MAP812_The Chi-Square Statistic_Samantha Ng

7/30/2019 MAP812_The Chi-Square Statistic_Samantha Ng

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The Chi-Square Statistic

2

Done by:Ng Shi Ying Samantha


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Characteristics of the chi-square test

The chi-square statistic is a non-parametric test used toevaluate hypotheses about the proportions or relationships

that exist within populations

Characteristics of variables Categorical (nominal) data (or at most, ordinal data)

Observations are independent

Two types of chi-square tests Chi-square test for Goodness-of-fit

Chi-square test for Independence


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Chi-Square Test for Goodness-of-Fit

Used to determine how well the obtained sample

proportions fit the population proportions specified by

the null hypothesis Observed frequencies (fo) vs. Expected frequencies (fe)

Assumptions One categorical variable, with two or more categories

A hypothesized proportion (equal or unequal)

No more than 20% of expected frequencies have counts less

than 5


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Chi-Square Test for Independence

Used to determine if there is a relationship between

two variables in a population

Assumptions

Two variables that are ordinal or nominal There are two or more categories in each variable


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Hypothesis Testing


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Steps for Hypothesis Testing

1. State hypotheses and select alpha level

2. Calculate degrees of freedom and locate critical region

3. Calculate chi-square statistic

4. State decision and conclusion

Report results for chi-square


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Stating Hypotheses

Test for Goodness-of-Fit

i. No preference among categories Two-tailed test

H0

: p1

= p2

(population proportions are equal among the

categories)

H1: p1 p2 (population proportions are not equal among the

categories)

One-tailed test H0: p1 = p2 (population proportions are equal among the

categories)

H1: p1 > p2 (population proportion is greater in one of the

categories)


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ii. No difference from known population

Two-tailed test H0: p1 = p2 (sample and known population proportions are

equal)

H1: p1 p2 (sample and known population proportions are not

equal) One-tailed test

H0: p1 = p2 (sample and known population proportions are

equal)

H1: p1 > p2 (sample proportion is greater than known

population proportion)


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Stating Hypotheses

Test for Independence

i. Variables are independent Two-tailed test

H0: p1 = p2 (population proportions are equal among the

categories)

H1: p1 p2 (population proportions are not equal among the

categories)

One-tailed test

H0: p1 = p2 (population proportions are equal among thecategories)

H1: p1 > p2 (population proportion is greater in one of the

categories)


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df and Critical Region

Test for Goodness-of-fit

df= C (no. of categories) -1


df= (C [no. of columns] -1) x(R [no. of rows] -1)


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dfand Critical Region

Critical region: Refer to table of critical values for

chi-square

e.g. Ifdf= 4, = .05:

2critical = 9.49


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Calculating the chi-square statistic

2statistic =

Test for Goodness-of-fit:f

o

(observed frequency in a category) =po

x no

fe (expected frequency in a category) =pe x ne

Test for Independence:f

c(observed frequency in a column) =p

cx n

c

fr (expected frequency in a row) =pr x nrfo (observed frequency in a category) =po x no

fe (expected frequency in a category) = (fc xfr)/n

(fofe)

2

fe


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State decision and conclusion

If2statistic > 2critical , reject Ho and conclude that there

are significant differences in the proportions

Reporting results for chi-square

E.g. The participants showed significant

preferences among the four orientations for

hanging the painting, 2 (3, n = 50) = 8.08,p < .05

(Gravetter & Wallnau, 2011)


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Effect Size for Test of Independence

2 x 2 matrix:

= (2/n)

> 2 x 2 matrix:

V = *2/n(df)]


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Examples

Example 1

Samantha wants to know if girls prefer flowers or chocolates as aValentines Day gift. She hypothesizes that girls will prefer

flowers as chocolates are fattening. She surveyed 500 girls to

find out if her hypothesis is true, and obtained the following

results:


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Example 1 (contd)

How many variables are being tested? One Is it a nominal variable?

Yes

Are you testing how well the obtained

sample proportions fit the population

proportions specified by the null

hypothesis?

Yes

Do more than 20% of

expected frequencies have

counts less than 5?

Yes Use GOODNESS-OF-FIT TEST


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Examples

Example 2

Samantha wants to know if there are gender differences inpreferred colour (Pink vs. Blue). She hypothesizes that Males will

prefer Blue while Females will prefer Pink. She surveyed 20

males and 20 females and obtained the following results:


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Example 2 (contd)

How many variables are being tested? Two Are they nominal variables?

Yes

Are you testing if there is a relationship

between two variables in a population?Yes

Are there two or more

categories in each

variable?

Yes Use INDEPENDENCE TEST


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Using SPSS


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Test for Goodness-of Fit

Example 1

Samantha wants to know if girls prefer flowers or chocolates as aValentines Day gift. She hypothesizes that girls will prefer

flowers as chocolates are fattening. She surveyed 500 girls to

find out if her hypothesis is true, and obtained the following

results:


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Steps for using SPSS:

1. Create variables and code for gift type.

2. Enter the data.


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3. Weight the cases

(a) Click Data > Weight Cases... (b) Select the "Weight

cases by" box andtransfer the "frequency"

variable into the

"Frequency Variable:"

box. Click OK.


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4. Start analysis

(a) Click Analyze > Nonparametric Tests (b) Transfer the "gift" variable into the

"Test> Legacy Dialogs > Chi-square Variable List:" box. Keep the "Allcategories equal" option selected in

the "Expected Values" area, as equal

proportions are assumed for each

category. Click OK.


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5. SPSS Output for Chi-Square Goodness-of-Fit Test

(a) This table provides the observed frequencies (Observed N) for each gift as well as

the expected frequencies (Expected N), which are the frequencies expected if the null

hypothesis is true. The difference between the observed and expected frequencies isprovided in the Residual column.

(b) This table provides the results of the Chi-Square Goodness-of-Fit test. We can see

from this table that the test statistic is statistically significant: 2(1) = 46.2, p < .0001.

We can, therefore, reject the null hypothesis and conclude that there are statistically

significant differences in the preference of the type of Valentines Day gift, with less

girls preferring chocolates(N = 174) compared to flowers (N = 326).

Gift

Observed N Expected N Residual

Flowers 326 250.0 76.0

Chocolate 174 250.0 -76.0

Total 500

Test Statistics

Gift

Chi-Square 46.208a

Df 1

Asymp. Sig. .000

a. 0 cells (.0%) have

expected frequencies

less than 5. The

minimum expected cell

frequency is 250.0.


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Example 2

Samantha wants to know if there are gender differences inpreferred colour (Pink vs. Blue). She hypothesizes that Males will

prefer Blue while Females will prefer Pink. She surveyed 20

males and 20 females and obtained the following results:


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Steps for using SPSS:

1. Create variables and code for Gender and Colour.

2. Enter the data.


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3. Start analysis

(a) Click Analyze > Descriptive Statistics (b) Transfer one of the variables

> Crosstabs... into the "Row(s):" box and the

other variable into the "Column(s):"

box. Click on "Display

clustered bar charts".


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(c) Click on the Statistics...button. (d) Click the Cells... button. Select

Select the "Chi-square" and "Phi and "Observed" from the "Counts" area

Cramer's V" options. Click Continue. and "Row", "Column" and "Total"

from the "Percentages" area. Click

Continue.

(e) Click OK to generate the output.


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4. SPSS Output for Chi-Square Independence Test:

(a) This table shows us that more males prefer blue while more females prefer pink:

Gender * Colour Crosstabulation

Colour

Total1 2

Gender 1 Count 2 18 20

% within

Gender

10.0% 90.0% 100.0%

% within Colour 12.5% 75.0% 50.0%

% of Total 5.0% 45.0% 50.0%

2 Count 14 6 20

% within

Gender

70.0% 30.0% 100.0%

% within Colour 87.5% 25.0% 50.0%

% of Total 35.0% 15.0% 50.0%

Total Count 16 24 40

% within

Gender

40.0% 60.0% 100.0%

% within Colour 100.0% 100.0% 100.0%

% of Total 40.0% 60.0% 100.0%


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(b) This table provides the results of the Chi-Square Independence test. We can

see from this table that the test statistic is statistically significant: 2(1) = 15.0,p

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MAP812_The Chi-Square Statistic_Samantha Ng