Chap09 2 sample test

Statistics for Managers Using Microsoft Excel, 4e © 2004 Prentice-Hall, Inc. Chap 9-1

Chapter 9

Two-Sample Tests

Statistics for ManagersUsing Microsoft® Excel

4th Edition


Chapter Goals

After completing this chapter, you should be able to: Test hypotheses for the difference between two

independent population means (standard deviations known or unknown)

Test two means from related samples for the mean difference

Complete a Z test for the difference between two proportions

Use the F table to find critical F values

Complete an F test for the difference between two variances


Two Sample Tests

Two Sample Tests

Population Means,

Independent Samples

Means, Related Samples

Population Variances

Group 1 vs. independent Group 2

Same group before vs. after treatment

Variance 1 vs.Variance 2

Examples:

Population Proportions

Proportion 1 vs. Proportion 2


Difference Between Two Means

Population means, independent

samples

σ1 and σ2 known

σ1 and σ2 unknown

Goal: Test hypotheses or form a confidence interval for the difference between two population means, μ1 – μ2

The point estimate for the difference is

X1 – X2

*


Independent Samples


samples

Different data sources Unrelated Independent

Sample selected from one population has no effect on the sample selected from the other population

Use the difference between 2 sample means

Use Z test or pooled variance t test

*

σ1 and σ2 known

σ1 and σ2 unknown


Difference Between Two Means


samples

σ1 and σ2 known

σ1 and σ2 unknown

*

Use a Z test statistic

Use S to estimate unknown σ , use a t test statistic and pooled standard deviation



samples

σ1 and σ2 known

σ1 and σ2 Known

Assumptions:

Samples are randomly and independently drawn

population distributions are normal or both sample sizes are 30

Population standard deviations are known

*σ1 and σ2 unknown



samples

σ1 and σ2 known …and the standard error of

X1 – X2 is

When σ1 and σ2 are known and both populations are normal or both sample sizes are at least 30, the test statistic is a Z-value…

2

22

1

21

XX n

σ

n

σσ

21

(continued)

σ1 and σ2 Known




samples

σ1 and σ2 known

2

22

1

21

2121

nσ

nσ

μμXXZ

The test statistic for

μ1 – μ2 is:

σ1 and σ2 Known


(continued)


Hypothesis Tests forTwo Population Means

Lower-tail test:

H0: μ1 μ2

H1: μ1 < μ2

i.e.,

H0: μ1 – μ2 0H1: μ1 – μ2 < 0

Upper-tail test:

H0: μ1 ≤ μ2

H1: μ1 > μ2

i.e.,

H0: μ1 – μ2 ≤ 0H1: μ1 – μ2 > 0

Two-tail test:

H0: μ1 = μ2

H1: μ1 ≠ μ2

i.e.,

H0: μ1 – μ2 = 0H1: μ1 – μ2 ≠ 0

Two Population Means, Independent Samples


Two Population Means, Independent Samples

Lower-tail test:

H0: μ1 – μ2 0H1: μ1 – μ2 < 0

Upper-tail test:

H0: μ1 – μ2 ≤ 0H1: μ1 – μ2 > 0

Two-tail test:

H0: μ1 – μ2 = 0H1: μ1 – μ2 ≠ 0

/2 /2

-z -z/2z z/2

Reject H0 if Z < -Z Reject H0 if Z > Z Reject H0 if Z < -Z/2

or Z > Z/2

Hypothesis tests for μ1 – μ2



samples

σ1 and σ2 known

2

22

1

21

21n

σ

n

σZXX

The confidence interval for

μ1 – μ2 is:

Confidence Interval, σ1 and σ2 Known




samples

σ1 and σ2 known

σ1 and σ2 Unknown

Assumptions:

Samples are randomly and independently drawn

Populations are normally distributed or both sample sizes are at least 30

Population variances are unknown but assumed equal




samples

σ1 and σ2 known

σ1 and σ2 Unknown(continued)


Forming interval estimates:

The population variances are assumed equal, so use the two sample standard deviations and pool them to estimate σ

the test statistic is a t value with (n1 + n2 – 2) degrees of freedom



samples

σ1 and σ2 known

σ1 and σ2 Unknown

The pooled standard deviation is

(continued)

1)n()1(n

S1nS1nS

21

222

211

p




samples

σ1 and σ2 known

σ1 and σ2 Unknown

Where t has (n1 + n2 – 2) d.f.,

and

21

2p

2121

n1

n1

S

μμXXt


μ1 – μ2 is:


1)n()1(n

S1nS1nS

21

222

2112

p

(continued)



samples

σ1 and σ2 known

21

2p2-nn21

n

1

n

1StXX

21


μ1 – μ2 is:

Where

1)n()1(n

S1nS1nS

21

222

2112

p


Confidence Interval, σ1 and σ2 Unknown


Pooled Sp t Test: Example

You are a financial analyst for a brokerage firm. Is there a difference in dividend yield between stocks listed on the NYSE & NASDAQ? You collect the following data:

NYSE NASDAQNumber 21 25Sample mean 3.27 2.53Sample std dev 1.30 1.16

Assuming both populations are approximately normal with equal variances, isthere a difference in average yield ( = 0.05)?


Calculating the Test Statistic

1.5021

1)25(1)-(21

1.161251.30121

1)n()1(n

S1nS1nS

22

21

222

2112

p

2.040

251

211

5021.1

02.533.27

n1

n1

S

μμXXt

21

2p

2121

The test statistic is:


Solution

H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)

H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)

= 0.05

df = 21 + 25 - 2 = 44Critical Values: t = ± 2.0154

Test Statistic: Decision:

Conclusion:

Reject H0 at = 0.05

There is evidence of a difference in means.

t0 2.0154-2.0154

.025

Reject H0 Reject H0

.025

2.040

2.040

251

211

5021.1

2.533.27t


Related Samples

Tests Means of 2 Related Populations Paired or matched samples Repeated measures (before/after) Use difference between paired values:

Eliminates Variation Among Subjects Assumptions:

Both Populations Are Normally Distributed Or, if Not Normal, use large samples

Related samples

D = X1 - X2


Mean Difference, σD Known

The ith paired difference is Di , whereRelated samples

Di = X1i - X2i

The point estimate for the population mean paired difference is D : n

DD

n

1ii

Suppose the population standard deviation of the difference scores, σD, is known

n is the number of pairs in the paired sample


The test statistic for the mean difference is a Z value:Paired

samples

n

σμD

ZD

D

Mean Difference, σD Known(continued)

WhereμD = hypothesized mean differenceσD = population standard dev. of differencesn = the sample size (number of pairs)


Confidence Interval, σD Known

The confidence interval for D isPaired samples

n

σZD D

Where n = the sample size

(number of pairs in the paired sample)


If σD is unknown, we can estimate the unknown population standard deviation with a sample standard deviation:

Related samples

1n

)D(DS

n

1i

2i

D

The sample standard deviation is

Mean Difference, σD Unknown


The test statistic for D is now a t statistic, with n-1 d.f.:Paired

samples

1n

)D(DS

n

1i

2i

D

n

SμD

tD

D

Where t has n - 1 d.f.

and SD is:

Mean Difference, σD Unknown(continued)


The confidence interval for D isPaired samples

1n

)D(DS

n

1i

2i

D

n

StD D

1n

where

Confidence Interval, σD Unknown


Lower-tail test:

H0: μD 0H1: μD < 0

Upper-tail test:

H0: μD ≤ 0H1: μD > 0

Two-tail test:

H0: μD = 0H1: μD ≠ 0

Paired Samples

Hypothesis Testing for Mean Difference, σD Unknown

/2 /2

-t -t/2t t/2

Reject H0 if t < -t Reject H0 if t > t Reject H0 if t < -t

or t > t Where t has n - 1 d.f.


Assume you send your salespeople to a “customer service” training workshop. Is the training effective? You collect the following data:

Paired Samples Example

Number of Complaints: (2) - (1)Salesperson Before (1) After (2) Difference, Di

C.B. 6 4 - 2 T.F. 20 6 -14 M.H. 3 2 - 1 R.K. 0 0 0 M.O. 4 0 - 4 -21

D = Di

n

5.67

1n

)D(DS

2i

D

= -4.2


Has the training made a difference in the number of complaints (at the 0.01 level)?

- 4.2D =

1.6655.67/

04.2

n/S

μDt

D

D

H0: μD = 0H1: μD 0

Test Statistic:

Critical Value = ± 4.604 d.f. = n - 1 = 4

Reject

/2

- 4.604 4.604

Decision: Do not reject H0

(t stat is not in the reject region)

Conclusion: There is not a significant change in the number of complaints.

Paired Samples: Solution

Reject

/2

- 1.66 = .01


Two Population Proportions

Goal: test a hypothesis or form a confidence interval for the difference between two population proportions,

p1 – p2

The point estimate for the difference is

Population proportions

Assumptions: n1p1 5 , n1(1-p1) 5

n2p2 5 , n2(1-p2) 5

21 ss pp




21

21

nn

XXp

The pooled estimate for the overall proportion is:

where X1 and X2 are the numbers from samples 1 and 2 with the characteristic of interest

Since we begin by assuming the null hypothesis is true, we assume p1 = p2

and pool the two ps estimates




21

21ss

n1

n1

)p1(p

ppppZ 21


p1 – p2 is a Z statistic:

(continued)

2

2s

1

1s

21

21

n

Xp ,

n

Xp ,

nn

XXp

21

where


Confidence Interval forTwo Population Proportions


2

ss

1

ssss n

)p(1p

n

)p(1pZpp 2211

21


p1 – p2 is:


Hypothesis Tests forTwo Population Proportions


Lower-tail test:

H0: p1 p2

H1: p1 < p2

i.e.,

H0: p1 – p2 0H1: p1 – p2 < 0

Upper-tail test:

H0: p1 ≤ p2

H1: p1 > p2

i.e.,

H0: p1 – p2 ≤ 0H1: p1 – p2 > 0

Two-tail test:

H0: p1 = p2

H1: p1 ≠ p2

i.e.,

H0: p1 – p2 = 0H1: p1 – p2 ≠ 0


Hypothesis Tests forTwo Population Proportions


Lower-tail test:

H0: p1 – p2 0H1: p1 – p2 < 0

Upper-tail test:

H0: p1 – p2 ≤ 0H1: p1 – p2 > 0

Two-tail test:

H0: p1 – p2 = 0H1: p1 – p2 ≠ 0

/2 /2

-z -z/2z z/2

Reject H0 if Z < -Z Reject H0 if Z > Z Reject H0 if Z < -Z

or Z > Z

(continued)


Example: Two population Proportions

Is there a significant difference between the proportion of men and the proportion of women who will vote Yes on Proposition A?

In a random sample, 36 of 72 men and 31 of 50 women indicated they would vote Yes

Test at the .05 level of significance


The hypothesis test is:H0: p1 – p2 = 0 (the two proportions are equal)

H1: p1 – p2 ≠ 0 (there is a significant difference between proportions)

The sample proportions are: Men: ps1 = 36/72 = .50

Women: ps2 = 31/50 = .62

.549122

67

5072

3136

nn

XXp

21

21

The pooled estimate for the overall proportion is:


(continued)


The test statistic for p1 – p2 is:


(continued)

.025

-1.96 1.96

.025

-1.31

Decision: Do not reject H0

Conclusion: There is not significant evidence of a difference in proportions who will vote yes between men and women.

1.31

501

721

.549)(1.549

0.62.50

n1

n1

)p(1p

ppppz

21

21ss 21

Reject H0 Reject H0

Critical Values = ±1.96For = .05


Hypothesis Tests for Variances

Tests for TwoPopulation Variances

F test statistic

H0: σ12 = σ2

2

H1: σ12 ≠ σ2

2Two-tail test

Lower-tail test

Upper-tail test

H0: σ12 σ2

2

H1: σ12 < σ2

2

H0: σ12 ≤ σ2

2

H1: σ12 > σ2

2

*


Hypothesis Tests for Variances

Tests for TwoPopulation Variances

F test statistic22

21

S

SF

The F test statistic is:

= Variance of Sample 1 n1 - 1 = numerator degrees of freedom

n2 - 1 = denominator degrees of freedom = Variance of Sample 2

21S

22S

*

(continued)


The F critical value is found from the F table

The are two appropriate degrees of freedom: numerator and denominator

In the F table, numerator degrees of freedom determine the column

denominator degrees of freedom determine the row

The F Distribution

where df1 = n1 – 1 ; df2 = n2 – 122

21

S

SF


F 0

Finding the Rejection Region

L22

21

U22

21

FS

SF

FS

SF

rejection region for a two-tail test is:

FL Reject H0

Do not reject H0

F 0

FU Reject H0Do not

reject H0

F 0 /2

Reject H0Do not reject H0 FU

H0: σ12 = σ2

2

H1: σ12 ≠ σ2

2

H0: σ12 σ2

2

H1: σ12 < σ2

2

H0: σ12 ≤ σ2

2

H1: σ12 > σ2

2

FL

/2

Reject H0

Reject H0 if F < FL

Reject H0 if F > FU


Finding the Rejection Region

F 0 /2

Reject H0Do not reject H0 FU

H0: σ12 = σ2

2

H1: σ12 ≠ σ2

2

FL

/2

Reject H0

(continued)

2. Find FL using the formula:

Where FU* is from the F table with

n2 – 1 numerator and n1 – 1

denominator degrees of freedom (i.e., switch the d.f. from FU)

*UL F

1F 1. Find FU from the F table

for n1 – 1 numerator and

n2 – 1 denominator

degrees of freedom

To find the critical F values:


F Test: An Example

You are a financial analyst for a brokerage firm. You want to compare dividend yields between stocks listed on the NYSE & NASDAQ. You collect the following data: NYSE NASDAQNumber 21 25Mean 3.27 2.53Std dev 1.30 1.16

Is there a difference in the variances between the NYSE & NASDAQ at the = 0.05 level?


F Test: Example Solution

Form the hypothesis test:

H0: σ21 – σ2

2 = 0 (there is no difference between variances)

H1: σ21 – σ2

2 ≠ 0 (there is a difference between variances)

Numerator: n1 – 1 = 21 – 1 = 20 d.f.

Denominator: n2 – 1 = 25 – 1 = 24 d.f.

FU = F.025, 20, 24 = 2.33

Find the F critical values for = .05:

Numerator: n2 – 1 = 25 – 1 = 24 d.f.

Denominator: n1 – 1 = 21 – 1 = 20 d.f.

FL = 1/F.025, 24, 20 = 1/2.41

= .41

FU: FL:


The test statistic is:

0

256.116.1

30.1

S

SF

2

2

22

21

/2 = .025

FU=2.33Reject H0Do not

reject H0

H0: σ12 = σ2

2

H1: σ12 ≠ σ2

2

F Test: Example Solution

F = 1.256 is not in the rejection region, so we do not reject H0

(continued)

Conclusion: There is not sufficient evidence of a difference in variances at = .05

FL=0.41

/2 = .025

Reject H0

F


Two-Sample Tests in EXCEL

For independent samples: Independent sample Z test with variances known:

Tools | data analysis | z-test: two sample for means

For paired samples (t test): Tools | data analysis… | t-test: paired two sample for means

For variances… F test for two variances:

Tools | data analysis | F-test: two sample for variances


Two-Sample Tests in PHStat


Sample PHStat Output

Input

Output


Sample PHStat Output

Input

Output

(continued)


Chapter Summary

Compared two independent samples Performed Z test for the differences in two means Performed pooled variance t test for the differences

in two means Formed confidence intervals for the differences

between two means Compared two related samples (paired

samples) Performed paired sample Z and t tests for the mean

difference Formed confidence intervals for the paired difference


Chapter Summary

Compared two population proportions Formed confidence intervals for the difference

between two population proportions Performed Z-test for two population proportions

Performed F tests for the difference between two population variances

Used the F table to find F critical values

(continued)

Chap09 2 sample test

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