Lesson 06 chapter 9 two samples test and Chapter 11 chi square test

Inductive StatisticsInductive Statistics

Dr. Ning DING

[email protected]

I.007 IBS, Hanze

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Table of Contents

Review:

Chapter 9: Testing Hypotheses: Two-Sample Tests~Basics~Independent Sample Test

~Large Samples~Small Samples

~Dependent Sample Test

~Differences between Proportions~Two-tailed Test~One-tailed Test

Chapter 11: Chi-Square ~Basics~Contingency Table 2-row~Contingency Table 3-row

Chapter 9 Testing Hypotheses: Two-Sample Tests: Basics-Independent

σ is known:σ is known:

σ is unknown:σ is unknown:

H0H1

Two-tailedtest

One-tailedtest

n <30 & σ is unknownn <30 & σ is unknown

Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice

Ch 9 No. 9-9 P.466

Step 1: Formulate hypothesesStep 1: Formulate hypotheses

9-99-9

Step 2: Find the Pooled Estimate of σ2Step 2: Find the Pooled Estimate of σ2

Step 3: Calculate the standard errorStep 3: Calculate the standard error

Step 4: Visualize and Find the t scoresStep 4: Visualize and Find the t scores

Gender Mean Standard Deviation

Sample size

Female 12.8 1.0667 10Male 11.625 1.4107 8

One-tailed Test

df = 16 area=0.10 t=1.746t=1.746

Review:Chapter 5

Chapter 6 Chapter 7

Chapter 8 Testing Hypothesis~Test for Mean* when σ is known* when σ is unknown AND n=<30~Test for Proportion

Chapter 9: Testing Hypotheses: Two-Sample Tests

~Basics

~Independent Sample Test~Large Samples~Small Samples


Chapter 9 Testing Hypotheses: Two-Sample Tests: Dependent Samples

9.2 Dependent SamplesReview:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test

~Differences between Proportions

~Two-tailed Test

~One-tailed Test

Chapter 11: Chi-Square

~Basics~Contingency

Table 2-row~Contingency

Table 3-row


9.2 Dependent Samples

Ch 9 Example P.468

Will the participant lose more than 17 pounds after the weight-reducing program? The survey data is:

Step 1: Formulate Hypotheses Step 1: Formulate Hypotheses

One-tailed Test

Example:Example:Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row



Ch 9 Example P.468

Step 2: Calculate the estimated standard deviation of the population differenceStep 2: Calculate the estimated standard deviation of the population difference

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row



Ch 9 Example P.468

Step 3: Find the Standard Error of the population differenceStep 3: Find the Standard Error of the population difference

Step 4: Calculate the t valueStep 4: Calculate the t value

Step 5: Visualize and get the t valuesStep 5: Visualize and get the t values

df = 10-1=9 area = 0.10

t=1.833t=1.833

One-tailed Test

reject H0

significant difference

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row

Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice

Ch 9 No. 9-15 P.474

Step 4: Visalize and Calculate the t valuesStep 4: Visalize and Calculate the t values

t=1.895t=1.895

9-159-15

Step 3: Find the Standard Error of the population differenceStep 3: Find the Standard Error of the population difference

Step 1: Formulate Hypotheses Step 1: Formulate Hypotheses

Step 2: Calculate the estimated standard deviation of the population differenceStep 2: Calculate the estimated standard deviation of the population difference

One-tailed Test

df=7 area=0.10 reject H0

sig difference

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row

Chapter 9 Testing Hypotheses: Two-Sample Tests: Proportion

Example:Example:

Ch 9 Example P.476

You are testing whether the two drugs cause different blood-pressure levels. The data is as below:

Step 1: Formulate HypothesesStep 1: Formulate Hypotheses

Step 3: Calculate the Standard Error of Proportion DifferenceStep 3: Calculate the Standard Error of Proportion Difference

Step 2: Calculate the Estimated Proportion DifferenceStep 2: Calculate the Estimated Proportion Difference

Two-tailed Test

Two-tailed Test

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row

Step 4: Visualize and get the z valuesStep 4: Visualize and get the z values

-1.96 +1.96

Accept H0

No significant difference


Ch 9 Example P.476

Two-tailed Test

Two-tailed Test

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row





One-tailed Test

One-tailed Test

Example:Example:

Ch 9 Example P.480

You are testing whether personal-appearance method causes fewer tax mistakes than mail method. The data is as below:

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row


One-tailed α=0.15 P=0.35 z= -1.04

Accept H0

No significant difference


Ch 9 Example P.480

One-tailed Test

One-tailed Test

-1.04

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row

Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice9-209-20

Ch 9 No. 9-20 P.483





z= -1.28z= -1.28

reject H0 sig difference

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row

Chapter 9 Testing Hypotheses: Two-Sample Tests: Practice9-219-21




Ch 9 No. 9-21 P.483


accept H0 no sig difference

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row

Chapter 11 Chi-Square Test-Basics

Contingency Table (Cross break table)

Male Female TotalJunior high school 40 60 100Senior high school 60 40 100Total 100 100 200

Rows * Columns == 2*2 tableRows * Columns == 2*2 table

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row




Rows * Columns == 2*2 tableRows * Columns == 2*2 table

Male Female TotalJunior high school 40(50) 60(50) 100Senior high school 60(50) 40(50) 100Total 100 100 200

Expected and Observed ValuesExpected and Observed Values

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row



1*1=11*1=1


Social Science

Nature Science

Sports

Junior high school 40 60 20Senior high school 60 40 40University 50 30 120

2*2=42*2=4

Degree of freedom= (row-1)*(column-1)Degree of freedom= (row-1)*(column-1) 2*2 table2*2 table

3*3 table3*3 table

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row

Chapter 11 Chi-Square Test–Calculate χ2

for 2-row Table

Step 1: Calculate the expected valuesStep 1: Calculate the expected values

Ch 11 Example P.570

Example:Example:Our employees’ attitude toward job-performance reviews. There are two review methods, the present one or the new one. Is the attitude dependent on geography? The survey looks like below:


for 2-row TableStep 2: Calculate the χ2Step 2: Calculate the χ2

2.7644

Ch 11 Example P.570

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row

Step 3: Find the critical χ2Step 3: Find the critical χ2


for 2-row Table

α = 0.10 2*4 df= 3 χ2=6.251

10%Acceptance

Region

χ2=2.7644

Accept H0

No sig difference among groups

Ch 11 Example P.570

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row


for 3-row TableFor a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below:

Ch 11 Example P.575

Step 1: Formulate the hypothesesStep 1: Formulate the hypotheses

H0: length of stay and insurance types are independent

H1: length of stay depends on insurance types

α=0.01

Days in HospitalCost Cover <5 5-10 >10 Total<25% 40 75 65 18025-50% 30 45 75 150>50% 40 100 190 330Total 110 220 330 660

Example:Example:

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row


for 3-row Table


For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below: Example:Example:

Ch 11 Example P.575

Step 2: Calculate the Expected Frequency For Any CellStep 2: Calculate the Expected Frequency For Any Cell

RT

CT

(30) (60)

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row


for 3-row Table

Ch 11 Example P.575

Step 3: Calculate the Chi-SquareStep 3: Calculate the Chi-Square

Days in Hospital

Cost Cover<5 5-10 >10 Total

<25% 40(30) 75(60) 65(90) 18025-50% 30(25) 45(50) 75(75) 150>50% 40(55) 100(110) 190(165) 330

Total 110 220 330 660

Example:Example:

For a national health insurance program, you believes that lengths of stays in hospitals are dependent on the types of health insurance that people have. The random data from the survey is as below:

3.33

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row


for 3-row Table

Ch 11 Example P.575


Days in Hospital

Cost Cover<5 5-10 >10 Total

<25% 40(30) 75(60) 65(90) 18025-50% 30(25) 45(50) 75(75) 150>50% 40(55) 100(110) 190(165) 330

Total 110 220 330 660

Example:Example:


3.33 3.75 6.94

1.00

4.09

0.500.91

0.003.79

=3.33+3.75+6.94+1.00+0.50+0.00+4.09+0.91+3.79=24.32

Chi-Square χ2 =24.32Chi-Square χ2 =24.32

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row


for 3-row Table

Ch 11 Example P.575

Step 4: Find the critical Chi-SquareStep 4: Find the critical Chi-Square




Example:Example:

1%

Acceptance Region

α = 0.01 3*3 df= 4

χ2 =13.277 Reject H0

dependent on each other

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row


for 3-row Table: Practice

Ch 11 No. 11-9/10 P.582

11-9/1011-9/10 Economy Weekly Chip SalesHigh Medium Low Total

At peak 20 7 3 30At Through 30 40 30 100Rising 20 8 2 30Falling 30 5 5 40Total 100 60 40 200





Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row



Ch 11 No. 11-9/10 P.582




H0: Sales and economy are independent

H1: sales depends on economy

α=0.10

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row



Ch 11 No. 11-9/10 P.582




Economy Weekly Chip SalesHigh Medium Low Total

At peak 20 (15) 7 (9) 3 (6) 30

At Through 30 (50) 40 (30) 30 (20) 100Rising 20 (15) 8 (9) 2 (6) 30Falling 30 (20) 5 (12) 5 (8) 40Total 100 60 40 200

Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row



Ch 11 No. 11-9/10 P.582

11-9/1011-9/10


= 34.60

10%

Acceptance Region

α = 0.10 4*3df= 6

χ2 =10.645


Reject H0

dependent on each other


At peak 20 (15) 7 (9) 3 (6) 30

At Through 30 (50) 40 (30) 30 (20) 100Rising 20 (15) 8 (9) 2 (6) 30Falling 30 (20) 5 (12) 5 (8) 40Total 100 60 40 200


At peak 1.67 0.44 1.50 30At Through 8.00 3.33 5.00 100Rising 1.67 0.11 2.67 30Falling 5.00 4.08 1.13 40Total 100 60 40 200


Review:


~Basics~Independent

Sample Test~Large

Samples~Small

Samples~Dependent

Sample Test


~Two-tailed Test

~One-tailed Test


~Basics~Contingency


Table 3-row

Table of Contents

Review:Chapter 8 Testing Hypothesis

Chapter 9: Testing Hypotheses: Two-Sample Tests~Basics~Independent Sample Test

~Large Samples~Small Samples


~Differences between Proportions~Two-tailed Test~One-tailed Test

Chapter 11: Chi-Square ~Basics~Contingency Table 2-row~Contingency Table 3-row

The Normal DistributionSPSS Tips

The data can be downloaded from:

Blackboard – Inductive Statsitics STA2—SPSS--Week 6 Chi-Square.sav

The Normal DistributionSPSS TipsAPPLE shop in Groningen wants to know whether our IBS students have an intention to buy an iPad and whether their interest depends on their nationalities. They have interviewed 64 Year TWO students and the data can be downloaded from Blackboard—STA2—SPSS –Chi square.sav.

The Normal DistributionSPSS TipsStep 1: Analyze Descriptive Statistics Crosstabs…

Step 1: Analyze Descriptive Statistics Crosstabs…


Step 2: Move the variable to “Rows” and “Columns” respectively.

Click on “Statistics” and choose: Chi-square.

Step 2: Move the variable to “Rows” and “Columns” respectively.

Click on “Statistics” and choose: Chi-square.

The Normal DistributionSPSS Tips Step 3:Click on “Cells” and choose: Observed and Expected.

Step 3:Click on “Cells” and choose: Observed and Expected.


Now you get the output!But how to interpret it?Now you get the output!But how to interpret it?


• In our example, the cross break table (Table 1) has shown the observed and expected values of the frequency concerning those who are willing to buy an iPad.

Table 1: Nationality and Intention to buy an iPad


Now you get the output!But how to interpret it?Now you get the output!But how to interpret it?

importantimportant


• We can reject at the .05 level the null hypothesis that population proportions are equal across the three categories, X2(3, N =64) = 0.645, p=.886.

Business

Lesson 06 chapter 9 two samples test and Chapter 11 chi square test