Transcript Chi Square Procedures Chapter 14 14.1 Test for Goodness of Fit     Suppose you open a 1.69 ounce bag of M&M’s and discover that out.

Chi Square
Procedures
Chapter 14
14.1 Test for Goodness of Fit




Suppose you open a 1.69 ounce bag of M&M’s and
discover that out of 57 total there are only 2 red
M&M’s. You know that Mars makes 13% reds, but in
your sample, your proportion is only 2/56 or 2%
You could perform the one proportion Z test to test the
hypothesis:
H0: ρ= .12 and Ha: ρ≠ .12
You could then perform additional tests for each of
the remaining colors, but this is inefficient.
Introducing, Chi Square! We compare the observed
counts for our sample with the counts that would be
expected
Example- Driving and cell
phones
A
study of 699 drivers who were using cell
phones when involved in a collision
examined this question. These drivers
made 26,798 calls during a 14 month
study period. Data classified by day of
the week:
Total
Sun Mon
Tues
Wed
Thurs Fri
Sat
699
20
126
159
136
12
 Are
133
113
accidents equally likely to occur on
any day of the week?





H0: Motor vehicle accidents involving cell phone use
are equally likely to occur on each of the seven days of
the week
Ha: The probabilities of a motor vehicle accident
involving cell phone use vary from day to day (that is,
they are not all the same)
(The more the observed counts differ from the
expected counts, the more evidence we have to reject
the null and conclude that the probabilities of an
accident involving cell phone use are not the same for
each day of the week)
The expected count for any categorical variable is
obtained by multiplying the expected proportion for
each category by the sample size.
We EXPECT the same proportion each day of the week,
or 1/7 of the total week accidents to occur on any
given day
Day
Observed
Count O
Expected
Count E
(1/7)*(699)
Sun
20
99.857
Mon
133
99.857
Tues
126
99.857
Wed
159
99.857
Thurs
136
99.857
Fri
113
99.857
Sat
12
99.857
Day
Observed
Count O
Expected
Count E
(1/7)*(699)
(O-E)2/E
Sun
20
99.857
63.86
Mon
133
99.857
11
Tues
126
99.857
6.84
Wed
159
99.857
35.03
Thurs
136
99.857
13.08
Fri
113
99.857
1.73
Sat
12
99.857
77.3
sum (χ2) =
208.84
The larger the differences between the observed and
expected values, the larger χ2 will be and the more evidence
there will be against the null.
Degrees of Freedom and
Conditions

Our Critical chi square values are based on
degrees of freedom, which are number of
categories – 1



(in this case, 7 – 1 which = 6) and there is a table in
our tables
For our example, for 6 df the critical value is 12.49.
Our calculated value of 208.84 was much bigger
so we reject the null
Conditions: You may use the test when all
individual expected counts are at least 1 and no
more than 20% of the expected counts are less
than 5 (good in our example)
Properties of Chi Square
distributions
Calculator
 Enter
the observed counts in L1
 Calculate the expected counts separately
and enter them in L2
 If you have a TI84 go to Stat, Test, χ2GOF
and enter info.

TI-83: If that isn’t an option, define L3 as (L1 –
L2)2/L2 Go to Math/List/Sum (L3) to
calculate χ2 To find the p-value, go to DISTR,
χ2CDF and enter (your test statistic, very large
number, df)
14.2 Inference for two way tables:
Homogeneity vs. Independence tests
 What
is important here for what APstat wants is
what you SAY about the result. This distinction rests
on HOW you get the data.
 There will always be 2 factors or variables ... you
might be looking at the connection between
males/females ... and if they agree or disagree with
some question. So, one variable or factor is gender
and the other is response.
Now, think of the data you could get.

A. You could take an SRS of say n=200 ... and AFTER the fact, find
out if they are M or F and if they said Y or N, and do the cross tabs
on the data. For neither of the two factors were you in control of
how many you get, how many M or F are in the sample, or how
many Y or N responses you get.
B. You could decide ahead of time to separate the people into
lists of M and F FIRST ... and take an SRS of say n=100 from each
grouping ... n=100 M and n=100 F. Then after that, you sample and
ask the question and get a Y or N from each of the 200 people.
You still make a cross tabs table ... with gender and response as
the factors ... BUT you know that you will have 100 M and you
know you will have 100 F ... so those values at the bottom or side
headings of the table are KNOWN, ahead of time. In this case,
you know what the overall results will be for gender ... what you
don't know is how they break down WITHIN each of the M or F
categories ... in terms of response of Y or N. So, here you control
n’s for one of the two factors ... but not both.
APstat says the first will be called a test of independence ... and the
second will be called a test of homogeneity.

Homogeneity asks whether the distribution of ONE
VARIABLE is the same in TWO (or more)
POPULATIONS. The design of the study involves
groups that have been sampled (or assigned)
separately, and the responses will address that
one variable in question. Because the group
members were identified in advance of
collecting the data, the table's column (or
perhaps row) totals are fixed in advance.

Independence asks whether there's an
association between TWO VARIABLES in ONE
POPULATION. The design of the study involves one
group that has been cross-categorized based on
responses to two variables. Only the table's total
number of respondents is known in advance; the
row and column totals appear only after the
data have been tallied.
 If
we wonder whether there's any association
between eye color and handedness, we'd
probably choose a random sample of
people (one population) and ask them what
color their eyes are and what hand they
write with (2 variables). That's a question of
independence.
 We
could examine a question about eye
color and sex either way.


1) Select one sample of people, then ask
about eye color and record the sex of each
respondent = independence.
2) Select separate samples of males and
females, then ask about eye color =
homogeneity.
Homogeneity of populations chi
square
 The
2 sample Z procedures of last chapter allow
us to compare the proportions of successes in
two groups. What if we want to compare more
than 2 groups?
 Example: Does background music influence wine
purchases (conditional distributions)

Researchers know that background music can
influence the mood and purchasing behavior of
customers so they played different music types and
recorded the numbers of bottles of each type of
wine purchased
Data then Column percents for
wine and music
Music
Wine
None
French
Italian
Total
French
30
39
30
99
Italian
11
1
19
31
Other
43
35
35
113
Total
84
75
84
243
Music
Wine
None
French
Italian
Total
French
35.7
52
35.7
40.7
Italian
13.1
1.3
22.6
12.8
Other
51.2
46.7
41.7
46.5
Total
100
100
100
100
Comparisons of different types
of wine sold for different music
conditions
s
Comparisons of types of wine
sold for different music
conditions
s
The problem of Multiple
comparisons


Researches expected that music would influence type
of wine purchased so music is the explanatory variable
and type of wine purchased is the response variable.
In general, easiest way to describe this kind of
relationship is to compare the conditional distributions
of the response variables for each value of the
explanatory variable


This is the column percents that give the conditional
distribution of purchases for each type of music played.
But this still doesn’t give us an accurate way of
analyzing b/c this becomes multiple comparisons!
Two-Way Tables
 First
step in comparing several proportions is to
arrange the data in a two-way table that gives
counts for both successes and failures. Our table
in the example is a 3x3 because it has 3 rows and 3
columns (not counting total). It shows the counts
for all 9 combinations of our variables.
A
table with r rows and c columns is an r x c table
Stating Hypothesis
 H 0:
The distribution of the response variable is the
same in all “c” populations
 For this example we are comparing three
populations:



1. bottles of wine sold when no music playing
2. bottles of wine sold when french music playing
3. bottles of wine sold when Italian music is playing
 We
have 3 independent samples of sizes 84, 75,
and 84 from each population.

H0: The proportions of each type of wine sold are
the same in all 3 populations.
Computing expected cell
counts
 Chi
Square Test: The Chi Square GOF test
we did in the first section is the same, just
that columns = 1.
 With homogeneity tests, df = (r-1)(c-1) but
everything else is calculated the same.
Conditions
a
Full example response for Wine
 Step
1: Populations and Parameters: We want to
use X2 to compare the distribution of types of
wine selected for each type of music. Our
hypotheses are


H0: The distributions of wine selected are the same in
all three populations of music types
Ha: The distributions of wine selected are not all the
same
 Step
2: Conditions: To use the chi-square test for
homogeneity of populations:


The data must come from independent SRS’s from
the populations of interest. We are willing to treat
the subjects in the three groups as SRS’s from their
respective populations.
All expected cell counts are greater than 1, and no
more than 20% are less than 5.
 Step



df (3-1)(3-1) = 4
X2 = 18.28 (will show you on calc shortly)
P value = .0019
 Step

3: Calculations:
4: Interpretation:
There is strong evidence to reject the null and
conclude that type of music being played has a
significant effect on wine sales.
Calculator…finally!
 Use

a matrix to store observed counts:
Enter observed counts in matrix [A]
 Press
2nd X-1 (MATRIX), to EDIT, choose 1: [A].
 Enter the observed counts from the two way table in the
matrix in the same locations.

Specify the chi-square test, the matrix where the
observed counts are found, and the matrix where
expected counts will be stored
 Press
STAT, TESTS, choose C: χ2-Test
 Choose calculate or draw.

If you want to see the expected counts, simply
display matrix [B]:
 2nd
MATRIX choose 2: [B]
Chi Square vs. Z test
 They
are the same! The 2 prop Z test from
section 13.2 and a chi square test (with 1
df) for a 2x2 table

P values will be the same
2. Chi Square test of
Association/Independence
 The
null hypothesis of “no difference
among treatments” takes the form of “no
association between two categorical
variables”
 Ex: Exclusive territory clause vs. success of
franchise (like mcD).
Exclusive
Territory
Success
Yes
No
Total
Yes
108
15
123
No
34
13
47
Total
142
28
170
 1:


 2:

 3:

 4:

Hypothesis
H0: There is no association between success and
exclusive territory
HA: There is an association between success and
exclusive territory
Conditions
To use the chi-square test of
association/independence, we must check that all
expected cell counts are at least 1 and that no more
than 20% are less than 5 (we’re good)
Calculations
The test statisticX2 is 4.9112, df = 1, P- value = .013
Interpretation
There is sufficient evidence of an association
between success and exclusive territory in the
population of franchise.
 *Association
does not imply causation!
How do I tell which Chi Square
test to do?

Examine the design of the study



The statement of the hypothesis differs on the sampling
design. It is all in the way you collect the data.



For Association/Independence: There is a single sample
from a single population
For homogeneity of populations- there is a sample from
each of two or more populations. Each individual is
classified based on a single categorical variable.
If I survey 1000 females selected at random and 1000 males
selected at random to determine political affiliation, then
this is a test of homogeneity.
If I call 2000 people at random, then ask their gender and
political affiliation, then this is a test for independence.
Asking one question--homogeneity, asking two questions-independence.