – 1) (r – 1)
 Calculate Fcal = MSC/MSE & MSR/MSE
 Calculate Fcrit = F(dfnum, dfden, α) where dfnum = c-1 or r-1, dfden = (c-1)(r-1)
 Compare Fcal & Fcrit and make your statistical & managerial decisions 18
Ludhiana

PRACTICE PROBLEM – TWO WAY ANOVA
 Three salesmen Kallu, Lallu & Mallu were assigned three cities
A,B & C. The data on sales for quarter ending June 2017
achieved by them is:
Is there any significant difference
in sales made by 3 of them?
Is there any significant difference
in sales made in 3 cities?
Check at 0.05 LOS?
(F1cal = 9.23, F1crit = 6.94)
(F2cal = 3.25, F2crit = 6.94)
(Rejected) & (Accepted)

19
Ludhiana
City Kallu Mallu Lallu
A 4 3 4
B 3 2 5
C 5 3 6

DECISION FLOW DIAGRAM -
ESTIMATION
20
Ludhiana
Start
Is n≥30
Is pop.
Known to
be normally
distributed
Use ‘Z’ table Stop
Use a Statistician
Is SD
known
?
Use ‘Z’
table
Stop
Use ‘t’
table
Stop

Chi Square & Anova

More Related Content

What's hot

Similar to Chi Square & Anova

More from Birinder Singh Gulati

Recently uploaded

Chi Square & Anova