Download - Chapter 10 10-1liush/ST/ch10.pdf · Chapter 10 10-1 Prof. Shuguang Liu Chapter 10 Two-Sample Tests Chap ... Goal: Test hypothesis or form a confidence interval for the difference

Chapter 10 10-1

Prof. Shuguang Liu

Chapter 10 Two-Sample Tests

Chap 10-1 Prof. Shuguang Liu

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

Chap 10-2

Prof. Shuguang Liu

Two-Sample Tests

Chap 10-3

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

Prof. Shuguang Liu

Difference Between Two Means

Chap 10-4

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

Prof. Shuguang Liu

Difference Between Two Means: Independent Samples

§  Different data sources •  Unrelated •  Independent

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

Chap 10-5


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.

DCOVA

Prof. Shuguang Liu

Hypothesis Tests for Two Population Means

Chap 10-6

Lower-tail test:

H0: µ1 ≥ µ2 H1: µ1 < µ2

i.e.,

H0: µ1 – µ2 ≥ 0 H1: µ1 – µ2 < 0

Upper-tail test:

H0: µ1 ≤ µ2 H1: µ1 > µ2

i.e.,

H0: µ1 – µ2 ≤ 0 H1: µ1 – µ2 > 0

Two-tail test:

H0: µ1 = µ2 H1: µ1 ≠ µ2

i.e.,

H0: µ1 – µ2 = 0 H1: µ1 – µ2 ≠ 0

Two Population Means, Independent Samples

DCOVA

Chapter 10 10-2

Prof. Shuguang Liu

Hypothesis tests for µ1 – µ2

Chap 10-7

Two Population Means, Independent Samples

Lower-tail test:

H0: µ1 – µ2 ≥ 0 H1: µ1 – µ2 < 0

Upper-tail test:

H0: µ1 – µ2 ≤ 0 H1: µ1 – µ2 > 0

Two-tail test:

H0: µ1 – µ2 = 0 H1: µ1 – µ2 ≠ 0

α α/2 α/2 α

-tα -tα/2 tα tα/2

Reject H0 if tSTAT < -tα Reject H0 if tSTAT > tα Reject H0 if tSTAT < -tα/2 or tSTAT > tα/2

DCOVA

Prof. Shuguang Liu

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

Chap 10-8


samples

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

Prof. Shuguang Liu

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

Chap 10-9


samples

•  The pooled variance is:

•  The test statistic is:

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

(continued)

( ) ( )1)n()1(nS1nS1nS

21

222

2112

p −+−

−+−=



( ) ( )

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−−−=

21

2p

2121STAT

n1

n1S

µµXXt

DCOVA

Prof. Shuguang Liu

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

Chap 10-10


samples

( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+±−

21

2p/221

n1

n1SXX αt

The confidence interval for µ1 – µ2 is:

Where tα/2 has d.f. = n1 + n2 – 2



DCOVA

Prof. Shuguang Liu

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 NASDAQ Number 21 25 Sample mean 3.27 2.53 Sample std dev 1.30 1.16

Chap 10-11

Assuming both populations are approximately normal with equal variances, is there a difference in mean yield (α = 0.05)?

DCOVA

Prof. Shuguang Liu

Pooled-Variance t Test Example: Calculating the Test Statistic

Chap 10-12

( ) ( ) ( ) ( )1.5021

1)25(1)-(211.161251.30121

1)n()1(n1n1n 22

21

222

2112 =

−+

−+−=

−+−

−+−=

SSSP

( ) ( ) ( )2.040

251

2115021.1

02.533.27

n1

n1S

µµXXt

21

2p

2121STAT =

⎟⎠

⎞⎜⎝

⎛+

−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+

−−−=

The test statistic is:

(continued)

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

DCOVA

Chapter 10 10-3

Prof. Shuguang Liu

Pooled-Variance t Test Example: Hypothesis Test Solution

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

Test Statistic:

Chap 10-13

Decision: Conclusion:

Reject H0 at α = 0.05

There is evidence of a difference in means.

t 0 2.0154 -2.0154

.025

Reject H0 Reject H0

.025

2.040

2.040

251

2115021.1

2.533.27tSTAT =

⎟⎠

⎞⎜⎝

⎛+

−=

DCOVA

Prof. Shuguang Liu

Excel Pooled-Variance t test Comparing NYSE & NASDAQ

Chap 10-14

Pooled-‐Variance t Test for the Difference Between Two Means (assumes equal popula=on variances)

Data Hypothesized Difference 0 Level of Significance 0.05

Popula*on 1 Sample Sample Size 21 =COUNT(DATA!$A:$A) Sample Mean 3.27 =AVERAGE(DATA!$A:$A) Sample Standard Devia=on 1.3 =STDEV(DATA!$A:$A)

Popula*on 2 Sample Sample Size 25 =COUNT(DATA!$B:$B) Sample Mean 2.53 =AVERAGE(DATA!$B:$B) Sample Standard Devia=on 1.16 =STDEV(DATA!$B:$B)

Intermediate Calcula*ons Popula=on 1 Sample Degrees of Freedom 20 =B7 -‐ 1 Popula=on 2 Sample Degrees of Freedom 24 =B11 -‐ 1 Total Degrees of Freedom 44 =B16 + B17 Pooled Variance 1.502 =((B16 * B9^2) + (B17 * B13^2)) / B18 Standard Error 0.363 =SQRT(B19 * (1/B7 + 1/B11)) Difference in Sample Means 0.74 =B8 -‐ B12 t Test Sta=s=c 2.040 =(B21 -‐ B4) / B20

Two-‐Tail Test Lower Cri=cal Value -‐2.015 =-‐TINV(B5, B18) Upper Cri=cal 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")




DCOVA

Prof. Shuguang Liu

Minitab Pooled-Variance t test Comparing NYSE & NASDAQ

Chap 10-15




Two-Sample T-Test and CI Sample N Mean StDev SE Mean 1 21 3.27 1.30 0.28 2 25 2.53 1.16 0.23 Difference = mu (1) - mu (2) Estimate for difference: 0.740 95% CI for difference: (0.009, 1.471) T-Test of difference = 0 (vs not =): T-Value = 2.04 P-Value = 0.047 DF = 44 Both use Pooled StDev = 1.2256

DCOVA

Prof. Shuguang Liu

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

( ) )471.1,009.0(3628.00154.274.0 n1

n1SXX

21

2p/221 =×±=⎟⎟

⎠

⎞⎜⎜⎝

⎛+±− αt

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

Chap 10-16

DCOVA

Prof. Shuguang Liu

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

Chap 10-17


samples

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

Prof. Shuguang Liu

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

1nnS

1nnS

nS

nS

2

2

2

22

1

2

1

21

2

2

22

1

21

−

⎟⎟⎠

⎞⎜⎜⎝

⎛

+−

⎟⎟⎠

⎞⎜⎜⎝

⎛

⎟⎟⎠

⎞⎜⎜⎝

⎛+

=ν

Chap 10-18


samples

(continued)

*



The test statistic is: ( ) ( )

2

22

1

21

2121STAT

nS

nS

µµXXt+

−−−=

tSTAT has d.f. ν =

DCOVA

Chapter 10 10-4

Prof. Shuguang Liu

Related Populations The 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

Chap 10-19

Related samples

Di = X1i - X2i

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Prof. Shuguang Liu

Related Populations The Paired Difference Test

The ith paired difference is Di , where

Chap 10-20

Related 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

Prof. Shuguang Liu

The Paired Difference Test: Finding tSTAT

§  The test statistic for µD is:

Chap 10-21

Paired samples

nSµDtD

STATD−

=

n  Where tSTAT has n - 1 d.f.

DCOVA

Prof. Shuguang Liu

The Paired Difference Test: Possible Hypotheses

Chap 10-22

Lower-tail test:

H0: µD ≥ 0 H1: µD < 0

Upper-tail test:

H0: µD ≤ 0 H1: µD > 0

Two-tail test:

H0: µD = 0 H1: µD ≠ 0

Paired Samples

α α/2 α/2 α

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

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

DCOVA

Prof. Shuguang Liu

The Paired Difference Confidence Interval

The confidence interval for µD is

Chap 10-23

Paired samples

1n

)D(DS

n

1i

2i

D −

−=∑=

nSD

2/αtD ±

where

DCOVA

Prof. Shuguang Liu

§  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:

Chap 10-24

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 = Σ Di

n

5.67

1n)D(D

S2

iD

=

−

−= ∑

= -4.2

DCOVA

Chapter 10 10-5

Prof. Shuguang Liu

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

Chap 10-25

- 4.2 D =

1.6655.67/04.2

n/Sµt

DSTAT

D −=−−

=−

=D

H0: µD = 0 H1: µD ≠ 0

Test Statistic:

t0.005 = ± 4.604 d.f. = n - 1 = 4

Reject

α/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

α/2

- 1.66 α = .01

DCOVA

Prof. Shuguang Liu

Paired t Test In Excel

Chap 10-26

Paired t Test

Data Hypothesized Mean Diff. 0 Level of Significance 0.05

Intermediate Calcula*ons Sample Size 5 =COUNT(I2:I6) Dbar -‐4.2 =AVERAGE(I2:I6) Degrees of Freedom 4 =B8 -‐ 1 SD 5.67 =STDEV(I2:I6) Standard Error 2.54 =B11/SQRT(B8) t-‐Test Sta=s=c -‐1.66 =(B9 -‐ B4)/B12

Two-‐Tail Test Lower Cri=cal Value -‐2.776 =-‐TINV(B5,B10) Upper Cri=cal 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 < 2.776 we do not reject the null hypothesis. Or Since p-value = 0.173 > 0.05 we do not reject the null hypothesis. Thus we conclude that there is Insufficient evidence to conclude there is a difference in the average number of complaints.

Prof. Shuguang Liu

Paired t Test In Minitab Yields The Same Conclusions

Chap 10-27

DCOVA Paired T-Test and CI: After, Before Paired T for After - Before N Mean StDev SE Mean After 5 2.40 2.61 1.17 Before 5 6.60 7.80 3.49 Difference 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

Prof. Shuguang Liu

Two Population Proportions

Chap 10-28

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

Prof. Shuguang Liu


Chap 10-29


21

21

nnXX

+

+=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

Prof. Shuguang Liu


Chap 10-30


( ) ( )

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−−−=

21

2121

11)1(nn

pp

ππppZSTAT

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

(continued)

2

22

1

11

21

21

nX , p

nX , p

nnXXp ==

+

+=where

DCOVA

Chapter 10 10-6

Prof. Shuguang Liu

Hypothesis Tests for Two Population Proportions

Chap 10-31


Lower-tail test:

H0: π1 ≥ π2 H1: π1 < π2

i.e.,

H0: π1 – π2 ≥ 0 H1: π1 – π2 < 0

Upper-tail test:

H0: π1 ≤ π2 H1: π1 > π2

i.e.,

H0: π1 – π2 ≤ 0 H1: π1 – π2 > 0

Two-tail test:

H0: π1 = π2 H1: π1 ≠ π2

i.e.,

H0: π1 – π2 = 0 H1: π1 – π2 ≠ 0

DCOVA

Prof. Shuguang Liu

Hypothesis Tests for Two Population Proportions

Chap 10-32


Lower-tail test:

H0: π1 – π2 ≥ 0 H1: π1 – π2 < 0

Upper-tail test:

H0: π1 – π2 ≤ 0 H1: π1 – π2 > 0

Two-tail test:

H0: π1 – π2 = 0 H1: π1 – π2 ≠ 0

α α/2 α/2 α

-zα -zα/2 zα zα/2

Reject H0 if ZSTAT < -Zα Reject H0 if ZSTAT > Zα Reject H0 if ZSTAT < -Zα/2

or ZSTAT > Zα/2

(continued)

DCOVA

Prof. Shuguang Liu

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

Chap 10-33

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Prof. Shuguang Liu

§  The hypothesis test is:

Chap 10-34

H0: π1 – π2 = 0 (the two proportions are equal) H1: π1 – π2 ≠ 0 (there is a significant difference between proportions)

n  The sample proportions are: n  Men: p1 = 36/72 = 0.50

n  Women: p2 = 35/50 = 0.70

582012271

50723536

21

21 .nnXXp ==

+

+=

+

+=

§  The pooled estimate for the overall proportion is:


(continued)

DCOVA

Prof. Shuguang Liu


Chap 10-35

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

7215821582

07050

11121

2121

.).(.

..

nn)p(p

ππppzSTAT

−=

⎟⎠

⎞⎜⎝

⎛+−

−−=

⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−−−=

Reject H0 Reject H0

Critical Values = ±1.96 For α = .05

DCOVA

Prof. Shuguang Liu

Two Proportion Test In Excel

Chap 10-36

DCOVA Z Test for Differences in Two Propor=ons

Data Hypothesized Difference 0 Level of Significance 0.05

Group 1 Number of items of interest 36 Sample Size 72

Group 2 Number of items of interest 35 Sample Size 50

Intermediate Calcula*ons Group 1 Propor=on 0.5 =B7/B8 Group 2 Propor=on 0.7 =B10/B11 Difference in Two Propor=ons -‐0.2 =B14 -‐ B15 Average Propor=on 0.582 =(B7 + B10)/(B8 + B11) Z Test Sta=s=c -‐2.20 =(B16-‐B4)/SQRT((B17*(1-‐B17))*(1/B8+1/B11))

Two-‐Tail Test Lower Cri=cal Value -‐1.96 =NORMSINV(B5/2) Upper Cri=cal 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

Chapter 10 10-7

Prof. Shuguang Liu

Two Proportion Test In Minitab Shows The Same Conclusions

Chap 10-37

DCOVA

Test and CI for Two Proportions Sample X N Sample p 1 36 72 0.500000 2 35 50 0.700000 Difference = p (1) - p (2) Estimate for difference: -0.2 95% CI for difference: (-0.371676, -0.0283244) Test for difference = 0 (vs not = 0): Z = -2.28 P-Value = 0.022

Prof. Shuguang Liu

Confidence Interval for Two Population Proportions

Chap 10-38


( )2

22

1

11/221 n

)p(1pn

)p(1pZpp −+

−±− α

The confidence interval for π1 – π2 is:

DCOVA

Prof. Shuguang Liu

Testing for the Ratio Of Two Population Variances

Chap 10-39

Tests for Two Population Variances

F test statistic

H0: σ12 = σ2

2 H1: σ1

2 ≠ σ22

H0: σ12 ≤ σ2

2 H1: σ1

2 > σ22

* 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

SS

Prof. Shuguang Liu

The F Distribution

§  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

Chap 10-40

df1 = n1 – 1 ; df2 = n2 – 1

22

21

SSFSTAT =

DCOVA

Prof. Shuguang Liu

Finding the Rejection Region

Chap 10-41

H0: σ12 = σ2

2 H1: σ1

2 ≠ σ22

H0: σ12 ≤ σ2

2 H1: σ1

2 > σ22

F 0 α

Fα Reject H0 Do not reject H0

Reject H0 if FSTAT > Fα

F 0

α/2

Reject H0 Do not reject H0 Fα/2

Reject H0 if FSTAT > Fα/2

DCOVA

Prof. Shuguang Liu

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 NASDAQ Number 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 α = 0.05 level?

Chap 10-42

DCOVA

Chapter 10 10-8

Prof. Shuguang Liu

F Test: Example Solution

§  Form the hypothesis test: (there is no difference between variances)

(there is a difference between variances)

Chap 10-43

n  Find the F critical value for α = 0.05:

n  Numerator d.f. = n1 – 1 = 21 –1 = 20

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

n  Fα/2 = F.025, 20, 24 = 2.33

DCOVA

22

211

22

210

σ : σH

σ : σH

≠

=

Prof. Shuguang Liu

F Test: Example Solution

§  The test statistic is:

Chap 10-44

0

256.116.130.1

2

2

22

21 ===SSFSTAT

α/2 = .025

F0.025=2.33 Reject H0 Do not

reject H0

H0: σ12 = σ2

2 H1: σ1

2 ≠ σ22

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

(continued)

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

F

DCOVA

Prof. Shuguang Liu

Two Variance F Test In Excel

Chap 10-45

DCOVA

F Test for Differences in Two Variables

Data Level of Significance 0.05

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

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

Intermediate Calcula*ons F Test Sta=s=c 1.256 1.255945 Popula=on 1 Sample Degrees of Freedom 20 =B6 -‐ 1 Popula=on 2 Sample Degrees of Freedom 24 =B9 -‐ 1

Two-‐Tail Test Upper Cri=cal 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 α = .05 because:

F statistic =1.256 < 2.327 Fα/2

or p-value = 0.589 > 0.05 = α.

Prof. Shuguang Liu

Two Variance F Test In Minitab Yields The Same Conclusion

Chap 10-46

DCOVA Test and CI for Two Variances Null hypothesis Sigma(1) / Sigma(2) = 1 Alternative hypothesis Sigma(1) / Sigma(2) not = 1 Significance level Alpha = 0.05 Statistics Sample N StDev Variance 1 21 1.300 1.690 2 25 1.160 1.346 Ratio of standard deviations = 1.121 Ratio of variances = 1.256 95% Confidence Intervals CI for Distribution CI for StDev Variance of Data Ratio Ratio Normal (0.735, 1.739) (0.540, 3.024) Tests Test Method DF1 DF2 Statistic P-Value F Test (normal) 20 24 1.26 0.589

Prof. Shuguang Liu

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 Chap 10-47 Prof. Shuguang Liu

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

Chap 10-48

(continued)