Statistics for Managers Using Microsoft Excel, 5e 2008 Prentice-Hall, Inc.Chap 10-1 Statistics for...

Statistics for Managers Using Microsoft Excel, 5e © 2008 Prentice-Hall, Inc. Chap 10-1

Statistics for ManagersUsing Microsoft® Excel

5th Edition

Chapter 10Two-Sample Tests


Learning Objectives

In this chapter, you learn how to use hypothesis testing for comparing the difference between:

The means of two independent populations The means of two related populations The proportions of two independent

populations The variances of two independent

populations


Two-Sample Tests Overview

Two Sample Tests

Independent Population

Means

Means, Related

Populations

Independent Population Variances

Group 1 vs. Group 2

Same group before vs. after treatment

Variance 1 vs.Variance 2

Examples

Independent Population Proportions

Proportion 1vs. Proportion 2


Two-Sample Tests

Independent Population Means

σ1 and σ2 known

σ1 and σ2 unknown

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

The point estimate for the difference between sample means:

X1 – X2


Two-Sample TestsIndependent Populations


σ1 and σ2 known

σ1 and σ2 unknown

Different data sources 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, pooled variance t test, or separate-variance t test




σ1 and σ2 known

σ1 and σ2 unknown

Use a Z test statistic

Use S to estimate unknown σ, use a t test statistic




σ1 and σ2 known

σ1 and σ2 unknown

Assumptions:

Samples are randomly and independently drawn

population distributions are normal




σ1 and σ2 known

σ1 and σ2 unknown

When σ1 and σ2 are known and both populations are normal, the test statistic is a Z-value and the standard error of X1 – X2 is

2

22

1

21

XX nσ

nσσ

21




σ1 and σ2 known

σ1 and σ2 unknown

2

22

1

21

2121

nσ

nσ

μμXXZ

The test statistic is:



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 Independent Populations, Comparing Means


Two-Sample TestsIndependent PopulationsTwo Independent Populations, Comparing Means

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

or Z > Za/2




σ1 and σ2 known

σ1 and σ2 unknown

Assumptions: Samples are randomly and independently drawn

Populations are normally distributed

Population variances are unknown but assumed equal




σ1 and σ2 known

σ1 and σ2 unknown

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




σ1 and σ2 known

σ1 and σ2 unknown

The pooled standard deviation is:

1)n()1(n

S1nS1nS21

222

211

p



Where t has (n1 + n2 – 2) d.f., and

21

2p

2121

n1

n1S

μμXXt


1)n()1(n

S1nS1nS21

222

2112

p


σ1 and σ2 known

σ1 and σ2 unknown



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, is there a difference in average yield ( = 0.05)?



1.50211)25(1)-(21

1.161251.301211)n()1(n

S1nS1nS22

21

222

2112

p

2.040

251

2115021.1

02.533.27

n1

n1S

μμXXt

21

2p

2121




H0: μ1 - μ2 = 0 i.e. (μ1 = μ2) H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2) = 0.05 df = 21 + 25 - 2 = 44 Critical Values: t = ± 2.0154 Test Statistic: 2.040

t0 2.0154-2.0154

.025

Reject H0 Reject H0

.025

Decision: Reject H0 at α = 0.052.040

Conclusion: There is evidence of a difference in the means.


Independent PopulationsUnequal Variance If you cannot assume population variances

are equal, the pooled-variance t test is inappropriate

Instead, use a separate-variance t test, which includes the two separate sample variances in the computation of the test statistic

The computations are complicated and are best performed using Excel




σ1 and σ2 known

σ1 and σ2 unknown

2

22

1

21

21nσ

nσXX Z

The confidence interval for μ1 – μ2 is:




σ1 and σ2 known

σ1 and σ2 unknown

21

2p2-nn21

n1

n1SXX

21t

The confidence interval for μ1 – μ2 is:

Where

1)n()1(n

S1nS1nS21

222

2112

p


Two-Sample TestsRelated Populations

D = X1 - X2

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


Two-Sample TestsRelated PopulationsThe ith paired difference is Di , where

n

DD

n

1ii

Di = X1i - X2i

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

Suppose the population standard deviation ofthe difference scores, σD, is known.


Two-Sample TestsRelated PopulationsThe test statistic for the mean difference is a Z

value:

nσ

μDZD

D

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


Two-Sample TestsRelated PopulationsIf σD is unknown, you can estimate the unknown population standard deviation with a sample standard deviation:

1n

)D(DS

n

1i

2i

D



1n

)D(DS

n

1i

2i

D

nS

μDtD

D

The test statistic for D is now a t statistic:

Where t has n - 1 d.f.

and SD is:



Lower-tail test:

H0: μD 0H1: μD < 0

Upper-tail test:

H0: μD ≤ 0H1: μD > 0

Two-tail test:

H0: μD = 0H1: μD ≠ 0

/2 /2

-t -t/2t t/2

Reject H0 if t < -ta Reject H0 if t > ta Reject H0 if t < -ta/2

or t > ta/2


Two-Sample TestsRelated Populations ExampleAssume you send your salespeople to a “customer service” training workshop. Has the training made a difference in the number of complaints? You collect the following data:

Salesperson Number of Complaints Difference, Di

(2-1)Before (1) After (2)

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


Two-Sample TestsRelated Populations Example

2.4n

DD

n

1ii

5.671n

)D(DS

2i

D

Salesperson Number of Complaints Difference, Di

(2-1)Before (1) After (2)

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


Two-Sample TestsRelated Populations ExampleHas the training made a difference in the number of complaints (at the α = 0.01 level)?

H0: μD = 0H1: μD 0

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

Test Statistic:

1.6655.67/04.2

n/Sμt

D

D

D


Two-Sample TestsRelated Populations Example

Reject

- 4.604 4.604

Reject

/2

- 1.66

Decision: Do not reject H0

(t statistic is not in the reject region)

Conclusion: There is no evidence of a significant change in the number of complaints

/2


Two-Sample TestsRelated Populations The confidence interval for μD (σ known) is:

nσDZD

Where n = the sample size (number of pairs in the paired sample)


Two-Sample TestsRelated Populations The confidence interval for μD (σ unknown) is:

1n

)D(DS

n

1i

2i

D

nStD D

1n

where


Two Population Proportions

Goal: Test a hypothesis or form a confidence interval for the difference between two independent population proportions, π1 – π2

Assumptions: n1π1 5 , n1(1-π1) 5

n2π2 5 , n2(1-π2) 5

The point estimate for the difference is p1 - p2



Since you begin by assuming the null hypothesis is true, you assume π1 = π2 and pool the two sample (p) estimates.

21

21

nnXXp

The pooled estimate for the overall proportion is:

where X1 and X2 are the number of successes in samples 1 and 2



21

2121

11)1(nn

pp

ppZ

The test statistic for p1 – p2 is a Z statistic:

2

22

1

11

21

21

nX ,

nX ,

nnXXp

PPwhere



Hypothesis for Population Proportions

Lower-tail test:

H0: π1 π2

H1: π1 < π2

i.e.,

H0: π1 – π2 0H1: π1 – π2 < 0

Upper-tail test:

H0: π1 ≤ π2

H1: π1 > π2

i.e.,

H0: π1 – π2 ≤ 0H1: π1 – π2 > 0

Two-tail test:

H0: π1 = π2

H1: π1 ≠ π2

i.e.,

H0: π1 – π2 = 0H1: π1 – π2 ≠ 0



Hypothesis for Population Proportions

Lower-tail test:

H0: π1 – π2 0H1: π1 – π2 < 0

Upper-tail test:

H0: π1 – π2 ≤ 0H1: π1 – π2 > 0

Two-tail test:

H0: π1 – π2 = 0H1: π1 – π2 ≠ 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


Two Independent Population Proportions: Example 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 of 72 men, 36 indicated they would vote Yes and, in a sample of 50 women, 31 indicated they would vote Yes

Test at the .05 level of significance


Two Independent Population Proportions: Example H0: π1 – π2 = 0 (the two proportions are equal) H1: π1 – π2 ≠ 0 (there is a significant difference

between proportions) The sample proportions are:

Men: p1 = 36/72 = .50

Women: p2 = 31/50 = .62

The pooled estimate for the overall proportion is:

.54912267

50723136

nnXXp

21

21


Two Independent Population Proportions: Example

The test statistic for π1 – π2 is:

1.31

501

721.549)(1.549

0.62.50

n1

n1)p(1p

z

21

2121

pp

Critical Values = ±1.96For = .05

.025

-1.96 1.96

.025

-1.31

Decision: Do not reject H0

Conclusion: There is no evidence of a significant difference in proportions who will vote yes between men and women.

Reject H0 Reject H0


Two Independent Population Proportions

2

22

1

1121 n

)(1n

)(1 ppppZpp

The confidence interval for π1 – π2 is:


Testing Population Variances

Purpose: To determine if two independent populations have the same variability.

H0: σ12 = σ2

2

H1: σ12 ≠ σ2

2

H0: σ12 σ2

2

H1: σ12 < σ2

2

H0: σ12 ≤ σ2

2

H1: σ12 > σ2

2

Two-tail test Lower-tail test Upper-tail test



22

21

SSF

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



The F critical value is found from the F table There 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



0

FL Reject H0

Do not reject H0

H0: σ12 σ2

2

H1: σ12 < σ2

2

Reject H0 if F < FL

0

FU Reject H0Do not reject H0

H0: σ12 ≤ σ2

2

H1: σ12 > σ2

2

Reject H0 if F > FU

Lower-tail test Upper-tail test



L22

21

U22

21

FSSF

FSSF

rejection region for a two-tail test is:F 0

/2

Reject H0Do not reject H0 FU

H0: σ12 = σ2

2

H1: σ12 ≠ σ2

2

FL

/2

Two-tail test



To find the critical F values:

1. Find FU from the F table for n1 – 1 numerator and n2 – 1 denominator degrees of freedom.

*UL F

1F 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)



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 25 Mean 3.27 2.53 Std dev 1.30 1.16

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



Form the hypothesis test: H0: σ2

1 – σ22 = 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

Numerator: n2 – 1 = 25 – 1 = 24 d.f.

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

FL = 1/F.025, 24, 20 = 0.41

FU: FL:


Testing Population Variances The test statistic is:

256.116.130.1

2

2

22

21

SSF

0 /2 = .025

FU=2.33Reject H0Do not

reject H0

FL=0.41

/2 = .025

Reject H0

F

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

Conclusion: There is insufficient evidence of a difference in variances at = .05


Chapter Summary

In this chapter, we have 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

Performed separate-variance t test


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

In this chapter, we have

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