Final 5 Two-sample T-test

8/8/2019 Final 5 Two-sample T-test

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TWO-SAMPLE T-TEST

Test of means for two

samples when

combined sample

sizes is small and

variances are

unknown but equal


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BASICS

HYPOTHESES

Q 1=Q 2

Q 1 { Q 2

Q1

>Q2

Q 1


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BASICS

CRITICAL VALUES/REJECTIONREGION

FROM THE T-DISTRIBUTION TABLE

Tail depends on alpha and H1

Degrees of freedom: df= n1 + n2 - 2Q 1 { Q 2 t > tE/2 or t < -tE/2

Q 1 >Q 2 t > tE

Q 1


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BASICS

TEST STATISTIC

21

2121

11

)(

nnS

xxt

p

!

QQ

2

)1()1(

21

2

22

2

11

!

nn

snsnSp


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Example 2:

A random sample of15 brand C softdrinks

showed a mean content of 237 ml withstandard deviation of 2.93 ml while a sample of

10 brand P softdrinks showed a mean of 240

ml with a standard deviation of 3.12 ml. Using

.05 level, is there a difference in the meancontent of the two brands of softdrinks?


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Brand

C

Mean = 237 ml

s = 2.93 ml

n =15

Brand P

Mean = 240 ml

s = 3.12 ml

n =10

Do they Differ?


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HYPOTHESES

Ho: Q 1=Q 2 The average

content of brand Csoftdrinks does not differ

significantly from the

average content of brand

P softdrinks.

H1:Q 1 { Q 2 The average

content of brand Csoftdrinks differs

significantly from the

average content of brand

P softdrinks.

They

dont have

the same

amount!

They have

the same

amount


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Test is two-tailed

E is divided into two Degree of freedom

df= n1+ n2 2

=15+10 2= 23

.025

-tcritical 0 tcritical

.025


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Critical values:

t = 2.069 Rejection Region:

t > 2.069 or

t < -2.069-2.069 0 2.069


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TEST STATISTIC: Find Sp first

2

)1()1(

21

2

22

2

11

!

nn

snsnSp

23)12

.

3)(9

()9

3.

2)(14(

22

!pS

01.3!pS

Brand C

Mean = 237 ml s = 2.93 ml

n =15

Brand P

Mean = 240 ml

s = 3.12 ml

n =10


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TEST STATISTIC: Solve for t

Brand C

Mean = 237 ml

s = 2.93 ml

n =15

Brand P

Mean = 240 ml

s = 3.12 ml

n =10

44.2

10

1

15

101.3

0)240237(!

!t

21p

2121

n

1

n

1S

)(xxt

QQ

!


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-2.44 is less than 2.069

Reject

-2.44 2.069

0

Reject

-2.069


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CONCLUSION

Reject Ho. The data provides enough evidence to

show that the average content of brand Cdiffers significantly with the average content of

Brand P softdrinks.

We have enough reason to believe that brand C

has less content than brand P.


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Example 1:

A study is conducted among female Chinese

Olympic swimmers and those of femaleAmerican Olympic swimmers in terms of the

circumference of the upper arm, in centimeters,

while relaxed. Test the hypothesis that Chinese

swimmers have larger upper arms thanAmerican swimmers at .01 level.


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data

CHINESE AMERICAN

N=10 N=11

MEAN= 26.3 cm MEAN= 24.8 cm

S=1.8 cm S=1.6 cm

Bigger?


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HYPOTHESES

Ho: Q 1=Q 2 The average

circumference ofChinese swimmers does

not differ significantly

from the average

circumference of

American swimmers

H1: Q 1 >Q 2 The average

circumference ofChinese swimmers is

significantly greater than

the average

circumference of

American swimmers


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Test is one-tailed

E is retained Degree of freedom

df= n1+ n2 2

=10+11 2=19

T table gives 2.539

0 ?


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Critical value:

t = 2.539 Rejection Region:

t > 2.539

0 2.539


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TEST STATISTIC: Find Sp first

2

)1()1(

21

2

22

2

11

!

nn

snsnSp

483.1!pS

Chinese

Mean = 26.3 s =1.8

n =10

American

Mean = 24.8

s =1.6

n =11

21110

)6.1)(111()8.1)(110(

S

22

p

!


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TEST STATISTIC: Solve for t

019.2

11

1

10

1483.1

0)8.243.26( !

!t

21p

2121

n

1

n

1S

)(xx

t

QQ

!

Chinese

Mean = 26.3 s =1.8

n =10

American

Mean = 24.8

s =1.6

n =11


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Reject

2.019 2.539


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CONCLUSION

Do Not Reject Ho. The data does not provide

enough evidence to show that the averagecircumference ofChinese swimmers is

significantly greater than the average

circumference ofAmerican swimmer


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Estimating Q1 Q2

21

2/2121

11)(

nnstxx p s! EQQ


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Example:

The following data shows the sales of rice bowls by twoadjacent stores in XUCanteen:

Booth 1: 103 94 110 87 98

Booth 2: 97 82 123 92 175 88 118

Compute for a 95% confidence interval for the differenceof their mean sales.


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Microsoft Excel Output

2.228t Critical two-tail

1.812t Critical one-tail

-0.824t Stat10.000df

0.000Hypothesized Mean Difference

652.063Pooled Variance

7.0005.000Observations1035.90576.300Variance

110.71498.400Mean

g2g1


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Your Assignment:

Work by Pairs:

From your electricity consumption/expenditures:

compare your homes EC/E with that of a

friends/classmates.


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Your Assignment:

From your electricity consumption/expenditures:

compare your homes EC/E with that of afriends/classmates.

1) Create a 95% confidence interval on the

difference of mean consumption

2) Create a 95% confidence interval on the

difference of mean expenditures


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Requirement:

Photocopy all billing statements for each

member and insert it with yoursolutions/findings in your designated folder

Each billing statement must be marked by the name

of the student researcher (you who obtained the

statement) Short bond paper


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Main Entries

Cover page Names, Section, Date of

Submission Raw Data

Summary Statistics

Interval Estimates

Interpretation

Documents

Final 5 Two-sample T-test