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# Quantitative Methods - Chapter 9: Chi-Square

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## Jean-Michel Sotiron

on 26 October 2016

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#### Transcript of Quantitative Methods - Chapter 9: Chi-Square

Quantitative Methods
Chi-Square
5. Compare x2 from our data to x2 from Table E:
Type 3: Three or More Chi-Square
VERY SIMILAR to the two-way test -- we are suggesting that:
Type 3: Three or More Chi-Square
5. Compare x2 from our data to x2 from the table:
Type 2: Two-Way Chi-Square
Like the one-way test, but we are suggesting that:
Type 2: Two-Way Chi-Square
A researcher assesses the political leanings of a random sample of 60 journalists to determine whether journalists are generally liberal. He found 15 conservatives, 18 social-democrats, and 27 liberals. Using a one-way chi-square, test the null hypothesis that journalists are conservative, moderate and liberal in equal amounts. Assume a level of significance of 0.05.

Remember the steps:

1. Draft the null hypothesis
Practice – Type 1: One-Way Chi-Square
Type 1: One-Way Chi-Square
CHI-SQUARE TEST
3. Obtain expected frequency for each cell
Type 3: Three or More Chi-Square
1. We draft our null hypothesis: (well, we already have it…)
Type 2: Two-Way Chi-Square
The chi-square (x2) value from the table is, once again, your critical “cut-off point”
if your chi-square from the data is less you RETAIN;
if your chi-square from the data is greater you REJECT (it’s in the REJECTION ZONE!)
5. Compare x2 from our data to x2 from the table:
Type 1: One-Way Chi-Square
1. We draft our null hypothesis:
Type 1: One-Way Chi-Square
Is there a tendency to assign certain responses over others?
Type 1: One-Way Chi-Square
Let’s Practice!
5. Set up a data in a summary table to determine chi-square value
Type 3: Three or More Chi-Square
4. Set up a data in a summary table to determine chi-square value
Type 2: Two-Way Chi-Square
3. Obtain expected frequency for each cell
Type 2: Two-Way Chi-Square
To determine whether frequencies we observe differ “significantly” from the frequencies we expect
Type 1: One-Way Chi-Square
It’s called CHI-SQUARE
Good news!
I know what you’re thinking…
Testing independent and dependent means for “significance” to reject or retain the null hypothesis.
Last two classes…
1. We draft our null hypothesis: (well, we already have it…)
Type 3: Three or More Chi-Square
df is calculated by multiplying
(# rows minus 1) by (# columns minus 1)
α is either .05 or .01
use .05
4. Calculate critical chi-square value using Table E (page 370)
Type 2: Two-Way Chi-Square
df is calculated by taking the # of categories minus 1
df = k-1
α is either .05 or .01  we tend to use .05
So, once again, we need to know α and df
4. Calculate the critical chi-square value using Table E (page 370)
Type 1: One-Way Chi-Square
Difference squared divided by fe
Difference between fo and fe
Expected frequency = fe
Observed frequency = fo
This final total (adding up all the differences squared and divided by fe -- is your chi-square value -- x2
3. Set up a data in a summary table to determine chi-square value
Type 1: One-Way Chi-Square
And difference squared
α is either .05 or .01
use .05
6. Calculate critical chi-square value using Table E (page 370)
Type 3: Three or More Chi-Square
df is calculated by multiplying
(# rows minus 1) by (# columns minus 1)
Two problems with that.
Assumes normal distribution
Assumes scale measurement
"What about MY abnormally distributed data that is nominal or ordinal?!"
test
There are 3 types:
one-way
two-way
three-or-more
(There are more...)
An example:
Imagine we want to determine whether, in preparing a multiple choice test, an instructor shows a tendency to assign certain responses over others…
How do we determine that?
Imagine this is our data –
This is what we call our
“Observed Frequencies”
The instructor shows no tendency to assign a particular correct response from A to E.

2. We calculate our “expected frequencies”
***Because if the null hypothesis applies, the same amount of answers should fall in each category (because they’re all equivalent)
In this case a total of 50 questions and 5 categories means 50/5 = 10
Expected Frequency = #Cases/#Categories
So, our expected frequency is 10 answers in each category (10 A’s; 10 B’s; 10 C’s, etc.)
5-1 = 4
If x2 from the data < x2 from the table = RETAIN null hypothesis
If x2 from the data > x2 from the table = REJECT null hypothesis
In our case:
4.6 9.488
Meaning, the degree of unevenness in responses was not sufficiently large to demonstrate an underlying preference – so we RETAIN our null hypothesis
RETAIN
What do you do?
2. Calculate “expected frequencies”
3. Set up summary table to determine chi-square value

4. Calculate chi-square from Table E
5. Compare chi-square from the data to chi-square from the table
Answer = 3.90, df = 2 Critical = 5.991
RETAIN the null hypothesis at 0.05
two populations do not differ
in terms of the frequency of occurrence of a given characteristic

Example – null hypothesis:
The relative frequency of
liberals
who are permissive parents
is the same as
the relative frequency of
conservatives
who are permissive parents.
NOW WHAT?
The relative frequency of liberals who are permissive parents is the same as the relative frequency of conservatives who are permissive parents.
2. We arrange our data in a 2 x 2 table to display “observed frequencies”
What is it?
** To find expected frequency = (row total)(column total)/N **
Upper left (Permissive Libs) =
Upper right (Permissive Cons) =
Lower left (Not Permissive Libs) =
Lower right (Not Permissive Cons) =
(25)(23)/43 = 13.37
(25)(20)/43 = 11.63
(18)(23)/43 = 9.63
(1)(20)/43 = 8.37
Let's put that in so you can see the magic
Help me out!
df = (r-1) (c-1)
df = (2-1) (2-1)
df = (1) (1)
df = 1
In our case:
2.66 < 3.841 = RETAIN
Meaning, the observed frequencies do not differ enough from the expected frequencies to indicate that an actual population difference exists.
What was x2 from our data?
What was x2 from Table E?
2.67
3.841
Three (or more) populations
do not differ
in terms of the frequency of occurrence of a
given characteristic
Example – null hypothesis:
The relative frequency of
permissive, moderate, and authoritarian
child-discipline is the same for
fathers, stepfathers, and live-in partners.
What is it again?
The relative frequency of permissive, moderate, and authoritarian child-discipline is the same for fathers, stepfathers, and live-in partners.
2. We arrange our data in a 3 x 3 table to display “observed frequencies”
Very familiar to what we did with 2 X2, eh?
** To find expected frequency = (row total)(column total)/N **
Upper left (Permissive Fathers) =
Upper middle (Permissive Step-fathers) =
Upper right (Permissive Live-in partners) =
Middle left (Moderate Fathers) =
Middle middle (Moderate Step-fathers) =
Middle right (Moderate Live-in partners) =
Lower left (Authoritarian Fathers) =
Lower middle (Authoritarian Step-fathers) =
Lower right (Authoritarian Live-in partners) =
(30)(32)/89 = 10.79
(30)(30)/89 = 10.11
(30)(27)/89 = 9.10
(28)(32)/89 = 10.07
(28)(30)/89 = 9.44
(28)(27) /89 = 8.49
(31)(32)/89 = 11.15
(31)(30)/89 = 10.45
(31)(27)/89 = 9.40
Still the same thing as Type 2!
df = (r-1) (c-1)
df = (3-1) (3-1)
df = (2) (2)
df = 4
7.57 < 9.488 = RETAIN
The observed frequencies do not differ enough from the expected frequencies to indicate that an actual population difference exists.
What is the X2 from our data?
What is the x2 from Table E?
Table E
Table E
7.57
9.488
~Hazardous-T
ASSUME A LEVEL OF SIGNIFICANCE OF .05
What's in store today
Roll call
Symbols today:
fo: observed frequency
fe: expected frequency
x2: chi-squared
Chi-square
Sum of...
observed frequencies
expected frequencies
Talk about the rest of the semester
Chi-square
Exercises
Departure
CH has a "k" sound
i as in guy
aka
fo
aka
fe
<
3 & 4 can be combined!
Find the row and column totals!
Full transcript