Chap 10-1 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-1 Chapter 10...

Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-1Chap 10-1

Chapter 10

Two-Sample Tests

Basic Business Statistics12th Edition

Chap 10-2Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-2

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 by

testing the ratio of the two variances


Two-Sample Tests

Two-Sample Tests

Population Means,

Independent Samples

Population Means, Related Samples

Population Variances

Group 1 vs. Group 2

Same group before vs. after treatment

Variance 1 vs.Variance 2

Examples:

Population Proportions

Proportion 1 vs. Proportion 2

DCOVA


Difference Between Two Means

Population means, independent

samples

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

The point estimate for the difference is

X1 – X2

*

σ1 and σ2 unknown, assumed equal

σ1 and σ2 unknown, not assumed equal

DCOVA


Difference Between Two Means: Independent Samples


samples*

Use Sp to estimate unknown σ. Use a Pooled-Variance t test.



Use S1 and S2 to estimate unknown σ1 and σ2. Use a Separate-Variance t test.

Different data sources Unrelated Independent

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

DCOVA


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

DCOVA


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

-ta -ta/2ta ta/2

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

or tSTAT > ta/2

Hypothesis tests for μ1 – μ2 DCOVA



samples

Hypothesis tests for µ1 - µ2 with σ1 and σ2 unknown and assumed equal

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

*σ1 and σ2 unknown, assumed equal


DCOVA



samples

• The pooled variance is:

• The test statistic is:

• Where tSTAT has d.f. = (n1 + n2 – 2)

(continued)

1)n()1(n

S1nS1nS

21

222

2112

p

*σ1 and σ2 unknown, assumed equal


Hypothesis tests for µ1 - µ2 with σ1 and σ2 unknown and assumed equal

21

2p

2121

STAT

n1

n1

S

μμXXt

DCOVA



samples

21

2p/221

n

1

n

1SXX t

The confidence interval for μ1 – μ2 is:

Where tα/2 has d.f. = n1 + n2 – 2

*

Confidence interval for µ1 - µ2 with σ1 and σ2 unknown and assumed equal



DCOVA


Pooled-Variance 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 meanyield ( = 0.05)?

DCOVA


Pooled-Variance t Test Example: Calculating the Test Statistic

1.5021

1)25(1)-(21

1.161251.30121

1)n()1(n

1n1n 22

21

222

2112

SS

SP

2.040

251

211

5021.1

02.533.27

n1

n1

S

μμXXt

21

2p

2121

STAT

The test statistic is:

(continued)

H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)H1: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)

DCOVA


Pooled-Variance t Test Example: Hypothesis Test 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

2.533.27tSTAT

DCOVA

Chap 10-14Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall

Excel Pooled-Variance t test Comparing NYSE & NASDAQ

Chap 10-14

Pooled-Variance t Test for the Difference Between Two Means(assumes equal population variances)

DataHypothesized Difference 0Level of Significance 0.05

Population 1 SampleSample Size 21=COUNT(DATA!$A:$A)Sample Mean 3.27=AVERAGE(DATA!$A:$A)Sample Standard Deviation 1.3=STDEV(DATA!$A:$A)

Population 2 SampleSample Size 25=COUNT(DATA!$B:$B)Sample Mean 2.53=AVERAGE(DATA!$B:$B)Sample Standard Deviation 1.16=STDEV(DATA!$B:$B)

Intermediate CalculationsPopulation 1 Sample Degrees of Freedom 20=B7 - 1Population 2 Sample Degrees of Freedom 24=B11 - 1Total Degrees of Freedom 44=B16 + B17Pooled Variance 1.502=((B16 * B9^2) + (B17 * B13^2)) / B18Standard Error 0.363=SQRT(B19 * (1/B7 + 1/B11))Difference in Sample Means 0.74=B8 - B12t Test Statistic 2.040=(B21 - B4) / B20

Two-Tail TestLower Critical Value -2.015=-TINV(B5, B18)Upper Critical Value 2.015=TINV(B5, B18)p-value 0.047=TDIST(ABS(B22),B18,2)

Reject the null hypothesis =IF(B27<B5,"Reject the null hypothesis","Do not reject the null hypothesis")

Decision:

Conclusion:



DCOVA


Minitab Pooled-Variance t test Comparing NYSE & NASDAQ

Chap 10-15

Decision:

Conclusion:



Two-Sample T-Test and CI

Sample N Mean StDev SE Mean1 21 3.27 1.30 0.282 25 2.53 1.16 0.23

Difference = mu (1) - mu (2)Estimate for difference: 0.74095% CI for difference: (0.009, 1.471)T-Test of difference = 0 (vs not =): T-Value = 2.04 P-Value = 0.047 DF = 44Both use Pooled StDev = 1.2256

DCOVA


Pooled-Variance t Test Example: Confidence Interval for µ1 - µ2

Since we rejected H0 can we be 95% confident that µNYSE > µNASDAQ?

95% Confidence Interval for µNYSE - µNASDAQ

Since 0 is less than the entire interval, we can be 95% confident that µNYSE > µNASDAQ

)471.1,009.0(3628.00154.274.0 n

1

n

1SXX

21

2p/221

t

DCOVA



samples

Hypothesis tests for µ1 - µ2 with σ1 and σ2 unknown, not assumed equal

Assumptions:

Samples are randomly and independently drawn

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

Population variances are unknown and cannot be assumed to be equal*



DCOVA



samples

(continued)

*



Hypothesis tests for µ1 - µ2 with σ1 and σ2 unknown and not assumed equal


2

22

1

21

2121

STAT

nS

nS

μμXXt

tSTAT has d.f. ν =

1n

nS

1n

nS

nS

nS

2

2

2

22

1

2

1

21

2

2

22

1

21

DCOVA


Related PopulationsThe Paired Difference Test

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

Di = X1i - X2i

DCOVA


Related PopulationsThe Paired Difference Test

The ith paired difference is Di , whereRelated samples

Di = X1i - X2i

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

DD

n

1ii

n is the number of pairs in the paired sample

1n

)D(DS

n

1i

2i

D

The sample standard deviation is SD

(continued)

DCOVA


The test statistic for μD is:Paired

samples

n

SμD

tD

STATD

Where tSTAT has n - 1 d.f.

The Paired Difference Test:Finding tSTAT

DCOVA


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

The Paired Difference Test: Possible Hypotheses

a a/2 a/2a

-ta -ta/2ta ta/2

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

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

DCOVA


The confidence interval for μD isPaired samples

1n

)D(DS

n

1i

2i

D

n

SD2/tD

where

The Paired Difference Confidence Interval

DCOVA


Assume 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:

Paired Difference Test: 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

DCOVA


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

μt

D

STATD

D

H0: μD = 0H1: μD 0

Test Statistic:

t0.005 = ± 4.604 d.f. = n - 1 = 4

Reject

/2

- 4.604 4.604

Decision: Do not reject H0

(tstat is not in the reject region)

Conclusion: There is insufficient evidence there is significant change in the number of complaints.

Paired Difference Test: Solution

Reject

/2

- 1.66 = .01

DCOVA


Paired t Test In Excel

Chap 10-26

Paired t Test

DataHypothesized Mean Diff. 0Level of Significance 0.05

Intermediate CalculationsSample Size 5=COUNT(I2:I6)Dbar -4.2=AVERAGE(I2:I6)Degrees of Freedom 4=B8 - 1SD 5.67=STDEV(I2:I6)Standard Error 2.54=B11/SQRT(B8)t-Test Statistic -1.66=(B9 - B4)/B12

Two-Tail TestLower Critical Value -2.776=-TINV(B5,B10)Upper Critical Value 2.776=TINV(B5,B10)p-value 0.173=TDIST(ABS(B13),B10,2)Do not reject the null Hypothesis

=IF(B18<B5,"Reject the null hypothesis","Do not reject the null hypothesis")

Data not shown is in column I

DCOVA

Since -2.776 < -1.66 < 2.776we do not reject the null hypothesis.

Or

Since p-value = 0.173 > 0.05 wedo not reject the null hypothesis.

Thus we conclude that there isInsufficient evidence to conclude there is a difference in the averagenumber of complaints.


Paired t Test In Minitab Yields The Same Conclusions

Chap 10-27

DCOVAPaired T-Test and CI: After, Before

Paired T for After - Before

N Mean StDev SE MeanAfter 5 2.40 2.61 1.17Before 5 6.60 7.80 3.49Difference 5 -4.20 5.67 2.54

95% CI for mean difference: (-11.25, 2.85)T-Test of mean difference = 0 (vs not = 0): T-Value = -1.66 P-Value = 0.173


Two Population Proportions

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

π1 – π2

The point estimate for the difference is

Population proportions

21 pp

DCOVA




21

21

nn

XX

p

The pooled estimate for the overall proportion is:

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

In the null hypothesis we assume the null hypothesis is true, so we assume π1

= π2 and pool the two sample estimates

DCOVA




21

2121

11)1(

nnpp

ππppZSTAT

The test statistic for π1 – π2 is a Z statistic:

(continued)

2

22

1

11

21

21

n

X , p

n

X , p

nn

XXp

where

DCOVA


Hypothesis Tests forTwo 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

DCOVA


Hypothesis Tests forTwo 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

a a/2 a/2a

-za -za/2za za/2

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

or ZSTAT > Z /2a

(continued)

DCOVA


Hypothesis Test 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 35 of 50 women indicated they would vote Yes

Test at the .05 level of significance

DCOVA


The hypothesis test is: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 = 0.50

Women: p2 = 35/50 = 0.70

5820122

71

5072

3536

21

21 .nn

XXp

The pooled estimate for the overall proportion is:


(continued)

DCOVA


The test statistic for π1 – π2 is:


(continued)

.025

-1.96 1.96

.025

-2.20

Decision: Reject H0

Conclusion: There is evidence of a difference in proportions who will vote yes between men and women.

202

501

721

5821582

07050

111

21

2121

.

).(.

..

nn)p(p

ππppzSTAT

Reject H0 Reject H0

Critical Values = ±1.96For = .05

DCOVA


Two Proportion Test In Excel

Chap 10-36

DCOVAZ Test for Differences in Two Proportions

DataHypothesized Difference 0Level of Significance 0.05

Group 1Number of items of interest 36Sample Size 72

Group 2Number of items of interest 35Sample Size 50

Intermediate CalculationsGroup 1 Proportion 0.5=B7/B8Group 2 Proportion 0.7=B10/B11Difference in Two Proportions -0.2=B14 - B15Average Proportion 0.582=(B7 + B10)/(B8 + B11)Z Test Statistic -2.20=(B16-B4)/SQRT((B17*(1-B17))*(1/B8+1/B11))

Two-Tail TestLower Critical Value -1.96=NORMSINV(B5/2)Upper Critical Value 1.96=NORMSINV(1 - B5/2)p-value 0.028=2*(1 - NORMSDIST(ABS(B18)))

Reject the null hypothesis =IF(B23 < B5,"Reject the null hypothesis","Do not reject the null hypothesis")

Decision: Reject H0

Conclusion: There is evidence of a difference in proportions who will vote yes between men and women.

Since -2.20 < -1.96

Or

Since p-value = 0.028 < 0.05

We reject the null hypothesis


Two Proportion Test In Minitab Shows The Same Conclusions

Chap 10-37

DCOVA

Test and CI for Two Proportions

Sample X N Sample p1 36 72 0.5000002 35 50 0.700000

Difference = p (1) - p (2)Estimate for difference: -0.295% CI for difference: (-0.371676, -0.0283244)Test for difference = 0 (vs not = 0): Z = -2.28 P-Value = 0.022


Confidence Interval forTwo Population Proportions


2

22

1

11/221 n

)p(1p

n

)p(1pZpp

The confidence interval for π1 – π2 is:

DCOVA


Testing for the Ratio Of Two Population Variances

Tests for TwoPopulation Variances

F test statistic

H0: σ12 = σ2

2

H1: σ12 ≠ σ2

2

H0: σ12 ≤ σ2

2

H1: σ12 > σ2

2

*Hypotheses FSTAT

= Variance of sample 1 (the larger sample variance)

n1 = sample size of sample 1

= Variance of sample 2 (the smaller sample variance)

n2 = sample size of sample 2

n1 –1 = numerator degrees of freedom

n2 – 1 = denominator degrees of freedom

Where:

DCOVA

21S

22S

22

21

S

S


The F critical value is found from the F table

There are two degrees of freedom required: numerator and denominator

The larger sample variance is always the numerator

When

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

The F Distribution

df1 = n1 – 1 ; df2 = n2 – 122

21

S

SFSTAT

DCOVA


Finding the Rejection Region

H0: σ12 = σ2

2

H1: σ12 ≠ σ2

2

H0: σ12 ≤ σ2

2

H1: σ12 > σ2

2

F 0

Fα Reject H0Do not

reject H0

Reject H0 if FSTAT > Fα

F 0

/2

Reject H0Do not reject H0 Fα/2

Reject H0 if FSTAT > Fα/2

DCOVA


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?

DCOVA


F Test: Example Solution

Form the hypothesis test:

(there is no difference between variances)

(there is a difference between variances)

Find the F critical value for = 0.05:

Numerator d.f. = n1 – 1 = 21 –1 = 20

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

Fα/2 = F.025, 20, 24 = 2.33

DCOVA

22

211

22

210

σ : σH

σ : σH



0

256.116.1

30.12

2

22

21

S

SFSTAT

/2 = .025

F0.025=2.33Reject H0Do not

reject H0

H0: σ12 = σ2

2

H1: σ12 ≠ σ2

2

F Test: Example Solution

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

(continued)

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

F

DCOVA


Two Variance F Test In Excel

Chap 10-45

DCOVA

F Test for Differences in Two Variables

DataLevel of Significance 0.05

Larger-Variance SampleSample Size 21Sample Variance 1.6900 =1.3^2

Smaller-Variance SampleSample Size 25Sample Variance 1.3456 =1.16^2

Intermediate CalculationsF Test Statistic 1.256 1.255945Population 1 Sample Degrees of Freedom 20 =B6 - 1Population 2 Sample Degrees of Freedom 24 =B9 - 1

Two-Tail TestUpper Critical Value 2.327 =FINV(B4/2,B14,B15)p-value 0.589 =2*FDIST(B13,B14,B15)

Do not reject the null hypothesis =IF(B19<B4,"Reject the null hypothesis","Do not reject the null hypothesis")

Conclusion:

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

F statistic =1.256 < 2.327 Fα/2

or

p-value = 0.589 > 0.05 = α.


Two Variance F Test In Minitab Yields The Same Conclusion

Chap 10-46

DCOVATest and CI for Two Variances

Null hypothesis Sigma(1) / Sigma(2) = 1Alternative hypothesis Sigma(1) / Sigma(2) not = 1Significance level Alpha = 0.05

StatisticsSample N StDev Variance1 21 1.300 1.6902 25 1.160 1.346

Ratio of standard deviations = 1.121Ratio of variances = 1.256

95% Confidence Intervals

CI forDistribution CI for StDev Varianceof Data Ratio RatioNormal (0.735, 1.739) (0.540, 3.024)

Tests TestMethod DF1 DF2 Statistic P-ValueF Test (normal) 20 24 1.26 0.589


Chapter Summary

Compared two independent samples Performed pooled-variance t test for the difference in

two means Performed separate-variance t test for difference in

two means Formed confidence intervals for the difference

between two means Compared two related samples (paired

samples) Performed paired t test for the mean difference Formed confidence intervals for the mean 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 test for the ratio of two population variances

(continued)

Chap 10-1 Copyright ©2012 Pearson Education, Inc. publishing as Prentice Hall Chap 10-1 Chapter 10...

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