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Chapter 9: Inferences for Two –Samples Yunming Mu Department of Statistics Texas A&M University

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Chapter 9: Inferences for Two –Samples

Yunming MuDepartment of StatisticsTexas A&M University

Outline

1 Overview

2 Inferences about Two Means: Independent and Small Samples

3 Inferences about Two Means: Independent and Large Samples

4 Inferences about Two Proportions

5 Inferences about Two Means: Matched Pairs

OverviewThere are many important and meaningful

situations in which it becomes necessary to compare two sets of sample data.

Definitions

Two Samples: IndependentThe sample values selected from one population are not related or somehow paired with the sample values selected from the other population.

If the values in one sample are related to the values in the other sample, the samples are dependent. Such samples are often referred to as matched pairs or paired samples.

Example

Population of all female college students

Sample of n2 = 21 females report average of 85.7 mph

Population of all male college students

Sample of n1 = 17 males report average of 102.1 mph

Do male and female college students differ with respect to their fastest reported driving speed?

Graphical summary of sample data

75 85 95 105 115 125 135 145

Fastest Driving Speed (mph)

Gender

female

male

Numerical summary ofsample data

Gender N Mean Median TrMean StDevfemale 21 85.71 85.00 85.26 9.39male 17 102.06 100.00 101.00 17.05

Gender SE Mean Minimum Maximum Q1 Q3female 2.05 75.00 105.00 77.50 92.50male 4.14 75.00 145.00 90.00 115.00

The difference in the sample means is 102.06 - 85.71 = 16.35 mph

The Question in Statistical Notation

Let µµµµM = the average fastest speed of all male students.and µµµµF = the average fastest speed of all female students.

Then we want to know whether µµµµM ≠≠≠≠ µµµµF.

This is equivalent to knowing whether µµµµM - µµµµF ≠≠≠≠ 0

All possible questions in statistical notation

In general, we can always compare two averages by seeing how their difference compares to 0:

This comparison… is equivalent to …

µµµµ1 ≠≠≠≠ µµµµ2 µµµµ1 - µµµµ2 ≠≠≠≠ 0

µµµµ1 > µµµµ2 µµµµ1 - µµµµ2 > 0

µµµµ1 < µµµµ2 µµµµ1 - µµµµ2 < 0

Set up hypotheses

• Null hypothesis: – H0: µµµµM = µµµµF [equivalent to µM - µF = 0]

• Alternative hypothesis:– Ha: µµµµM ≠≠≠≠ µµµµF [equivalent to µM - µF ≠ 0]

Inferences about Two Means:Independent and Small Samples

Assumptions:

1. The two samples are independent.

2. Both samples are normal or the two sample sizes are small, n1 < 30 and n2 < 30

3. Both variances are unknown but equal. Assume variances are equal only if neither sample standard deviation is more than twicethat of the other sample standard deviation.

Pooled Two-Sample T Test and T Interval

Confidence IntervalsNormal Samples w/ Unknown Equal

Variance

1 2

2/ 2, 2 1 2

2 22 1 1 2 2

1 2

(1/ 1/ )

( 1) ( 1)

( 2)

n n p

p

E t S n n

n S n SS

n n

α + −= +

− + −=+ −

(x1 - x2) - E < (µ1 - µ2) < (x1 - x2) + E

where

Leaded vs Unleaded

Each of the cars selected for the EPA study was tested and the number of miles per gallon for each was obtained and recorded (Leaded=1 and Unleaded=2).

Leaded (1) Unleaded(2)

n 11 10

x 17.2 19.9

S 2.1 2

95% Confidence Interval

1 2/ 2, 2 0.025,11 10 2

2 22

0.05, 2.093

(11 1)2.1 (10 1)2.04.216

(11 10 2)

1 117.2 19.9 2.093* 4.216( )

11 10

n n

p

t t

S

αα + − + −= = =

− + −= =+ −

− ± +

(-4.58, -0.82)

Pooled Two-Sample T TestsNormal Samples w/ unknown Variance

1 2 1 2

21 2

( )

(1/ 1/ )p

X Xt

S n n

µ µ− − −=+

P-value: Use t distribution with n1+n2-2 degrees of freedom and find the P-value by following the same procedure for t tests summarized in Ch 8.

Critical values: Based on the significance level αααα, use for upper tail tests, use

for lower tail tests and use for two tailed tests.

1 2, 2n ntα + −

1 2, 2n ntα + −−1 2/ 2, 2| |n ntα + −

Leaded vs Unleaded

Claim: µµµµ1 < µµµµ2

Ho : µµµµ1 = µµµµ2

H1 : µµµµ1 < µµµµ2

αααα = 0.01

t

Fail to reject H0Reject H0

-1.729

0.05,19 1.729t− = −

Leaded vs Unleaded

Pooled Two-Sample T Test

1 2 1 2

21 2

( ) 17.2 19.9 03.01

4.216(1/11 1/10)(1/ 1/ )p

X Xt

S n n

µ µ− − − − −= = = −++

Claim: µµµµ1 < µµµµ2

Ho : µµµµ1 = µµµµ2 H1 : µµµµ1 < µµµµ2 αααα = 0.05

Leaded vs Unleaded

Claim: µµµµ1 < µµµµ2

Ho : µµµµ1 = µµµµ2

H1 : µµµµ1 < µµµµ2

αααα = 0.01

t

Fail to reject H0Reject H0

-1.729sample data:t = - 3.01

Reject Null

There is significant evidence to support the claim that the leaded

cars have a lower mean mpg than unleaded cars

P-value=0.0077(=area of red region)

Assumptions:

1. The two samples are independent.

2. Both samples are normal or the two sample sizes are small, n1 < 30 and n2 < 30

3. Both variances are unknown but unequal

Two-Sample T Test and T Interval

Confidence IntervalsNormal Samples w/ Unknown Unequal

Variance

2 21 2

/ 2,1 2

v

S SE t

n nα= +

(x1 - x2) - E < (µ1 - µ2) < (x1 - x2) + E

where

2 2 21 2 1 2

1 24 41 2 1 2

1 2

[( ) ( ) ], ,

( ) ( )1 1

se se S Sv se se

se se n nn n

+= = =+

− −

(round v down to the nearest integer)

Unpooled Two Sample T-TestNormal Samples w/ Known Variance

1 2 1 2

2 22 1 1 2

( )

/ /

X Xt

S n S n

µ µ− − −=+

P-value: Use t distribution with v degrees of freedom and find the P-value by following the same procedure for t tests summarized in Ch 8.

Critical values: Based on the significance level αααα, use for upper tail tests, use

for lower tail tests and use for two tailed tests.

, vt α

,vtα−/ 2,| |vtα

Example

We compare the density of two different types of brick. Assuming normality of the two densities distributions and unequal unknown variances, test if there is a difference in the mean densities of two different types of brick.

Type I brick Type 2 brick

n 6 5

x 22.73 21.95

S 0.10 0.24

1x

Unpooled Two-Sample T-Test

1 2 1 2

2 2 2 21 1 2 2

0.025,6

( ) 22.73 21.95 06.792

/ / 0.1 / 6 0.24 / 5

6,| | | 2.446 |

X Xt

S n S n

v t

µ µ− − − − −= = =+ +

≈ =

Ho : µµµµ1 = µµµµ2 H1 : µµµµ1 ≠≠≠≠ µµµµ2 αααα = 0.05

P-Value = 0.001; Reject the null and conclude that there is significant difference in the mean densities of the two types of brick

Two-sample t-test in Minitab

• Select Stat. Select Basic Statistics. • Select 2-sample t to get a Pop-Up window.• Click on the radio button before Samples in one

Column. Put the measurement variable in Samples box, and put the grouping variable in Subscripts box.

• Specify your alternative hypothesis.• If appropriate, select Assume Equal Variances.• Select OK.

Pooled two-sample t-test

Two sample T for Fastest

Gender N Mean StDev SE Meanfemale 21 85.71 9.39 2.0male 17 102.1 17.1 4.1

95% CI for mu (female) - mu (male ): ( -25.2, -7.5)T-Test mu (female) = mu (male ) (vs not =): T = -3.75

P = 0.0006DF = 36

Both use Pooled StDev = 13.4

(Unpooled) two-sample t-test

Two sample T for Fastest

Gender N Mean StDev SE Meanfemale 21 85.71 9.39 2.0male 17 102.1 17.1 4.1

95% CI for mu (female) - mu (male ): ( -25.9, -6.8)T-Test mu (female) = mu (male ) (vs not =): T = -3.54

P = 0.0017DF = 23