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©2006 Thomson/South-Western 1
Chapter 10 –Chapter 10 –
Hypothesis Hypothesis Testing for the Testing for the Mean of a Mean of a PopulationPopulation
Slides prepared by Jeff HeylLincoln University
©2006 Thomson/South-Western
Concise Managerial StatisticsConcise Managerial Statistics
KVANLIPAVURKEELING
KVANLIPAVURKEELING
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©2006 Thomson/South-Western 2
Hypothesis Testing on the Hypothesis Testing on the MeanMean
Null hypothesis Null hypothesis ((HHoo)): A statement : A statement (equality or inequality) concerning a (equality or inequality) concerning a population parameter; the researcher population parameter; the researcher wishes to discredit this statement.wishes to discredit this statement.
Alternative hypothesis Alternative hypothesis ((HHaa)): A : A statement in contradiction to the null statement in contradiction to the null hypothesis; the researcher wishes to hypothesis; the researcher wishes to support this statement.support this statement.
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©2006 Thomson/South-Western 3
Type I and Type II ErrorsType I and Type II Errors
= probability of rejecting the Ho when Ho is true (Type I error)
= probability of failing to rejecting the Ho when Ho is false (Type II error)
Actual Situation
Conclusion Ho True Ho False
Fail to Reject Ho Correct decision Type II errorType II error
Reject Ho Type I errorType I error Correct decision
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©2006 Thomson/South-Western 4
Hypothesis Testing ProcessHypothesis Testing Process
Determine the HDetermine the Hoo and H and Haa
Determine the significance levelDetermine the significance level
Compare the sample mean (variance) Compare the sample mean (variance) to the hypothesized mean (variance)to the hypothesized mean (variance)
Decide whether to fail to reject or Decide whether to fail to reject or reject Hreject Hoo
Determine what the decision means in Determine what the decision means in reference to the problemreference to the problem
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©2006 Thomson/South-Western 5
Height ExampleHeight Example
HHoo: : = 5.9 = 5.9
HHaa: : ≠≠ 5.9 5.9
= .05 == .05 = P(rejecting H P(rejecting Hoo when H when Hoo is true) is true) critical value critical value = ± 1.96= ± 1.96
RejectReject HHoo if if > 1.96 > 1.96XX - 5.9 - 5.9
/ / nnXX - 5.9 - 5.9
/ / nn
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Height ExampleHeight Example
Figure 10.1
Z-k 0 k
|Z| > k
.025 .025
Area = .5 - .025 = .475
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©2006 Thomson/South-Western 7
Height ExampleHeight Example
Figure 10.2
X5.79’ µx = 5.9’
distance = .11’ = 2.38 x .046
x ≈ = .046’.4 75
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©2006 Thomson/South-Western 8
Computed ValueComputed Value
BecauseBecause - 2.38 < - 1.96, - 2.38 < - 1.96, we reject Hwe reject Hoo thus wethus we
conclude that the average population male conclude that the average population male height is not equal toheight is not equal to 5.9 5.9
ZZ = = = -2.38 = = = = -2.38 = ZZ **XX - 5.9 - 5.9
/ / nn5.79 - 5.95.79 - 5.9
.4 / 75.4 / 75
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Height ExampleHeight Example
Figure 10.3
Z-k 0
Area = .01
k
Area = .5 - .005 = .495
.005
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©2006 Thomson/South-Western 10
Height ExampleHeight Example
Figure 10.4
Z * = -2.38
Z-2.575 2.5750
Area = .01
Reject Ho if Z * falls here
Reject Ho if Z * falls here
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©2006 Thomson/South-Western 11
Hypothesis TestingHypothesis Testing5 Step Procedure5 Step Procedure
3.3. Define the rejection region Define the rejection region 4.4. Calculate the test statisticCalculate the test statistic5.5. Give a conclusion in terms of the problemGive a conclusion in terms of the problem
1.1. Set up the null and alternative hypothesisSet up the null and alternative hypothesis
ZZ = =XX - µ - µoo
/ / nn
2.2. Define the test statisticDefine the test statistic
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©2006 Thomson/South-Western 12
Everglo Light ExampleEverglo Light Example
1.1. Define the hypotheses Define the hypotheses
HHoo: µ = 400: µ = 400 H Haa: µ ≠ 400: µ ≠ 400
3.3. Define the rejection region Define the rejection region
reject Hreject Hoo if Z if Z > 1.645> 1.645 or Z or Z < -1.645< -1.645
2.2. Define the test statisticDefine the test statistic
ZZ = = XX - 400 - 400
/ / nn
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©2006 Thomson/South-Western 13
Everglo Light ExampleEverglo Light Example
Figure 10.5
Z-1.645 1.6450
Area = = .1
Area = .5 - .05 = .45
2
= .05 2
= .05
Z * = 2.5
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©2006 Thomson/South-Western 14
Everglo Light ExampleEverglo Light Example
5.5. State the conclusion State the conclusion
There is sufficient evidence to conclude There is sufficient evidence to conclude that the average lifetime of Everglo bulbs that the average lifetime of Everglo bulbs is not 400 hoursis not 400 hours
4.4. Calculate the value of the test statisticCalculate the value of the test statistic
ZZ ** ≈ = = 2.5 ≈ = = 2.5202088
420 - 400420 - 40040 / 2540 / 25
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©2006 Thomson/South-Western 15
Confidence Intervals and Confidence Intervals and Hypothesis TestingHypothesis Testing
When testing HWhen testing Hoo: µ = µ: µ = µoo versus H versus Haa: µ ≠ µ: µ ≠ µoo using the five-step procedure and a using the five-step procedure and a significance level, significance level, , H, Hoo will be rejected if will be rejected if and only if µand only if µoo lies outside the lies outside the (1 - (1 - ) • 100) • 100 confidence interval for confidence interval for µµ
X X -- k to X k to X + + kk nn
nn
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©2006 Thomson/South-Western 16
Power of a Statistical TestPower of a Statistical Test
== P(fail to reject H P(fail to reject Hoo when H when Hoo is false) is false)
1- 1- = = P(rejecting H P(rejecting Hoo when H when Hoo is false) is false)
1- 1- = = the power of the test the power of the test
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Probability of Rejecting HProbability of Rejecting Hoo
Figure 10.6Figure 10.6
XX405405 413.16413.16
BB
Area = 1 - Area = 1 - when µ = 405when µ = 405
Area = Area = = .10 = .10
386.84386.84 400400
AA
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©2006 Thomson/South-Western 18
Power of TestsPower of Tests
2.2. Power of testPower of test = = PP((ZZ > > zz11) + ) + PP((ZZ < < zz22))
1.1. DetermineDetermine
zz11 = = ZZ/2/2 - -µ - µµ - µoo
/ / nn
zz11 = - = -ZZ/2/2 - -µ - µµ - µoo
/ / nn
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©2006 Thomson/South-Western 19
Power CurvePower Curve
Power curve for test of Ho versus Ha using five-step procedure
Power curve for test of Ho versus Ha using any other procedure
400 405
.1655.10
Figure 10.7
µ
1.0
1 - = P(rejecting Ho)
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©2006 Thomson/South-Western 20
One Tailed Test for One Tailed Test for µµ
HHoo: µ ≥ 32.5: µ ≥ 32.5 HHaa: µ < 32.5: µ < 32.5
1.1. Define HDefine Hoo and H and Haa prior to observation prior to observation
ZZ = = XX - µ - µoo
/ / nn
2.2. Define the test statisticDefine the test statistic
3. Reject if3. Reject if
HHoo if Zif Z = < -1.645 = < -1.645XX - 32.5 - 32.5
/ / nnExample 10.4Example 10.4
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One Tailed Test for One Tailed Test for µµ
5.5. Study supports the claim that the Study supports the claim that the average mileage for the Bullet is less average mileage for the Bullet is less than than 32.532.5 mpg – supports claim of mpg – supports claim of false advertisingfalse advertising
ZZ * * = = -2.70 = = -2.7030.4 - 32.530.4 - 32.5
5.5 / 505.5 / 50
4.4. The value of the test statistic isThe value of the test statistic is
Example 10.4Example 10.4
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©2006 Thomson/South-Western 22
One Tailed Test for µ One Tailed Test for µ
Figure 10.8Figure 10.8
-1.645-1.645 00ZZ
Area = .5 - .05 = .45Area = .5 - .05 = .45
Area = Area = = .05 = .05
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©2006 Thomson/South-Western 23
Statistical Software Statistical Software ExampleExample
Z Z * = 1.531* = 1.531
Figure 10.9Figure 10.9
ZZ2.332.3300
Area = .49Area = .49
Area = Area = = .01 = .01
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Statistical Software ExampleStatistical Software Example
Figure 10.10Figure 10.10
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Statistical Software ExampleStatistical Software Example
Figure 10.11Figure 10.11
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©2006 Thomson/South-Western 26
Power of TestsPower of Tests
2.2. Power of test = PPower of test = P((Z Z > > zz11))
1.1. DetermineDetermine
zz11 = = ZZ/2/2 - -µ - µµ - µoo
/ / nn
HHoo: µ ≤ µ: µ ≤ µoo verses verses HHaa: µ > µ: µ > µoo
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©2006 Thomson/South-Western 27
Power of TestsPower of Tests
2.2. Power of testPower of test = = PP((ZZ < < zz22))
1.1. DetermineDetermine
zz11 = - = -ZZ/2/2 - -µ - µµ - µoo
/ / nn
HHoo: µ ≥ µ: µ ≥ µoo verses verses HHaa: µ < µ: µ < µoo
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©2006 Thomson/South-Western 28
Power of TestsPower of TestsTests on a Population Mean (Tests on a Population Mean ( Known) Known)
Two-Tailed TestTwo-Tailed Test
HHoo: µ = µ: µ = µoo
HHaa: µ ≠ µ: µ ≠ µoo
rejectreject HHoo if | if |ZZ *| > *| > ZZ/2/2
One-Tailed TestOne-Tailed Test
HHoo: µ ≤ µ: µ ≤ µoo
HHaa: µ > µ: µ > µoo
rejectreject HHoo if if ZZ * > * > ZZ
HHoo: µ ≥ µ: µ ≥ µoo
HHaa: µ < µ: µ < µoo
rejectreject HHoo if if ZZ * < -* < -ZZ
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Determining the p-ValueDetermining the p-Value
The p-value is the value of The p-value is the value of at at which the hypothesis test which the hypothesis test procedure changes conclusions procedure changes conclusions based on a given set of data. It is based on a given set of data. It is the largest value of the largest value of for which for which you will fail to reject Hyou will fail to reject Hoo
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Determining the p-ValueDetermining the p-Value
Figure 10.12
-1.9
6 0Z
-2.5
75
1.96
2.57
5
Z * = -2.38
Area = .025
Area =.005
Area = .025
Area = .005
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Determining the p-ValueDetermining the p-Value
Figure 10.13
0Z
-2.38 2.38
Area = p value
Area = .5 - .4913= .0087
Area = .4913 (Table A.4)
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©2006 Thomson/South-Western 32
Procedure for Finding the Procedure for Finding the p-Valuep-Value
For HFor Haa: µ ≠ µ: µ ≠ µoo
p = 2p = 2 •• ( (area outside Zarea outside Z **))For HFor Haa: µ > µ: µ > µoo
p = area to the right of Zp = area to the right of Z **
For HFor Haa: µ < µ: µ < µoo
p = area to the left of Zp = area to the left of Z **
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Determining the p-ValueDetermining the p-Value
Figure 10.14
0Z
Z * = 1.53
p = area= .5 - .4370= .0630
Area = .4370
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©2006 Thomson/South-Western 34
Interpreting the p-ValueInterpreting the p-Value
Classical ApproachClassical Approachreject Hreject Hoo if p-value if p-value < <
fail to reject Hfail to reject Hoo is p-value is p-value ≥ ≥ General rule of thumbGeneral rule of thumb
reject Hreject Hoo if p-value is small if p-value is small ((p p < .01)< .01)
fail to reject Hfail to reject Hoo is p-value is large is p-value is large ((p p > .1)> .1)
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Interpreting the p-ValueInterpreting the p-Value
Small Small pp
Reject Reject HHoo
.01.01
Large Large pp
Fail to reject Fail to reject HHoo
pp.1.1
InconclusiveInconclusive
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Another InterpretationAnother Interpretation1.1. For a two-tailed test where HFor a two-tailed test where Hoo: µ ≠ µ: µ ≠ µoo,, the p-value is the p-value is
the probability that the value of the test statistic, Zthe probability that the value of the test statistic, Z *, *, will be at least as large (in absolute value) as the will be at least as large (in absolute value) as the observed Zobserved Z *, if *, if µµ is in fact equal to is in fact equal to µµoo
2.2. For a one-tailed test where HFor a one-tailed test where Haa: µ > µ: µ > µoo,, the p-value is the p-value is
the probability that the value of the test statistic, Zthe probability that the value of the test statistic, Z *, *, will be at least as large as the observed Zwill be at least as large as the observed Z *, if *, if µµ is in is in fact equal to fact equal to µµoo
3.3. For a one-tailed test where HFor a one-tailed test where Haa: µ < µ: µ < µoo,, the p-value is the p-value is
the probability that the value of the test statistic, Zthe probability that the value of the test statistic, Z *, *, will be at least as large as the observed Zwill be at least as large as the observed Z *, if *, if µµ is in is in fact equal to fact equal to µµoo
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Statistical Software ExampleStatistical Software Example
Figure 10.15Figure 10.15
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Statistical Software ExampleStatistical Software Example
Figure 10.16Figure 10.16
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Statistical Software ExampleStatistical Software Example
Figure 10.17Figure 10.17
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Practical Versus StatisticalPractical Versus Statistical
Figure 10.18
ZZ * = -2.69
Area = p value= .0036 (from Table A.4)
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Hypothesis Testing on the Hypothesis Testing on the Mean of a Normal Population Mean of a Normal Population
(( Unknown) Unknown)
Normal populationNormal population Population standard deviation unknownPopulation standard deviation unknown Small sampleSmall sample Student t distributionStudent t distribution
Nonparametric procedureNonparametric procedure
tt = = XX - µ - µoo
ss / / nn
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Small Sample TestSmall Sample Test
(a)
Normal population
Symmetric (nonnormal) population Skewed
population
(b)
Figure 10.19
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Clark Products ExampleClark Products Example1. When a question is phrased “Is there 1. When a question is phrased “Is there
evidence to indicate that ...,” what follows evidence to indicate that ...,” what follows is the alternative hypothesisis the alternative hypothesis
HHoo: µ = 10 : µ = 10 andand HHaa: µ ≠ 10: µ ≠ 10
2.2. The test statistic here isThe test statistic here is
tt = = XX - µ - µoo
ss / / nn
3.3. Using a significance level of .05 and Figure 10.20, Using a significance level of .05 and Figure 10.20, the corresponding two-tailed procedure is tothe corresponding two-tailed procedure is to
rejectreject HHoo if | if |tt| > | > tt.025,17.025,17 = 2.11 = 2.11
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Clark Products ExampleClark Products Example
Figure 10.20
Z-2.11 2.11
Area = .025 Area = .025
Reject Ho here Reject Ho here
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Clark Products ExampleClark Products Example
Figure 10.21
t * = 1.83
t
2.11
1.740
Area = .05
Area = .025
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5. There is insufficient evidence to indicate 5. There is insufficient evidence to indicate that the average output voltage is different that the average output voltage is different from 10 voltsfrom 10 volts
4.4. The value of the test statistic isThe value of the test statistic is
tt ** = = 1.83 = = 1.83 10.331 - 1010.331 - 10
.767 / 18.767 / 18
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Clark Products ExampleClark Products Example
Figure 10.22Figure 10.22
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Clark Products ExampleClark Products Example
Figure 10.24Figure 10.24
Figure 10.23Figure 10.23
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Small-Sample Tests on aSmall-Sample Tests on aNormal Population MeanNormal Population Mean
Two-tailed testTwo-tailed test
HHoo: µ = µ: µ = µoo
HHaa: µ ≠ µ: µ ≠ µoo
rejectreject HHoo if | if |tt *| > *| > tt/2, /2, nn-1-1
One tail testOne tail test
HHoo: µ = µ: µ = µoo
HHaa: µ < µ: µ < µoo
rejectreject HHoo if if tt * < t* < t, n-1, n-1
HHoo: µ = µ: µ = µoo
HHaa: µ > µ: µ > µoo
rejectreject HHoo if if tt * > * > tt, , nn-1-1