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Introduction to ProbabilityIntroduction to Probabilityand Statisticsand StatisticsTenth EditionTenth Edition

Chapter 10Inference fromSmall Samples

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IntroductionIntroduction When the sample size is small, the

estimation and testing procedures of Chapter 8 are not appropriate.

There are equivalent small sample test andestimation procedures for

, the mean of a normal population 1 2 , the difference between two

population means 2, the variance of a normal populationThe ratio of two population variances.

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The Sampling DistributionThe Sampling Distributionof the Sample Meanof the Sample Mean

When we take a sample from a normal population, the sample meanhas a normal distribution for any sample size n, and

has a standard normal distribution. But if is unknown, and we must use s to estimate it, the resulting

statisticis not normalis not normal .

nxz

/ = normal!not is

/ nsx

x

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Students t DistributionStudents t Distribution Fortunately, this statistic does have a sampling distribution that is

well known to statisticians, called the Students t distribution,Students t distribution, with nn -1 degrees of freedom.-1 degrees of freedom.

ns

xt

/

=

We can use this distribution to createestimation testing procedures for the populationmean .

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Properties of StudentsProperties of Students t t

Shape depends on the sample size n or the

degrees of freedom,degrees of freedom, nn -1.-1. As n increases the shapes of the t and z

distributions become almost identical.

Mound-shapedMound-shaped andsymmetric about 0.More variable thanMore variable than z z ,with heavier tails

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Using theUsing the t t -Table-Table Table 4 gives the values of t that cut off certain

critical values in the tail of the t distribution. Index df df and the appropriate tail area aa to find

t t aa ,,the value of t with area a to its right.

For a random sample of size n =10, find a value of t that cutsoff .025 in the right tail.

Row = df = n 1 = 9

t .025 = 2.262

Column subscript = a = .025

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Small Sample InferenceSmall Sample Inference

for a Population Meanfor a Population Mean The basic procedures are the same as those usedfor large samples. For a test of hypothesis:

.1on withdistributi-ta

on basedregionrejection aor values- using/

statistictesttheusing

tailedor two one :H versus:HTest

0

a00

=

=

=

ndf

pns

xt

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For a 100(1 )% confidence intervalfor the population mean :

.1on withdistributi-ta of tailin the

/2 area off cutsthat of valuetheis where 2/

2/

=

ndf

t t n

st x

Small Sample InferenceSmall Sample Inferencefor a Population Meanfor a Population Mean

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ExampleExample

A sprinkler system is designed so that the averagetime for the sprinklers to activate after beingturned on is no more than 15 seconds. A test of 5

systems gave the following times:17, 31, 12, 17, 13, 25

Is the system working as specified? Test using = .05.

specified) as ng(not worki 15:H

specified) as (working 15:H

a

0

>=

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ExampleExampleFirst, calculate the sample mean and standard deviation,using your calculator or the formulas in Chapter 2.

387.75

6

1152477

1

)(

167.196115

222

=

=

=

==

=

nn

xx

s

nx

x i

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ExampleExampleCalculate the test statistic and find the rejection region for

=.05.

5161 38.16/387.7

15167.19

/

:freedom of Degrees :statisticTest

0

====

=

= ndf nsx

t

Rejection Region: Reject H 0 if t >2.015. If the test statistic falls inthe rejection region, its p-valuewill be less than = .05.

Rejection Region: Reject H 0 if t >2.015. If the test statistic falls inthe rejection region, its p-valuewill be less than = .05.

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ConclusionConclusionCompare the observed test statistic to the rejection region,

and draw conclusions.

.015.2 if HReject

:RegionRejection

38.1 :statisticTest

0 >

=

t

t

Conclusion: For our example, t = 1.38 does not fall in therejection region and H 0 is not rejected. There is insufficientevidence to indicate that the average activation time is greater than 15.

15:H 15:H

a

0

>=

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Approximating theApproximating thepp-value-value

You can only approximate the p-valuefor the test using Table 4.

Since the observed valueof t = 1.38 is smaller than t .10 = 1.476,

p-value > .10.

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The exactThe exact pp-value-value You can get the exact p-valueusing some calculators or a computer.

One-Sample T: TimesTest of mu = 15 vs > 15

95%

Lower

Variable N Mean StDev SE Mean Bound T P

Times 6 19.1667 7.3869 3.0157 13.0899 1.38 0.113

One-Sample T: TimesTest of mu = 15 vs > 15

95%

Lower

Variable N Mean StDev SE Mean Bound T P

Times 6 19.1667 7.3869 3.0157 13.0899 1.38 0.113

p-value = .113 which

is greater than .10 aswe approximatedusing Table 4.

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Testing the DifferenceTesting the Difference

between Two Meansbetween Two Means

normal. bemust spopulation twothesmall, are sizes sample theSince

.22

and21

variancesand 2

and 1

meanswith2 and 1 spopulation from

drawn are 2

and 1

size of samples randomtindependen 9,Chapter in As

nn

To test:

H0: 1 2 = D0 versus H a: one of threewhere D 0 is some hypothesized difference,usually 0.

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between Two Meansbetween Two MeansThe test statistic used in Chapter 9

does not have either a z or a t distribution, and

cannot be used for small-sample inference.We need to make one more assumption, thatthe population variances, although unknown,are equal.

2

22

1

21

21z

ns

ns

xx

+

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between Two Meansbetween Two MeansInstead of estimating each population varianceseparately, we estimate the common variancewith

2)1()1(

21

222

2112

++=

nnsnsn

s

+

=

21

2

021

11

nns

Dxxt has a t distributionwith n1+n2-2 degreesof freedom.

And the resultingtest statistic,

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Estimating the DifferenceEstimating the Difference

between Two Meansbetween Two MeansYou can also create a 100(1- )% confidenceinterval for 1- 2.

2)1()1(

with21

222

2112

++=

nnsnsn

s

+

21

22/21

11)(

nnst xx

Remember the three

assumptions:1. Original populations

normal

2. Samples random andindependent

3. Equal populationvariances.

Remember the three

assumptions:1. Original populations

normal

2. Samples random andindependent

3. Equal populationvariances.

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ExampleExample Two training procedures are compared by

measuring the time that it takes trainees toassemble a device. A different group of trainees aretaught using each method. Is there a difference in thetwo methods? Use = .01.

Time toAssemble

Method 1 Method 2

Sample size 10 12

Sample mean 35 31

Sample Std Dev 4.9 4.5

0:H 210 =

+

=

21

2

21

11

0

:statisticTest

nns

xxt

0:H 21a

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ExampleExample Solve this problem by approximating the p-

value usingTable 4.

Time toAssemble Method 1 Method 2

Sample size 10 12

Sample mean 35 31


99.1

121

101

942.21

3135

:statisticTest

=

+

=

t

942.2120

)5.4(11)9.4(92

)1()1(

:Calculate

22

21

222

2112

=+=+

+=nn

snsns

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ExampleExample

value)-(21

)99.1(

)99.1()99.1( :value-

pt P

t P t P p

=>

.025 < ( p-value) < .05

.05 < p-value < .10

Since the p-value is

greater than = .01, H 0 is not rejected. There isinsufficient evidence toindicate a difference inthe population means.

.025 < ( p-value) < .05

.05 < p-value < .10

Since the p-value isgreater than = .01, H 0 is not rejected. There isinsufficient evidence toindicate a difference inthe population means.

df = n1 + n2 2 = 10 + 12 2 = 20df = n1 + n2 2 = 10 + 12 2 = 20

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between Two Meansbetween Two MeansIf the population variances cannot be assumedequal, the test statistic

has an approximate t distribution with degreesof freedom given above. This is most easilydone by computer.

2

22

1

21

21

ns

ns

xxt

+

1)/(

1)/(

2

22

22

1

21

21

2

2

2

2

1

2

1

+

+

nns

nns

nsnsdf

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The Paired-DifferenceThe Paired-DifferenceTestTest

Sometimes the assumption of independentsamples is intentionally violated, resulting in amatched-pairsmatched-pairs or paired-difference testpaired-difference test.By designing the experiment in this way, we caneliminate unwanted variability in the experimentby analyzing only the differences,

d d ii == xx11ii xx22iito see if there is a difference in the twopopulation means, 1 2.

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ExampleExample

One Type A and one Type B tire are randomly assignedto each of the rear wheels of five cars. Compare theaverage tire wear for types A and B using a test of hypothesis.

Car 1 2 3 4 5

Type A 10.6 9.8 12.3 9.7 8.8

Type B 10.2 9.4 11.8 9.1 8.3

0:H

0:H

21a

210

=

But the samples are notindependent. The pairs of responses are linked becausemeasurements are taken on thesame car.

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The Paired-DifferenceThe Paired-Difference

TestTest

.1on withdistributi-ta

on basedregionrejection aor value- theUse. s,difference theof deviation standard andmean

theare and pairs, of number where

/0

statistictesttheusing

0 :Htestwe0:HtestTo d0210

=

=

=

==

ndf

pd

sd n

nsd t

i

d

d

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ExampleExampleCar 1 2 3 4 5

Type A 10.6 9.8 12.3 9.7 8.8

Type B 10.2 9.4 11.8 9.1 8.3

Difference .4 .4 .5 .6 .5

0:H

0:H

21a

210

=

( ).0837

.48 Calculate

=

=

==

1

22

nnd

d s

n

d d

ii

d

i8.12

5/0837.

048.

/

0

:statisticTest

=

=

=ns

d t

d

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ExampleExampleCar Car 1 2 3 4 5

Type A 10.6 9.8 12.3 9.7 8.8

Type B 10.2 9.4 11.8 9.1 8.3

Difference .4 .4 .5 .6 .5

Rejection region: Reject H0

if t > 2.776 or t < -2.776.

Conclusion: Since t = 12.8, H 0 is rejected. There is a

difference in the average tirewear for the two types of tires.

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Some NotesSome Notes

You can construct a 100(1- )% confidenceinterval for a paired experiment using

Once you have designed the experiment bypairing, you MUST analyze it as a pairedexperiment. If the experiment is not designed as apaired experiment in advance, do not use thisprocedure.

n

st d d 2/

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Inference ConcerningInference Concerning

a Population Variancea Population VarianceSometimes the primary parameter of interestis not the population mean but rather thepopulation variance 2. We choose a randomsample of size n from a normal distribution.The sample variance s2 can be used in itsstandardized form:

2

2

2)1(

sn

=

which has a Chi-Square distribution with n - 1degrees of freedom.

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a Population Variancea Population VarianceTable 5 gives both upper and lower criticalvalues of the chi-square statistic for a given df .

For example, the value of chi-square that cuts off .05 in the upper tail of thedistribution with df = 5 is 2 =11.07.

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a Population Variancea Population Variance

.1on withdistributi square-chi a

on basedregionrejection awith)1(

statistictesttheuse we

tailedor two one :H versus:HtestTo

20

22

a20

20

=

=

=

ndf

sn

2

)2/1(

22

2

2/

2 )1()1(

:interval Confidence

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ExampleExampleA cement manufacturer claims that his cementhas a compressive strength with a standarddeviation of 10 kg/cm 2 or less. A sample of n =10 measurements produced a mean and standarddeviation of 312 and 13.96, respectively.

A test of hypothesis:

H0: 2 = 10 (claim iscorrect)

Ha: 2 > 10 (claim is

wrong)

A test of hypothesis:

H0: 2 = 10 (claim iscorrect)

Ha: 2 > 10 (claim iswrong)

uses the test statistic:uses the test statistic:

5.17100

)96.13(910

)1( 22

22 === sn

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Approximating theApproximating thep-p- valuevalue

91with)5.17( :value- 2 ==> ndf P p

.025 < p-value < .05Since the p-value is lessthan = .05, H 0 is notrejected. There issufficient evidence toreject the manufacturersclaim.

.025 < p-value < .05Since the p-value is lessthan = .05, H 0 is notrejected. There is

sufficient evidence toreject the manufacturersclaim.

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Two Population VariancesTwo Population VariancesWe can make inferences about the ratio of two population variances in the form a ratio.We choose two independent random samplesof size n1 and n2 from normal distributions.If the two population variances are equal, thestatistic

22

21

ss

F =

has an F distribution with df 1 = n1 - 1 and df 2 =

n2 - 1 degrees of freedom.

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Two Population VariancesTwo Population VariancesTable 6 gives only upper critical values of theF statistic for a given pair of df 1 and df 2.

For example, the value of F that cuts off .05 in theupper tail of the

distribution with df 1 = 5and df 2 = 8 is F =3.69.

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Inference ConcerningInference ConcerningTwo Population VariancesTwo Population Variances

.1 and 1on withdistributi anon basedregionrejection awith

.variancessample twotheof larger theis where

statistictesttheuse wetailedor two one :H versus:HtestTo

2211

2

122

21

a22

210

==

=

=

ndf ndf F

sss

F

12

21

,22

21

22

21

,22

21 1

:interval Confidence

df df df df

F ss

F ss

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ExampleExample

An experimenter has performed a labexperiment using two groups of rats. He wantsto test H 0: 1 = 2, but first he wants tomake sure that the population variances areequal. Standard (2) Experimental (1)

Sample size 10 11

Sample mean 13.64 12.42


22

21a

22

210 :H versus:H

:y testPreliminar

=

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ExampleExampleStandard (2) Experimental (1)

Sample size 10 11


22

21a

2

2

2

10

:H

:H

=

36.63.28.5

:statis ticTest

2

2

22

21 ===

ss

F

We designate the sample with the larger standarddeviation as sample 1, to force the test statisticinto the upper tail of the F distribution.

We designate the sample with the larger standarddeviation as sample 1, to force the test statisticinto the upper tail of the F distribution.

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ExampleExample

22

21a

2

2

2

10

:H :H

= 36.63.28.5

:statisticTest

2

2

22

21 ===

ss

F

The rejection region is two-tailed, with = .05, but we onlyneed to find the upper critical value, which has /2 = .025 toits right.

From Table 6, with df 1=10 and df 2 = 9, we reject H 0 if F >3.96.

CONCLUSION: Reject H 0. There is sufficient evidence toindicate that the variances are unequal. Do not rely on the

assumption of equal variances for your t test!

The rejection region is two-tailed, with = .05, but we onlyneed to find the upper critical value, which has /2 = .025 toits right.

From Table 6, with df 1=10 and df 2 = 9, we reject H 0 if F >

3.96.

CONCLUSION: Reject H 0. There is sufficient evidence toindicate that the variances are unequal . Do not rely on theassumption of equal variances for your t test!

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Key ConceptsKey ConceptsI. Experimental Designs for Small SamplesI. Experimental Designs for Small Samples

1. Single random sample: The sampled populationmust be normal.2. Two independent random samples: Both sampled

populations must be normal.a. Populations have a common variance 2.b. Populations have different variances

3. Paired-difference or matched-pairs design: Thesamples are not independent.

K CK C

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Key ConceptsKey ConceptsII. Statistical Tests of SignificanceII. Statistical Tests of Significance

1. Based on the t , F , and 2

distributions2. Use the same procedure as in Chapter 93. Rejection region critical values and significance levels:based on the t, F, and 2 distributions with the appropriate

degrees of freedom4. Tests of population parameters: a single mean, thedifference between two means, a single variance, and theratio of two variances

III. Small Sample Test StatisticsIII. Small Sample Test StatisticsTo test one of the population parameters when the sample sizesare small, use the following test statistics:

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Key ConceptsKey Concepts

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