**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

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3 & 4 can be combined!

Find the row and column totals!