Business 205

Review

CorrelationMS5

Preview

Chi-Square

Types of Tests

Parametric Tests Assume normal and homogeneity of variance Require parameters

Z-test, T-test, 2-Group T, ANOVA

Non-parametric Tests Few (if any) assumptions about the population

distribution Rarely state a hypotheses in terms of a specific

parameter Chi-Square

Chi-Square

Tests for “goodness of fit”

Looks at sample data to test hypothesis about the shape or proportions of a population distribution

How well does the sample proportions fit the population proportions?

2

Chi-Square Hypotheses

Regular Hypothesis Specifies

preferences Specifies

differences in population

Null Hypothesis Specifies no

preferences Specifies no

difference in population

A B C

≠ 1/3 ≠ 1/3 ≠ 1/3

A B C

1/3 1/3 1/3

One-Sample Chi-Square χ2

Compares the goodness of fit of the data to that of the null hypothesis

Compares observed frequencies against expected frequencies

Looks at the difference between 1 IV with multiple levels

Observed Frequencies

Number of individuals from the sample who are classified in a particular category

How many times it occurs

Expected Scores/Frequencies

The frequency value that is predicted from the null hypothesis and the sample size.

N

Scores Individual 2 SampleOne

One-Sample Chi-Square

score expected

score expected - score observed 22

1 Sample Chi-Square Example

Student Observed Scores

1 29

2 19

3 18

4 25

5 17

6 10

7 15

8 11

18N

Scores Individual Scores Expected

Fill out the chart

Observed Scores (O)

Expected Scores (E)

O - E (O - E)2 (O - E)2

E

29 18 11 121 6.72

19 18 1 1 .06

18 18 0 0 0

25 18 7 49 2.72

17 18 -1 1 .06

10 18 -8 64 3.56

15 18 -3 9 .50

11 18 -7 49 2.72

Chi-Square

Chi-Square = (6.72+.06+0+2.72+.06+3.56+.50+2.72) =

16.34Critical Chi-Square value at an alpha

of .05 is 3.84. Accept that your sample is not from the

population because 16.34 is larger than 3.84.

Two Sample Chi-Square TestChi-Square Test for Independence

2 FactorsMany levelsTests whether or not there is a

relationship between 2 variables

Two Sample Chi-Square TestChi-Square Test for Independence

N

totalRow x alColumn tot Frequency Expected

df = (number of rows - 1)(number of columns - 1)

score expected

score expected - score observed ceIndependenfor Test Square-Chi

2

Two Sample Chi-Square TestChi-Square Test for Independence

Three different drug treatments are used to control hypertension. At the end of treatment, the investigator classifies patients as having either a favorable or unfavorable response to the medication. Your hypothesis is that there is a different between treatments.

Two Sample Chi-Square TestChi-Square Test for Independence

Treatment 1 Treatment 2 Treatment 3 Total in rows

Favorable 70 160 168

Unfavorable 30 40 32

Total in columns N = 500

1.) Total up each column2.) Total up each row3.) Find the expected score for each cell

N

totalRow x alColumn totE

Two Sample Chi-Square TestChi-Square Test for Independence

Treatment 1

Treatment 2

Treatment 3

Total in rows

Favorable 70(E = 79.6)

160(E = 159.2)

168(E = 159.2)

398

Unfavorable 30(E = 20.4)

40(E = 40.8)

32(E = 40.8)

102

Total in columns

100 200 200 N = 500

Two Sample Chi-Square TestChi-Square Test for Independence

Observed

Expected O-E (O-E)2 (O-E)2/E

70 (398)(100)/500 = 79.6 -9.6 92.16 1.15

30 (102)(100)/500 = 20.4 9.6 92.16 4.15

160 (398)(200)/500 = 159.2

.8 .64 .00

40 (102)(200)/500 = 40.8 -.8 .64 .01

168 (398)(200)/500 = 159.2

8.8 77.4 .48

32 (102)(200)/500 = 40.8 -8.8 77.4 1.89

Two Sample Chi-Square TestChi-Square Test for Independence

Chi-Square = (1.15+4.15+.00+.01+.48+1.89) = 8.02

Critical value for an alpha of .05 and at 2 df = 5.99

Ours falls above the critical region so we can accept the hypothesis and say that the treatments differ.

Reporting findingsChi-Square Test for Independence

The chi-square value is in the critical region. Therefore, we can reject the null hypothesis. There is a relationship between drug treatments and responses towards medication (2, n =500) = 8.02, p <.05.2

Chi-Square In Class Example

Hotel 1 Hotel 2 Hotel 3 Total in rows

Favorable 50 100 85

Unfavorable 10 20 16

Total in columnsN =

H1: There is a difference between hotels.

Excel: Chi-Square

DO NOT USE:CHIDIST( ) Only returns a probabilityFor Single Chi-Squares, you must program

that in by hand

CHITEST( ) Only returns a probabilityFor a test of independence (2 sample) you

must program that in by hand