Two-Sample Tests

Chap 9-1

Two-Sample Tests

Chap 9-2

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

Chap 9-3

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

*

Chap 9-4

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

Chap 9-5

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

Chap 9-6


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

Chap 9-7


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


Chap 9-8


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)

Chap 9-9

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

Chap 9-10

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

a a/2 a/2a

-za -za/2za za/2

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

or Z > Za/2

Hypothesis tests for μ1 – μ2

Chap 9-11


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


Chap 9-12


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


Chap 9-13


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

Chap 9-14


samples

σ1 and σ2 known

σ1 and σ2 Unknown

The pooled standard deviation is

(continued)

1)n()1(n

S1nS1nS

21

222

211

p


Chap 9-15


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

The test statistic for μ1 – μ2 is:


1)n()1(n

S1nS1nS

21

222

2112

p

(continued)

Chap 9-16


samples

σ1 and σ2 known

21

2p2-nn21

n

1

n

1StXX

21

The confidence interval for μ1 – μ2 is:

Where

1)n()1(n

S1nS1nS

21

222

2112

p


Confidence Interval, σ1 and σ2 Unknown

Chap 9-17

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)?

Chap 9-18

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:

Chap 9-19

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 a = 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

Chap 9-20

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

Chap 9-21

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

Chap 9-22

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)

Chap 9-23

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)

Chap 9-24

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

Chap 9-25

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)

Chap 9-26

The confidence interval for D isPaired samples

1n

)D(DS

n

1i

2i

D

n

StD D

1n

where

Confidence Interval, σD Unknown

Chap 9-27

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

a a/2 a/2a

-ta -ta/2ta ta/2

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

or t > t /2a Where t has n - 1 d.f.

Chap 9-28

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

Chap 9-29

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

Chap 9-30

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

Chap 9-31



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

Chap 9-32



21

21ss

n1

n1

)p1(p

ppppZ 21

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

(continued)

2

2s

1

1s

21

21

n

Xp ,

n

Xp ,

nn

XXp

21

where

Chap 9-33

Confidence Interval forTwo Population Proportions


2

ss

1

ssss n

)p(1p

n

)p(1pZpp 2211

21

The confidence interval for p1 – p2 is:

Chap 9-34

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

Chap 9-35

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

a a/2 a/2a

-za -za/2za za/2

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

or Z > Z /2a

(continued)

Chap 9-36

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

Chap 9-37

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)

Chap 9-38

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

Chap 9-39

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

*

Chap 9-40

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)

Chap 9-41

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

Chap 9-42

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

Chap 9-43

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:

Chap 9-44

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?

Chap 9-45

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:

Chap 9-46

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

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Two-Sample Tests