Chapter 14: Chi-Square Procedures. 14.1 – Test for Goodness of Fit.

Chapter 14: Chi-Square Procedures

14.1 – Test for Goodness of Fit

Chi-square test for goodness of fit:

Used to determine if what outcome you expect to happen actually does happen

Observed Counts: Count of actual results

Expected Counts: Count of expected results

To find the expected counts multiply the proportion you expect times the sample size

(2)

Note: Sometimes the probabilities will be expected to be the same and sometimes they will be expected to be different

Ho: All of the proportions are as expected


HA: One or more of the proportions are different from expected

( )∑ −=

22

E

EOχ


df = k – 1 categories

Properties of the Chi-distribution:

• Always positive (being squared)

• Skewed to the right

• Distribution changes as degrees of freedom change

• Area is shaded to the right

Properties of the Chi-distribution:

• Always positive (being squared)

• Skewed to the right

• Distribution changes as degrees of freedom change


Conditions:

all expected counts are ≥ 5

Calculator Tip! Goodness of Fit

L1: Observed

L2: Expected

L3: (L1 – L2)2 / L2

List – Math – Sum – L3

Calculator Tip! P(2 > #)

2nd dist - 2cdf(2, big #, df)

Example #1A study yields a chi-square statistic value of 20 (2 = 20). What is the P value of the test if the study was a goodness-of-fit test with 12 categories?

Example #1A study yields a chi-square statistic value of 20 (2 = 20). What is the P value of the test if the study was a goodness-of-fit test with 12 categories?

< P <0.025 0.05

Or by calc:

2cdf(20, 10000000, 11) =

2nd dist - 2cdf(2, big #, df)

0.04534

Example #2Find the expected values and calculate the 2 of the 96 rolls of the die. Then find the probability.

( )E

EO 2−

Face Value

1 2 3 4 5 6

Observed 19 15 10 14 17 21

Expected 16 16 16 16 16 16

(19 – 16)2

16= 0.5625

0.5625

(15 – 16)2

16= 0.0625

0.0625 2.25 0.25 0.0625 1.5625

Example #2Find the expected values and calculate the 2 of the 96 rolls of the die. Then find the probability.

( )E

EO 2−

Face Value

1 2 3 4 5 6

Observed 19 15 10 14 17 21

Expected 16 16 16 16 16 16

0.5625 0.0625 2.25 0.25 0.0625 1.5625

0.5625 + 0.0625 + 2.25 + 0.25 + 0.0625 + 1.5625 =

( )22 O E

Eχ

−= ∑ = 4.75

P(2 > 4.75) =

df = 6 – 1 = 5

P(2 > 4.75) =

df = 6 – 1 = 5

P(2 > 4.75) > 0.25

Or by calc:

2cdf(4.75, 10000000, 5) = 0.4471

Example #3The number of defects from a manufacturing process by day of the week are as follows:

The manufacturer is concerned that the number of defects is greater on Monday and Friday. Test, at the 0.05 level of significance, the claim that the proportion of defects is the same each day of the week.

Monday Tuesday Wednesday Thursday Friday

# 36 23 26 25 40

P: The true proportion of defects from a manufacturing process per day

H:

The proportion of defects from a manufacturing process is the same Mon-Fri

Ho:

The proportion of defects from a manufacturing process is not the same Mon-Fri (on one day or more)

HA:

A:


# 36 23 26 25 40

Expected

150 errors total.

150 5

= 30

30 30 30 30 30

(All expected counts > 5)

N: Chi-Square Goodness of Fit

T:

2 =


# 36 23 26 25 40

Expected 30 30 30 30 30

(O – E)2

E= = 7.533(36 – 30)2

30+ (23 – 30)2

30+ …

O: P(2 > 7.533) =

df = 5 – 1 = 4

P(2 > 7.533) =

df = 5 – 1 = 4

0.10 < P(2 > 7.533) > 0.15

Or by calc:

2cdf(7.533, 10000000, 4) = 0.1102

O:

M:

P ___________

0.1102 0.05

>

Accept the Null

S:

There is not enough evidence to claim the proportion of defects from a manufacturing process is not the same Mon-Fri (on one day or more)

14.2 – Inference for Two-Way Tables

To compare two proportions, we use a 2-Proportion Z Test. If we want to compare three or more proportions, we need a new procedure.

Two – Way Table:

Organize the data for several proportions

R rows and C columns

Dimensions are r x c

To calculate the expected counts, multiply the row total by the column total, and divide by the table total:

Expected count = Row total x Column total table total

Degrees of Freedom: (r – 1)(c – 1)

Chi-Square test for Homogeneity:

Ho: The proportions are the same among 2 or more populations

Ha: The proportions are different among 2 or more populations

Expected Counts are ≥ 5

SRS

Conditions:

Compare two or more populations on one categorical variable

Chi-Square test for Association/Independence:

Ho: There is no association between two categorical variables

Ha: There is an association

Expected Counts are ≥ 5

SRS

Conditions:

Two categorical variables collected from a single population

Calculator Tip! Test for Homogeneity/Independence

2nd – matrx – edit – [A] – rxc – Table info

Stat – tests – 2–test

Observed: [A]Expected: [B]Calculate

Then:

Note: Expected: [B] is done automatically!

Example #1The table shows the number of people in each grade of high school who preferred a different color of socks.

a. What is the expected value for the number of 12th graders who prefer red socks?

Expected count = Row total x Column total table total

20 x 14 56

=

2014

1818

121515 56

= 5

Example #1The table shows the number of people in each grade of high school who preferred a different color of socks.

b. Find the degrees of freedom.

(r – 1)(c – 1)

(3 – 1)(4 – 1)

(2)(3)

6

Example #2An SRS of a group of teens enrolled in alternative schooling programs was asked if they smoked or not. The information is classified by gender in the table. Find the expected counts for each cell, and then find the chi-square statistic, degrees of freedom, and its corresponding probability.

Expected Counts: Row total x Column total table total

70 x 79 217

79 138

70

147217

147 x 79 217

70 x 138 217

147 x 138 217

=

=

=

=

25.484

53.516

44.516

93.484

2 = (O – E)2

E=

Expected Counts:

Smoker Non-Smoker

Male

Female

25.484

53.516

44.516

93.484

0.56197

(23 – 25.484)2

25.48+ (47 – 44.516)2

44.516+ (56 – 53.516)2

53.516+ (91 – 93.484)2

93.484

2 = (O – E)2

E=

Expected Counts:

Smoker Non-Smoker

Male

Female

25.484

53.516

44.516

93.484

0.56197

(r – 1)(c – 1) = (2 – 1)(2 – 1) = (1)(1) = 1Degrees of Freedom:

P(2 > 0.56197) =

2 = (O – E)2

E=

Expected Counts:

Smoker Non-Smoker

Male

Female

25.484

53.516

44.516

93.484

0.56197


P(2 > 0.56197) = More than 0.25 OR: 0.4535

Example #3At a school a random sample of 20 male and 16 females were asked to classify which political party they identified with.

Democrat Republican Independent

Male 11 7 2

Female 7 8 1

Are the proportions of Democrats, Republicans, and Independents the same within both populations? Conduct a test of significance at the α = 0.05 level.

P: The proportion of Democrats, Republicans, and Independents that are Male and Female.

H:

Ho: The proportions are the same among males and females and their political party

HA: The proportions are different among males and females and their political party

A: SRS (says)

Expected Counts 5


Male 11 7 2

Female 7 8 1

Row total x Column total table total

20 x 18 36

16 x 18 36

20 x 15 36

16 x 15 36

=

=

=

=

10

8

8.33

6.67

20 x 3 36

16 x 3 36

=

=

1.67

1.33

18 15 3

20

1636

Not all are expected counts are 5, proceed with caution!

N: Chi-Square test for Homogeneity

T:

2 = (O – E)2

E= 0.855

(11 – 10)2

10+ (7 – 8)2

8


Male 11 7 2

Female 7 8 1

Male 10 8.33 1.67

Female 8 6.67 1.33

Expected

+ (7 – 8.33)2

8.33+ (8 – 6.67)2

6.67+ (2 – 1.67)2

1.67+ (1 – 1.33)2

1.33

O:


P(2 > 0.855) =

More than 0.25

OR:

0.6521

O:


P(2 > 0.855) =

P(2 > 0.855) =

M:

P ___________

0.6521 0.05

>

Accept the Null

S:

There is not enough evidence to claim the proportions are different among males and females and their political party

Example #4The following chart represents the score distribution on the AP Exams for different subjects at a certain high school. Is there evidence that the score distribution is dependent from the subject?

P: Determine if AP scores are independent from the subject areas.

H:

Ho: AP scores and AP test are independent.

HA: AP scores and AP test are not independent.

A: SRS (says)

Expected Counts 5

Row total x Column total table total

52 68 30

3442441812

150

34 x 52 150

42 x 52 150

34 x 68 150

42 x 68150

=

=

=

=

11.787

14.56

15.413

19.04

34 x 30 150

42 x 30150

=

=

6.8

8.4

52 68 30

3442441812

150

44 x 52 150

44 x 68 150

= =15.253 19.947 44 x 30 150

= 8.8

18 x 52 150

12 x 52 150

18 x 68 150

12 x 68150

=

=

=

=

6.24

4.16

8.16

5.44

18 x 30 150

12 x 30150

=

=

3.6

2.4

52 68 30

3442441812

150

Not all are expected counts are 5, proceed with caution!

N: Chi-Square test for Independence

T:

5 11.787 15.413 6.8

4 14.56 19.04 8.4

3 15.253 19.947 8.8

2 6.24 8.16 3.6

1 4.16 5.44 2.4

Expected

2 = (O – E)2

E= 5.227698

O:


P(2 > 5.227698) =

More than 0.25

OR:

0.732985

O: P(2 > 5.227698) =

P(2 > 5.227698) =


M:

P ___________

0.732985 0.05

>

Accept the Null

S:

There is not enough evidence to claim the AP scores and AP test are dependent

Chapter 14: Chi-Square Procedures. 14.1 – Test for Goodness of Fit.

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