type ii error in hypo test.pdf

8/14/2019 type II Error in hypo test.pdf

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1Slide

Chapter 9, Part BHypothesis Tests

2Slide

Learning objectives

1. Able to do hypothesis test about Population Proportion

2. Calculatethe Probability of Type II Errors

3. Understand power of the test

4. Determinethe Sample Size for Hypothesis Tests About aPopulation Mean


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3Slide

n The equality part of the hypotheses always appears

in the null hypothesis.

n In general, a hypothesis test about the value of a

population proportionp must take one of the

following three forms (wherep0 is the hypothesized

value of the population proportion).

A Summary of Forms for Null and AlternativeHypotheses About a Population Proportion

One-tailed(lower tail)

One-tailed(upper tail)

Two-tailed

0 0:H p p0 0:H p p

0:aH p p> 0:aH p p>0:aH p p< 0:aH p p 5 and n(1 p) > 5

L.O.Test for pPowerSample size


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5Slide

n Rejection Rule: p Value Approach

H0: pz

Reject H0 if z < -z

Reject H0 if z < -z/2 or z >z/2

H0: p>p0

H0: p=p0

Tests About a Population Proportion

Reject H0 ifp value <

n Rejection Rule: Critical Value Approach


6Slide

n Example: National Safety Council

For a Christmas and New Year!s week, the

National Safety Council estimated that

500 people would be killed and 25,000

injured on the nation!s roads. The

NSC claimed that 50% of the

accidents would be caused by

drunk driving.

Two-Tailed Test About aPopulation Proportion



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7Slide

A sample of 120 accidents showed that

67 were caused by drunk driving. Use

these data to test the NSC!s claim with

= .05.


n Example: National Safety Council


8Slide


1. Determine the hypotheses.

2. Specify the level of significance.

3. Compute the value of the test statistic.

= .05

n p Value and Critical Value Approaches

0 : .5H p =0 : .5H p =: .5aH p : .5aH p

= = =0

(67/120) .5 1.28

.045644p

p pz

= = =0

(67/120) .5 1.28

.045644p

p pz

0 0(1 ) .5(1 .5)

.045644120pp p

n

= = =

0 0(1 ) .5(1 .5)

.045644120p

p p

n

= = =a commonerror is using

in thisformulapp



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9Slide

n pValue Approach

4. Compute thep -value.

5. Determine whether to reject H0.

Becausepvalue = .2006 > = .05, we cannot reject H0.


For z = 1.28, cumulative probability = .8997

pvalue = 2(1 .8997) = .2006


10Slide


n Critical Value Approach

5. Determine whether to reject H0.

For /2 = .05/2 = .025, z.025 = 1.96

4. Determine the critical value and rejection rule.

Reject H0 if z < -1.96 or z > 1.96

Because 1.278 > -1.96 and < 1.96, we cannot reject H0.



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11Slide

In-class exercise

n #35 (p.368) (two-tail test)

n #36 (p.368) (one-tail test)


12Slide

Statistical power Type II errorL.O.Test for pPowerSample size

n As said earlier (Part a.), the conclusion that theresearch hypothesis is truecomes from the test thatcontradict (or reject) the null hypothesis.

n When a test fails to reject a null hypothesis (i.e.,acceptsH

0), type II error () needs to be checked.

n A Type II error is acceptingH0 when it is false.

n When a test rejectH0, possible error is Type I error

that can be controlled easily.

n Article by Sawyer and Ball (1981)


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13Slide

Calculating the Probability of a Type II Errorin Hypothesis Tests About a Population Mean

1. Formulate the null and alternative hypotheses.

3. Using the rejection rule, solve for the value of thesample mean corresponding to the critical value ofthe test statistic.

2. Using the critical value approach, use the level ofsignificance to determine the critical value andthe rejection rule for the test.


14Slide

Calculating the Probability of a Type II Errorin Hypothesis Tests About a Population Mean

4. Use the results from step 3 to state the values of the

sample mean that lead to the acceptance of H0; thisdefines the acceptance region.

5. Using the sampling distribution of for a value of

satisfying the alternative hypothesis, and the acceptance

region from step 4, compute the probability that the

sample mean will be in the acceptance region. (This is

the probability of making a Type II error at the chosen

level of .)

xx



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15Slide

n Example: Metro EMS (revisited)

The EMS director wants toperform a hypothesis test, with a.05 level of significance, to determine

whether or not the service goal of 12 minutes or less isbeing achieved.

Recall that the response times fora random sample of 40 medicalemergencies were tabulated. Thesample mean is 13.25 minutes.The population standard deviationis believed to be 3.2 minutes.

Calculating the Probabilityof a Type II Error


16Slide

=

121.645

3.2 / 40

xz

=

121.645

3.2 / 40

xz

+ =

3.212 1.645 12.8323

40x

+ =

3.212 1.645 12.8323

40x

4. We will accept H0 when x < 12.8323

3. Value of the sample mean that identifiesthe rejection region:

2. Rejection rule is: Reject H0 if z > 1.645

1. Hypotheses are: H0: < 12 and Ha: > 12




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12.0001 1.645 .9500 .050012.4 0.85 .8023 .1977

12.8 0.06 .5239 .476112.8323 0.00 .5000 .500013.2 -0.73 .2327 .767313.6 -1.52 .0643 .935714.0 -2.31 .0104 .9896

12.8323

3.2 / 40z

=

12.8323

3.2 / 40z

=

Values of 1-

5. Probabilities that the sample mean will bein the acceptance region:



18Slide


n Calculating the Probability of a Type II Error

l When the true population mean is far abovethe null hypothesis value of 12, there is a low

probability that we will make a Type II error.

l When the true population mean is close tothe null hypothesis value of 12, there is a highprobability that we will make a Type II error.

Observations about the preceding table:

Example: = 12.0001, = .9500

Example: = 14.0, = .0104



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19Slide

Power of the Test

n The probability of correctly rejecting H0 when it isfalse is called the power of the test.

n For any particular value of , the power is 1 .

n We can show graphically the power associatedwith each value of ; such a graph is called apower curve. (See next slide.)


20Slide

Power Curve

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

11.5 12.0 12.5 13.0 13.5 14.0 14.5

ProbabilityofCorrectly

RejectingNullHypothesis

H0 False



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21Slide

In-class exercise

n #46 (p.374)

n #50 (p.375)


22Slide

n The specified level of significance determines the

probability of making a Type I error.

n By controlling the sample size, the probability ofmaking a Type II error is controlled.

Determining the Sample Size fora Hypothesis Test About a Population Mean



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23Slide

0

a

xx

xx

Samplingdistributionof whenH0 is trueand = 0

xx

Samplingdistributionof when

H0 is falseand a > 0

xx

Reject H0

c

c

H0: < 0Ha: > 0

=

x n

=

x nNote:



24Slide

where

z = z value providing an area of in the tail

z = z value providing an area of in the tail

= population standard deviation

0 = value of the population mean in H0a = value of the population mean used for the

Type II error

nz z

a

=+

( )

( )

2 2

02

nz z

a

=+

( )

( )

2 2

02


Note: In a two-tailed hypothesis test, use z /2 not z



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25Slide

Relationship Among , , and n

n Once two of the three values are known, the othercan be computed.

n For a given level of significance , increasing thesample size n will reduce .

n For a given sample size n, decreasing will increase, whereas increasing will decrease b.


26Slide


n Let!s assume that the director of medical

services makes the following statements about the

allowable probabilities for the Type I and Type II

errors:

"If the mean response time is = 12 minutes, I amwilling to risk an = .05 probability of rejecting H0.

"If the mean response time is 0.75 minutes over thespecification ( = 12.75), I am willing to risk a = .10probability of not rejecting H0.



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27Slide


= .05, = .10

z = 1.645, z = 1.28

0 = 12, a = 12.75

= 3.2

2 2 2 2

2 2

0

( ) (1.645 1.28) (3.2)155.75 156

( ) (12 12.75)a

z zn

+ += = =

2 2 2 2

2 2

0

( ) (1.645 1.28) (3.2)155.75 156

( ) (12 12.75)a

z zn

+ += = =


28Slide

In-class exercise

n #54 (p.379)

n #55 (p.379)



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29Slide

End of Chapter 9, Part B

type ii error in hypo test.pdf

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