Chi Squared Tests

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Chi Squared Tests

• Chapter 16

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16.1 Introduction

• Two statistical techniques are presented, to
  analyze nominal data.
• A goodness-of-fit test for the multinomial
  experiment.
• A contingency table test of independence.
• Both tests use the c2 as the sampling
  distribution of the test statistic.

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16.2 Chi-Squared Goodness-of-Fit Test

• The hypothesis tested involves the probabilities
  p1, p2, , pk.of a multinomial distribution.
• The multinomial experiment is an extension of the
  binomial experiment.
• There are n independent trials.
• The outcome of each trial can be classified into
  one of k categories, called cells.
• The probability pi that the outcome fall into
  cell i remains constant for each trial. Moreover,
  p1 p2 pk 1.
• Trials of the experiment are independent

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16.2 Chi-squared Goodness-of-Fit Test

• We test whether there is sufficient evidence to
  reject a pre-specified set of values for pi.
• The hypothesis

• The test builds on comparing actual frequency and
  the expected frequency of occurrences in all the
  cells.

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The multinomial goodness of fit test - Example

• Example 16.1
• Two competing companies A and B have enjoy
  dominant position in the market. The companies
  conducted aggressive advertising campaigns.
• Market shares before the campaigns were
• Company A 45
• Company B 40
• Other competitors 15.

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The multinomial goodness of fit test - Example

• Example 16.1 continued
• To study the effect of the campaign on the market
  shares, a survey was conducted.

• 200 customers were asked to indicate their
  preference regarding the product advertised.
• Survey results
• 102 customers preferred the company As product,
• 82 customers preferred the company Bs product,
• 16 customers preferred the competitors product.

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The multinomial goodness of fit test - Example

• Example 16.1 continuedCan we conclude at 5
  significance level that the market shares were
  affected by the advertising campaigns?

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The multinomial goodness of fit test - Example

• Solution
• The population investigated is the brand
  preferences.
• The data are nominal (A, B, or other)
• This is a multinomial experiment (three
  categories).
• The question of interest Are p1, p2, and p3
  different after the campaign from their values
  before the campaign?

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The multinomial goodness of fit test - Example

• The hypotheses are
• H0 p1 .45, p2 .40, p3 .15
• H1 At least one pi changed.

The expected frequency for each category (cell)
if the null hypothesis is true is shown below
What actual frequencies did the sample return?
90 200(.45)
102
82
80 200(.40)
30 200(.15)
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The multinomial goodness of fit test - Example

• The statistic is
• The rejection region is

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The multinomial goodness of fit test - Example

• Example 16.1 continued

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The multinomial goodness of fit test - Example

• Example 16.1 continued

c2 with 2 degrees of freedom
Conclusion Since 8.18 gt 5.99, there is
sufficient evidence at 5 significance level to
reject the null hypothesis. At least one of the
probabilities pi is different. Thus, at least two
market shares have changed.
P value
Alpha
5.99
8.18
Rejection region
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Required conditions the rule of five

• The test statistic used to perform the test is
  only approximately Chi-squared distributed.
• For the approximation to apply, the expected cell
  frequency has to be at least 5 for all the cells
  (npi ³ 5).
• If the expected frequency in a cell is less than
  5, combine it with other cells.

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16.3 Chi-squared Test of a Contingency Table

• This test is used to test whether
• two nominal variables are related?
• there are differences between two or more
  populations of a nominal variable
• To accomplish the test objectives, we need to
  classify the data according to two different
  criteria.

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Contingency table c2 test Example

• Example 16.2
• In an effort to better predict the demand for
  courses offered by a certain MBA program, it was
  hypothesized that students academic background
  affect their choice of MBA major, thus, their
  courses selection.
• A random sample of last years MBA students was
  selected. The following contingency table
  summarizes relevant data.

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Contingency table c2 test Example
The observed values
There are two ways to address the problem
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Contingency table c2 test Example

• Solution
• The hypotheses are
• H0 The two variables are independent
• H1 The two variables are dependent

Since ei npi but pi is unknown, we need to
estimate the unknown probability from the data,
assuming H0 is true.
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Estimating the expected frequencies
Undergraduate
MBA Major
Degree
Accounting
Finance
Marketing
Probability
60
BA
60
60/152
BENG
31
31/152
39
BBA
39
39/152
Other
22
22/152
61
44
152
152
61
44
47
152
Probability
61/152
44/152
47/152
Under the null hypothesis the two variables
are independent P(Accounting and BA)
P(Accounting)P(BA)
61/15260/152.
The number of students expected to fall in the
cell Accounting - BA is eAcct-BA n(pAcct-BA)
152(61/152)(60/152) 6160/152 24.08
The number of students expected to fall in the
cell Finance - BBA is eFinance-BBA
npFinance-BBA 152(44/152)(39/152) 4439/152
11.29
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The expected frequencies for a contingency table

• The expected frequency of cell of raw i and
  column j in the contingency table is calculated by

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Calculation of the c2 statistic

• Solution continued

Undergraduate
MBA Major
Degree
Accounting
Finance
Marketing
31 24.08
BA
31 (24.08)
13 (17.37)
16 (18.55)
60
BENG
8 (12.44)
16 (8.97)
7 (9.58)
31
31 24.08
BBA
12 (15.65)
10 (11.29)
17 (12.06)
39
7 6.80
5 6.39
Other
10 (8.83)
5 (6.39)
7 (6.80)
22
31 24.08
61
44
47
152
7 6.80
5 6.39
31 24.08
The expected frequency
7 6.80
5 6.39
31 24.08
7 6.80
5 6.39
(31 - 24.08)2 24.08
(5 - 6.39)2 6.39
(7 - 6.80)2 6.80
c2
14.70
.
.
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Contingency table c2 test Example

• Solution continued
• The critical value in our example is

• Conclusion
• Since c2 14.70 gt 12.5916, there is
  sufficient evidence to infer at 5 significance
  level that students undergraduate degree
  and MBA students courses selection are
  dependent.

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Using the computer
Select the Chi squared / raw data Option from
Data Analysis Plus under tools. See Xm16-02
Define a code to specify each nominal value.
Input the data in columns one column for each
category.
Code Undergraduate degree 1 BA 2 BENG 3
BBA 4 OTHERS MBA Major 1
ACCOUNTING 2 FINANCE 3 MARKETING
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Required condition Rule of five

• The c2 distribution provides an adequate
  approximation to the sampling distribution under
  the condition that eij gt 5 for all the cells.
• When eij lt 5 rows or columns must be added such
  that the condition is met.

Example
4 (5.1) 7 (6.3) 4 (3.6)
18 (17.9) 23 (22.3) 12 (12.8)
14 4 12.8 5.1 16 7 16 6.3 8
4 9.2 3.6
We combine column 2 and 3
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16.5 Chi-Squared test for Normality

• The goodness of fit Chi-squared test can be used
  to determined if data were drawn from any
  distribution.
• The general procedure
• Hypothesize on the parameter values of the
  distribution we test (i.e. m m0, s s0 for
  the normal distribution).
• For the variable tested X specify disjoint ranges
  that cover all its possible values.
• Build a Chi squared statistic that (aggregately)
  compares the expected frequency under H0 and the
  actual frequency of observations that fall in
  each range.
• Run a goodness of fit test based on the
  multinomial experiment.

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15.5 Chi-Squared test for Normality

• Testing for normality in Example 12.1
• For a sample size of n50 (see Xm12-01) ,the
  sample mean was 460.38 with standard error of
  38.83. Can we infer from the data provided that
  this sample was drawn from a normal distribution
  with m 460.38 and s 38.83? Use 5
  significance level.

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c2 test for normality
Solution First let us select z values that define
each cell (expected frequency gt 5 for each
cell.) z1 -1 P(z lt -1) p1 .1587 e1
np1 50(.1587) 7.94 z2 0 P(-1 lt zlt 0)
p2 .3413 e2 np2 50(.3413) 17.07 z3 1
P(0 lt z lt 1) p3 .3413 e3 17.07
P(z gt 1) p4 .1587 e4 7.94
The cell boundaries are calculated from the
corresponding z values under H0.
The expected frequencies can now be determined
for each cell.
e2 17.07
e3 17.07
z1 (x1 - 460.38)/38.83 -1 x1 421.55
e4 7.94
e1 7.94
499.21
460.38
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c2 test for normality

• The test statistic

(10 - 7.94)2 7.94
c2
(19 - 17.07)2 17.07
(13 - 17.07)2 17.07
(8 - 7.94)2 7.94
1.72



f3 19
e3 17.07
e2 17.07
f2 13
f1 10
f4 8
e4 7.94
e1 7.94
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c2 test for normality

• The test statistic

(10 - 7.94)2 7.94
c2
(13 - 17.07)2 17.07
(19 - 17.07)2 17.07
(8 - 7.94)2 7.94
1.72

• The rejection region

• Conclusion There is insufficient evidence to
  conclude at 5 significance level that the data
  are not normally distributed.