4_uji Hipotesis

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Copyright 2003 McGraw-Hill Ryerson Limited. All rights reserved.


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Define null and alternative hypothesis

and hypothesis testing

Define Type Iand Type IIerrors

Describe the five-step hypothesis testing procedure

Distinguish between a one-tailedand

a two-tailed test of

hypothesis

When you have completed this chapter, youwill be able to:


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Conductatest of hypothesisabout a

population mean

Conductatest of hypothesisabout a

population proportion

Explain the relationship betweenhypothesis

testingand confidence interval estimation

Compute the probability ofa Type II error,

and power of a test


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TerminologyHypothesis

is a statement about a population distribution such that:

Examples

Examples the mean monthly income for allsystems analysts is $3569.

the mean monthly income for allsystems analysts is $3569.

35% of all customers buying coffee

at Tim Hortons return within a week.

35% of all customers buying coffee

at Tim Hortons return within a week.

(i) it is either true or false, but never both, and

(ii) with full knowledgeof thepopulation data,it is possible to identify, with certainty,

whether it is true or false.


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Terminology

is the complementof the alternative hypothesis.

We accept the null hypothesis as the defaulthypothesis. It is not rejected unless there is

convincingsample evidence against it.

Null Hypothesis Ho

Alternative Hypothesis H1is the statementthat

we are interested in proving

.It is usually a research hypothesis.


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State the decision rule

State the decision rule

Identify the test statistic

Identify the test statistic

Do NOT reject H0

Do NOT reject H0 Reject H0 and acceptH1

Reject H0 and acceptH1

Compute the value of the test statistic

and make a decision

Computethe value of the test statisticand make a decision

Step 1

Step 1

Select the level of significance

Select the level of significanceStep 2

Step 2

Step 3

Step 3

Step 4

Step 4

Step 5

Step 5

Hypothesis Testing

Hypothesis Testing

State the null and alternate hypotheses

State the null and alternate hypotheses


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When a decision is based on analysis ofsampledata

and not the entirepopulation data, it is not possible

to make a correct decision all the time.

Our objective is to try to keep the probabilityofmaking a wrong decision

as small as possible!


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Lets look at the Canadian legal system for an analogy...Lets look at the Canadian legal system for an analogy...

1. the accused person is innocent

2. the accused person is guilty

Two hypotheses:

After hearing from both the prosecution and the defence,

a decision is made, declaring the accused either:After hearing from both the prosecution and the defence,

a decision is made, declaring the accused either:

Innocent! But do the courts alwaysmake the right

decision?

Guilty!


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Person isinnocent

Person isguilty

Person isdeclared

not

guilty

Person isdeclared

guilty

CorrectDecisionCorrectDecision

Correct

Decision

Correct

DecisionError

Error

H0: person is innocent H1: person is guilty

H0 is true

H1 is trueType II Error

Type I Error

Court Decision

Reality


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Terminology

Level of Significanceis the probability ofrejecting the nullhypothesis

when it is actually true, i.e. Type I Error

accepting the null hypothesis when it isactually false.

Type II Error


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Terminology

Test Statisticis a value, determined from sample information,

used to determine

whether or notto rejectthe null hypothesis.

Critical Value

is the dividing point betweenthe region wherethe null hypothesis isrejectedand the

region where it isnot rejected.


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TestsTests


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0Critical z

=rejection

region

1- =

acceptance

region


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0

=

rejectionregion

1- =

acceptance

region

z /2-z /2 /2 /2


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A test is one-tailedwhen the alternatehypothesis, H1, states a direction.

H1:The mean yearly commissions earned byfull-time realtors is more than $65,000. (>$65,000)

H1: The mean speedof trucks traveling on the 407 in

Ontario is less than 120 kilometres per hour. (


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5% Level of Significance =.05=.05Reject Ho when z>1.65Reject Ho when z>1.65

0

= 5%rejection

region

1- = 95%acceptance

region

1.651.65


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A test is two-tailedwhen no directionisspecifiedin the alternatehypothesis, H1

H1: The mean time Canadian families live in aparticular home is not equal to 10 years. ( 10)

H1: The average speedof trucks travelling on the

407 in Ontario is different than 120 kph.

( 120)

H1

: The percentage of repeat customers within a

week at Tim Hortons is not50%. .50

ExamplesExamples

Tests of Significance


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5% Level of Significance

Reject Ho whenz>1.96orz< -1.96Reject Ho whenz>1.96orz< -1.96

= 5%rejection

region

= 95%acceptance

region

0.0250.025

1.96 & -1.96 are called critical values1.96 & -1.96 are called critical values


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Testing for the Population Mean:

Large Sample,

Population Standard Deviation

Known


Large Sample,

Population Standard Deviation

Known

Test Statistic to be used:

n/

=

Xz


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Large Sample,

Population Standard Deviation Known


Large Sample,

Population Standard Deviation Known

The processors of eye drop medication indicate on the

label that the bottle contains 16 mlof medication.

The standard deviation of the process is 0.5 ml.

A sample of 36 bottles from the last hours

production revealed a mean weight of 16.12 mlper bottle.

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At the .05 significance levelis the process out of control?

That is, can we conclude that the mean amount per bottle

is different from 16 ml?

At the .05 significance levelis the process out of control?That is, can we conclude that the mean amount per bottle

is different from 16 ml?


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Hypothesis TestHypothesis Test

State the null and alternate hypothesesState the null and alternate hypothesesStep 1Step 1

Select the level of significanceSelect the level of significanceStep 2Step 2

Identify the test statisticIdentify the test statisticStep 3Step 3

State the decision ruleState the decision ruleStep 4Step 4

Compute the test statistic

and make a decision


and make a decision

Step 5Step 5

H0: = 16

H1: 16 = 0.05

Because we know the standarddeviation, the test statistic is Z

Reject H0 ifz > 1.96 orz< -1.96

44.1=365.0

00.1612.16 ==n

Xz

Do not reject the null hypothesis.

We cannot conclude


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Large Sample,

Population Standard Deviation Unknown


Large Sample,


Rocks Discount Store chain issues its own credit card.

Lisa, the credit manager, wants to find out if the

meanmonthly unpaid balance is

more than $400.

Should Lisa conclude that the populationmean is

greater than $400, or is it reasonable to assume that

the difference of $7 ($407-

$400) is due to chance?

A random check of 172 unpaidbalances revealed the

sample mean to be $407

and the sample standard deviation

to be $38.

The level ofsignificance is set at .05.The level ofsignificance is set at .05.


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When the sample is large, i.e. over 30, you can

use the z-distribution as your test statistic.

Remember, use the best that you have!Remember, use the best that you have!

(Just replace the sample standarddeviation for the

population standarddeviation)


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and make a decision


and make a decision

Step 5Step 5

H0: = 400

H1: > 400 = 0.05Because the sample is large,

we use the test statistic Z

Reject H0 ifz> 1.645

42.2==

n

Xz

Reject the hypothesis. H0 . Lisa can conclude

that the mean unpaid balance is greater than

=17238$

400$407$


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Small Sample,



SmallSample,


ns

Xt

/

=


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The current production rate for producing 5 amp fuses at

Neds Electric Co. is 250 per hour.


Small Sample,



SmallSample,


A new machine has been purchased and installed that, accordingto the supplier, will increasethe production rate!

A sample of 10 randomly selected hours from last month revealed

the mean hourly production on thenew machine was 256units,

with a sample standarddeviation of 6per hour.

At the .05 significance level,

can Ned conclude that the new machine is

faster?

At the .05 significance level,

can Ned conclude that the new machine is

faster?


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and make a decision


and make a decisionStep 5Step 5

H0: = 250

H1: > 250 = 0.05Because the sample is small and

is unknown, we use the t-test

Reject H0 if t> 1.833

162.3== n

Xt

Reject the hypothesis. H0 . Ned can conclude

that the new machine will increase the

= 106250256

10 -1 = 9 degrees of freedom


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A P-Valueis the probability,(assuming that the null hypothesis is true)

of finding a value of the test statistic

at least as extremeas the computed valuefor the test!

If the P-Value issmallerthan thesignificance level,

H0 is rejected.

If the P-Value is largerthan the significance level,

H0 is notrejected.


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Since P-value is smaller than of 0.05, reject H0.The

population mean is greaterthan $400.

Rocks Discount Store chainissues its own credit card.Lisa, the credit manager,wants to find out if themeanmonthly unpaid

balance is more than $400.The levelofsignificance is set at .05.

Arandom check of 172

unpaidbalances revealedthe sample mean to be $407

and the sample standarddeviation to be $38.

Should Lisaconclude that the

population mean is greaterthan $400?

Rocks Discount Store chainissues its own credit card.Lisa, the credit manager,wants to find out if themeanmonthly unpaid

balance is more than $400.The level

ofsignificance is set at .05.A

random check of 172unpaidbalances revealed

the sample mean to be $407and the sample standarddeviation to be $38.

Should Lisaconclude that the

population mean is greaterthan $400?

= 0.05

P(z 2.42) =

Previouslydetermined

.5 - .4922

= .0078

42.2==ns

Xz


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The processors of eye drop

medication indicate on thelabel

that the bottle contains 16 mlof

medication. The standard

deviation of the process is 0.5 ml.A sample of 36 bottles from last

hours production revealed a

mean weight of 16.12 mlper

bottle. At the .05 significance levelis the process out of control?

That is, can we conclude that the

mean amount per bottle

is different from 16

ml?

= 0.05 = 0.05Previouslydetermined

P-Value =2p(z |computed value|)P-Value =

2p(z |computed value|)= 2p(z |1.44|)= 2(.5 - .4251)= 2(.0749)= .1498

= 2p(z |1.44|)= 2(.5- .4251)= 2(.0749)= .1498

Since .1498 > .05, do not reject H0.

Since .1498 > .05, do not reject H0.

44.1==n

Xz


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Interpreting theWeight of Evidence against Ho

InterpretingtheWeight ofEvidence against Ho

If the P-value is less than If the P-value is less than

.10 we have some evidence thatH

o

is not true

.05 we have strongevidence thatHo is not true

.01we have very strongevidence that

Ho is not true

.001we have extremely strongevidencethat Ho is not true


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If the P-value is less thanIf the P-value is less than

.10 we have some evidence.05 we have strongevidence

.01we have very strongevidence.001we have extremely strongevidence

that Ho is not true

Since P-value is .0078Since P-value is .0078

we have

very strongevidence

to conclude that the

population mean

is greater than

we have

very strongevidenceto conclude that the

population mean

is greater than


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is the fraction orpercentage that indicates the

part of thepopulation or

sample having a

particular trait of interest

is the fractionor

percentage that indicates the

part of thepopulation or

sample having a

particular trait of interest

A Proportion

is denoted byp

is found by:

Sample Proportion

sampledNumber

samplein thesuccessesofNumber=p


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Testing a

Single Population Proportion:

Testing a

Single Population Proportion:


is the symbol for sample proportion

is the symbol for population proportionp

p

p0 represents a population proportion of interest

npp

ppz

)1(

00

0=


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In the past, 15%ofthe mail ordersolicitations for a certain charity

resulted in a financial contribution.

At the .05 significance levelcan it be concluded that the

new letter is more effective?

A new solicitation letter that has been draftedis sent to a sample of 200 people and

45 respondedwith a contribution.


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and make a decision


and make a decisionStep 5Step 5

= 0.05We will use the z-test

Reject the hypothesis. More than 15% are

responding with a pledge, therefore, the new letter ismore e ective!

H1: p > .15

H0: p = .15

Reject H0 ifz > 1.645

ppz

npp )1( =

200)15.1(.15 200

45

15.= 97.2=


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Relationship Between Hypothesis Testing

Procedure and Confidence Interval

Estimation

RelationshipBetween Hypothesis Testing

Procedure andConfidence Interval

Estimation

Case 1:Case 1:

Our decision rule can be restated as:

Do not reject H0

if 0 lies in the (1- )confidence interval

estimate of the population mean,

TEST


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0

=

rejection

region

1- =

Confidence

Interval

region

Do not reject Ho when zfallsin the confidence interval estimate


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Relationship Between

Hypothesis Testing Procedure and

Confidence Interval Estimation

RelationshipBetween

Hypothesis TestingProcedure and

Confidence IntervalEstimation

Case 2:Case 2: Lower-tailed test


Do not reject H0

if 0 is less than or equaltothe (1- )

upper confidence boundfor , computed fromthe sample data.

R l i hi B



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0

=

rejectionregion

1- =confidence

level region

Do not reject

Relationship BetweenHypothesis Testing Procedure and


RelationshipBetweenHypothesis TestingProcedure and


R l i hi B


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Hypothesis Testing Procedure and


RelationshipBetween

Hypothesis TestingProcedure and


Case 3:Case 3: Upper-tailed test


Do not reject H0 if 0 is greater than or equaltothe (1- ) lower confidence boundfor ,

computed from the sample data.


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0

=

rejection

region

1- =acceptance

region

T II E


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Level of Significanceis the probability ofrejecting the nullhypothesis

when it is actually true,i.e. Type I Error

accepting the null hypothesis when it isactually false.

Type II Error

Type II ErrorType II Error

C l l i h P b bili


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Calculating the Probability

of a Type II Error

Calculating the Probability

of a Type IIError

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A batch of 5000 light bulbs either belong

to a superior type, with a mean life of 2400

hours, orto an inferior type,

with a mean life of 2000 hours.(By default,

the bulbs will be sold as the

inferior type.)

A batch of5000 light bulbs either belong

to a superior type, with a mean life of 2400

hours, orto an inferior type,

with a mean life of 2000 hours.(By default,

the bulbs will be sold as the

inferior type.)

Suppose we select a sample of 4 bulbs.

Find the probability of a

Type II error.


Find the probability of a

Type II error.

Both bulb distributions are normal, with a

standard deviation of 300 hours. = 0.025.Both bulb distributions are normal, with a

standard deviation of300 hours. = 0.025.


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H0: = 2000H1: = 2400 = 0.025

As populations are normal, is known, we use the z-test

Reject H0 if the computed z > 1.96,

or stated

another way,If the computed value x baris greater than xu = 2000+1.96(300/n), REJECT H0 in favour of H1

Superior: =2400 Inferior: =2000 =300 =0.025

Superior: =2400 Inferior: =2000 =300 =0.025


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Suppose H0 is false and H1 is true.

i.e. the true value of is 2400,

then x bar is approximatelynormally distributedwith a mean of

2400 and a standard deviation of /n= 300/n

Suppose H0 is false and H1 is true.

i.e. the true value of is 2400,

then x bar is approximatelynormally distributedwith a mean of

2400 and a standard deviation of /n= 300/n

is the probability of not rejecting Ho

is the probability that the value ofxbar

obtained will be less than or equal to xu

The probability of a Type II Error

XuX


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Then x bar has a mean of 2400 and a

sd of 300/4 = 150


Then x bar has a mean of 2400 and a

sd of 300/

4 = 150

Xu = 2000+1.96(300/4) = 2294A1 = 0.2611,

giving us a

left tail area

of 0.24

A1 = 0.2611,

giving us a

left tail area

of 0.24

70666.04300

24002294 ===n

Xz


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The probability of a Type II error is 0.24i.e. =0.24The probability of a Type II error is 0.24i.e. =0.24


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If we decrease the value of (alpha), the value zincreases and the critical value xumoves to the right,

and therefore the value of (beta) increases.

Conversely, if we increasethe valueof (alpha), xu moves to the left, therebydecreasingthe value of (beta)

For a given value of (alpha), the value of (beta)can be decreasedby increasingthe sample size.

P f T t


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Power of a TestPower of a Test

is defined as the probability ofrejecting H0when H0 is false,or

the probability ofcorrectly identifyinga true alternative hypothesis

it is equal to(1- )In previous example, = 0.24

Therefore, the tests power is 1-0.24 = 0.76

In previous example, = 0.24Therefore, the tests power is 1-0.24 = 0.76

10 2


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This completes Chapter 10This completes Chapter 10

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4_uji Hipotesis