©2003 Thomson/South-Western 1
Chapter 8 –HypothesisTesting for theMean and Varianceof a PopulationChapter 8 –HypothesisTesting for theMean and Varianceof a Population
Slides prepared by Jeff Heyl, Lincoln UniversitySlides prepared by Jeff Heyl, Lincoln University
©2003 South-Western/Thomson Learning™
Introduction to Business Statistics, 6eIntroduction to Business Statistics, 6e
Kvanli, Pavur, KeelingKvanli, Pavur, Keeling
©2003 Thomson/South-Western 2
Hypothesis Testing on theMeanHypothesis Testing on theMean
Null hypothesis (Ho): A statementconcerning a population parameterNull hypothesis (Ho): A statementconcerning a population parameter
Alternative hypothesis: A statementin contradiction of the nullhypothesisAlternative hypothesis: A statementin contradiction of the nullhypothesis
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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 rejecting the Ho when Ho is true(Type I error)
=probability of failing to rejecting the Ho when Hois false (Type II error) =probability of failing to rejecting the Ho when Hois false (Type II error)
Actual SituationActual Situation
Conclusion Ho TrueHo FalseConclusion Ho TrueHo False
Fail to Reject Ho Correct decision Type II errorFail to Reject Ho Correct decision Type II error
Reject Ho Type I error Correct decisionReject Ho Type I error Correct decision
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Hypothesis Testing ProcessHypothesis Testing Process
Determine the Ho and HaDetermine the Ho and Ha
Determine the significance levelDetermine the significance level
Compare the sample mean (variance)to the hypothesized mean (variance)Compare the sample mean (variance)to the hypothesized mean (variance)
Decide whether to fail to reject orreject HoDecide whether to fail to reject orreject Ho
Determine what the decision means inreference to the problemDetermine what the decision means inreference to the problem
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Height ExampleHeight Example
Ho: = 5.9Ho: = 5.9
Ha: 5.9Ha: 5.9
= .05 = P(rejecting Ho when Ho is true)critical value = ± 1.96 = .05 = P(rejecting Ho when Ho is true)critical value = ± 1.96
Reject Ho if > 1.96Reject Ho if > 1.96
X - 5.9X - 5.9
/ n/ n
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Height ExampleHeight Example
ZZ
.025.025
-k-k
00
|Z| > k|Z| > k
kk
.025.025
Area = .5 - .025 = .475Area = .5 - .025 = .475
Figure 8.1Figure 8.1
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Height ExampleHeight Example
Figure 8.2Figure 8.2
XX
distance = .14’ =2.53 x .055distance = .14’ =2.53 x .055
5.76’5.76’
x ≈ = .055’x ≈ = .055’
.48.48
75 75
µx = 5.9’µx = 5.9’
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Computed ValueComputed Value
Because - 2.53 < - 1.96, we reject Ho thus weBecause - 2.53 < - 1.96, we reject Ho thus we
conclude that the average population maleconclude that the average population male
height is not equal to 5.9height is not equal to 5.9
Z = ≈ = = -2.53 = Z*Z = ≈ = = -2.53 = Z*
X - 5.9X - 5.9
/ n / n
X - 5.9X - 5.9
s / ns / n
5.76 - 5.95.76 - 5.9
.48 / 75.48 / 75
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Height ExampleHeight Example
Figure 8.3Figure 8.3
XX
Area = .5 - .005 = .495Area = .5 - .005 = .495
-k-k
00
Area = .01Area = .01
kk
.005.005
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Height ExampleHeight Example
Figure 8.4Figure 8.4
ZZ
Reject Ho if Z*falls hereReject Ho if Z*falls here
-2.575-2.575
Z* = -2.53Z* = -2.53
2.5752.575
00
Area = .01Area = .01
Reject Ho if Z*falls hereReject Ho if Z*falls here
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Hypothesis Testing5 Step ProcedureHypothesis Testing5 Step Procedure
3.Define the rejection region3.Define the rejection region
4.Calculate the test statistic4.Calculate the test statistic
5.Give a conclusion in terms of the problem5.Give a conclusion in terms of the problem
1.Set up the null and alternative hypothesis1.Set up the null and alternative hypothesis
Z = ≈Z = ≈
X - µoX - µo
/ n / n
X - µoX - µo
s / ns / n
2.Define the test statistic2.Define the test statistic
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Everglo Light ExampleEverglo Light Example
1.Define the hypotheses1.Define the hypotheses
Ho: µ = 400 Ha: µ ≠ 400Ho: µ = 400 Ha: µ ≠ 400
3.Define the rejection region3.Define the rejection region
reject Ho if Z > 1.645 or Z < -1.645reject Ho if Z > 1.645 or Z < -1.645
2.Define the test statistic2.Define the test statistic
Z = ≈Z = ≈
X - 400X - 400
/ n / n
X - 400X - 400
s / ns / n
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Everglo Light ExampleEverglo Light Example
Figure 8.5Figure 8.5
ZZ
-1.645-1.645
Z* = 2.59Z* = 2.59
1.6451.645
00
Area = = .1Area = = .1
Area = .5 - .05 = .45Area = .5 - .05 = .45
22
= .05= .05
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Everglo Light ExampleEverglo Light Example
5.State the conclusion5.State the conclusion
There is sufficient evidence to concludethat the average lifetime of Everglo bulbsis not 400 hoursThere is sufficient evidence to concludethat the average lifetime of Everglo bulbsis not 400 hours
4.Calculate the value of the test statistic4.Calculate the value of the test statistic
Z* ≈ = = 2.59Z* ≈ = = 2.59
1111
4.254.25
411 - 400411 - 400
42.5 / 10042.5 / 100
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Confidence Intervals andHypothesis TestingConfidence Intervals andHypothesis Testing
When testing Ho: µ = µo versus Ha: µ ≠ µousing the five-step procedure and asignificance level, , Ho will be rejected ifand only if µo lies outside the (1 - ) • 100confidence interval for µWhen testing Ho: µ = µo versus Ha: µ ≠ µousing the five-step procedure and asignificance level, , Ho will be rejected ifand only if µo lies outside the (1 - ) • 100confidence interval for µ
X - k to X + kX - k to X + k
n n
n n
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Power of a Statistical TestPower of a Statistical Test
= P(fail to reject Ho when Ho is false) = P(fail to reject Ho when Ho is false)
1- = P(rejecting Ho when Ho is false)1- = P(rejecting Ho when Ho is false)
1- = the power of the test1- = the power of the test
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Probability of Rejecting HoProbability of Rejecting Ho
Figure 8.6Figure 8.6
391.775391.775
400400
403403
408.225408.225
AA
BB
XX
Area = = .10Area = = .10
Area = 1 - Area = 1 -
when µ = 403when µ = 403
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Power of TestsPower of Tests
2.Power of test = P(Z > z1) + P(Z < z2)2.Power of test = P(Z > z1) + P(Z < z2)
1.Determine1.Determine
z1 = Z/2 -z1 = Z/2 -
µ - µoµ - µo
/ n / n
z1 = -Z/2 -z1 = -Z/2 -
µ - µoµ - µo
/ n / n
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Power CurvePower Curve
Power curve for test of Hoversus Ha using five-stepprocedurePower curve for test of Hoversus Ha using five-stepprocedure
Power curve for test of Hoversus Ha using any otherprocedurePower curve for test of Hoversus Ha using any otherprocedure
µµ
400 403400 403
1.01.0
.1617.1617
.10.10
1 - =P(rejecting Ho)1 - =P(rejecting Ho)
Figure 8.7Figure 8.7
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One Tailed Test for µOne Tailed Test for µ
Ho: µ ≥ 32.5Ha: µ < 32.5Ho: µ ≥ 32.5Ha: µ < 32.5
1.Define Ho and Ha prior to observation1.Define Ho and Ha prior to observation
Z = ≈Z = ≈
X - µoX - µo
/ n / n
X - µoX - µo
s / ns / n
2.Define the test statistic2.Define the test statistic
3. Reject if3. Reject if
Ho if Z ≈ < -1.645Ho if Z ≈ < -1.645
X - 32.5X - 32.5
s / ns / n
Example 8.4Example 8.4
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One Tailed Test for µOne Tailed Test for µ
5.Study supports the claim that theaverage mileage for the Bullet is lessthan 32.5 mpg – supports claim offalse advertising5.Study supports the claim that theaverage mileage for the Bullet is lessthan 32.5 mpg – supports claim offalse advertising
Z* ≈ = -2.80Z* ≈ = -2.80
30.4 - 32.530.4 - 32.5
5.3 / 505.3 / 50
4.The value of the test statistic is4.The value of the test statistic is
Example 8.4Example 8.4
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One Tailed Test for µOne Tailed Test for µ
Figure 8.8Figure 8.8
-1.645-1.645
00
Area = .5 - .05 = .45Area = .5 - .05 = .45
Area = = .05Area = = .05
ZZ
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Excel ExampleExcel Example
ZZ
Z* = 1.54Z* = 1.54
Area = .49Area = .49
Area = = .01Area = = .01
2.332.33
00
Figure 8.9Figure 8.9
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Excel ExampleExcel Example
Figure 8.10Figure 8.10
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Excel ExampleExcel Example
Figure 8.11Figure 8.11
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Power of TestsPower of Tests
2.Power of test = P(Z > z1)2.Power of test = P(Z > z1)
1.Determine1.Determine
z1 = Z/2 -z1 = Z/2 -
µ - µoµ - µo
/ n / n
Ho: µ ≤ µo verses Ha: µ > µoHo: µ ≤ µo verses Ha: µ > µo
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Power of TestsPower of Tests
2.Power of test = P(Z < z2)2.Power of test = P(Z < z2)
1.Determine1.Determine
z1 = -Z/2 -z1 = -Z/2 -
µ - µoµ - µo
/ n / n
Ho: µ ≥ µo verses Ha: µ < µoHo: µ ≥ µo verses Ha: µ < µo
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Power of TestsPower of Tests
Large-Sample Tests on a Population MeanLarge-Sample Tests on a Population Mean
Two-Tailed TestTwo-Tailed Test
Ho: µ = µoHo: µ = µo
Ha: µ ≠ µoHa: µ ≠ µo
reject Ho if |Z*| > Z/2reject Ho if |Z*| > Z/2
One-Tailed TestOne-Tailed Test
Ho: µ ≤ µoHo: µ ≤ µo
Ha: µ > µoHa: µ > µo
reject Ho if Z* > Zreject Ho if Z* > Z
Ho: µ ≥ µoHo: µ ≥ µo
Ha: µ < µoHa: µ < µo
reject Ho if Z* < -Zreject Ho if Z* < -Z
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Determining the p-ValueDetermining the p-Value
The p-value is the value of atwhich the hypothesis testprocedure changes conclusionsbased on a given set of data. It isthe largest value of for whichyou will fail to reject HoThe p-value is the value of atwhich the hypothesis testprocedure changes conclusionsbased on a given set of data. It isthe largest value of for whichyou will fail to reject Ho
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Determining the p-ValueDetermining the p-Value
Figure 8.12Figure 8.12
-1.96-1.96
00
Area = .025Area = .025
Area =.005Area =.005
ZZ
-2.575-2.575
Z* = -2.53Z* = -2.53
Area = .025Area = .025
Area = .005Area = .005
1.961.96
2.5752.575
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Determining the p-ValueDetermining the p-Value
Figure 8.13Figure 8.13
00
ZZ
-2.53-2.53
Area = .4943 (Table A-4)Area = .4943 (Table A-4)
Area= .5 - .4943Area= .5 - .4943
= .0057= .0057
2.532.53
Area = p valueArea = p value
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Procedure for Finding thep-ValueProcedure for Finding thep-Value
For Ha: µ ≠ µop = 2 * (area outside Z*)For Ha: µ ≠ µop = 2 * (area outside Z*)
For Ha: µ > µop = area to the right of Z*For Ha: µ > µop = area to the right of Z*
For Ha: µ < µop = area to the left of Z* For Ha: µ < µop = area to the left of Z*
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Determining the p-ValueDetermining the p-Value
Figure 8.14Figure 8.14
00
ZZ
Area = .025Area = .025
p= areap= area
= .5 - .4382= .5 - .4382
= .0618= .0618
Z* = 1.54Z* = 1.54
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Interpreting the p-ValueInterpreting the p-Value
Classical ApproachClassical Approach
reject Ho if p-value < reject Ho if p-value <
fail to reject Ho is p-value ≥ fail to reject Ho is p-value ≥
General rule of thumbGeneral rule of thumb
reject Ho if p-value is small (p < .01)reject Ho if p-value is small (p < .01)
fail to reject Ho is p-value is large (p > .1)fail to reject Ho is p-value is large (p > .1)
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Interpreting the p-ValueInterpreting the p-Value
Small pSmall p
Reject HoReject Ho
.01.01
Large pLarge p
Fail to reject HoFail to reject Ho
pp
.1.1
InconclusiveInconclusive
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Another InterpretationAnother Interpretation
1.For a two-tailed test where Ho: µ ≠ µo, the p-value isthe probability that the value of the test statistic,Z*, will be at least as large (in absolute value) asthe observed Z*, if µ is in fact equal to µo1.For a two-tailed test where Ho: µ ≠ µo, the p-value isthe probability that the value of the test statistic,Z*, will be at least as large (in absolute value) asthe observed Z*, if µ is in fact equal to µo
2.For a one-tailed test where Ha: µ > µo, the p-valueis the probability that the value of the test statistic,Z*, will be at least as large as the observed Z*, if µis in fact equal to µo2.For a one-tailed test where Ha: µ > µo, the p-valueis the probability that the value of the test statistic,Z*, will be at least as large as the observed Z*, if µis in fact equal to µo
3.For a one-tailed test where Ha: µ < µo, the p-value isthe probability that the value of the test statistic,Z*, will be at least as large as the observed Z*, if µis in fact equal to µo3.For a one-tailed test where Ha: µ < µo, the p-value isthe probability that the value of the test statistic,Z*, will be at least as large as the observed Z*, if µis in fact equal to µo
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Excel ExampleExcel Example
Figure 8.15Figure 8.15
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Excel ExampleExcel Example
Figure 8.16Figure 8.16
ZZ
p value = twice this areap value = twice this area
3.063.06
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Excel ExampleExcel Example
Figure 8.17Figure 8.17
Frequency HistogramFrequency Histogram
3535
3030
2525
2020
1515
1010
55
00
10.12 and10.14 and10.16 and10.18 and10.20 and10.22 and10.24 and10.26 and10.28 and10.30 and10.12 and10.14 and10.16 and10.18 and10.20 and10.22 and10.24 and10.26 and10.28 and10.30 and
underunderunderunderunderunderunderunderunderunderunderunderunderunderunderunderunderunderunderunder
10.1410.1610.1810.2010.2210.2410.2610.2810.3010.3210.1410.1610.1810.2010.2210.2410.2610.2810.3010.32
Class LimitsClass Limits
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Practical Versus StatisticalPractical Versus Statistical
Figure 8.18Figure 8.18
Area= p valueArea= p value
= .004 (from Table A-4)= .004 (from Table A-4)
ZZ
Z* = -2.65Z* = -2.65
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Hypothesis Testing on theMean of a Normal Population:Small SampleHypothesis Testing on theMean of a Normal Population:Small Sample
Normal populationNormal population
Population standard deviation unknownPopulation standard deviation unknown
Small sampleSmall sample
Student t distributionStudent t distribution
Nonparametric procedureNonparametric procedure
t =t =
X - µoX - µo
s / ns / n
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Small Sample TestSmall Sample Test
AA
NormalpopulationNormalpopulation
Symmetric(nonnormal)populationSymmetric(nonnormal)population
SkewedpopulationSkewedpopulation
BB
Figure 8.19Figure 8.19
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Clark Products ExampleClark Products Example
1. When a question is phrased “Is thereevidence to indicate that ...,” what followsis the alternative hypothesis1. When a question is phrased “Is thereevidence to indicate that ...,” what followsis the alternative hypothesis
Ho: µ = 10 and Ha: µ ≠ 10Ho: µ = 10 and Ha: µ ≠ 10
2.The test statistic here is2.The test statistic here is
t =t =
X - µoX - µo
s / ns / n
3.Using a significance level of .05 and Figure 8.20,the corresponding two-tailed procedure is to3.Using a significance level of .05 and Figure 8.20,the corresponding two-tailed procedure is to
reject Ho if |t| > t.025,17 = 2.11reject Ho if |t| > t.025,17 = 2.11
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Clark Products ExampleClark Products Example
Area= .025Area= .025
ZZ
-2.11-2.11
Reject Ho hereReject Ho here
2.112.11
Area = .025Area = .025
Reject Ho hereReject Ho here
Figure 8.20Figure 8.20
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Clark Products ExampleClark Products Example
Figure 8.21Figure 8.21
ZZ
2.112.11
Area = .05Area = .05
Area = .025Area = .025
1.741.74
t* = 1.83t* = 1.83
00
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Clark Products ExampleClark Products Example
5. There is insufficient evidence to indicatethat the average output voltage is differentfrom 10 volts5. There is insufficient evidence to indicatethat the average output voltage is differentfrom 10 volts
4.The value of the test statistic is4.The value of the test statistic is
t* = = 1.83t* = = 1.83
10.331 - 1010.331 - 10
.767 / 18.767 / 18
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Clark Products ExampleClark Products Example
Figure 8.22Figure 8.22
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Small-Sample Tests on aNormal Population MeanSmall-Sample Tests on aNormal Population Mean
Two-tailed testTwo-tailed test
Ho: µ = µoHo: µ = µo
Ha: µ ≠ µoHa: µ ≠ µo
reject Ho if |t*| > t/2, n-1reject Ho if |t*| > t/2, n-1
One tail testOne tail test
Ho: µ = µoHo: µ = µo
Ha: µ < µoHa: µ < µo
reject Ho if t* < t, n-1reject Ho if t* < t, n-1
Ho: µ = µoHo: µ = µo
Ha: µ > µoHa: µ > µo
reject Ho if t* > t, n-1reject Ho if t* > t, n-1
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Chi-Square DistributionChi-Square Distribution
22
Area = aArea = a
a, dfa, df
22
Figure 8.23Figure 8.23
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Chi-Square DistributionChi-Square Distribution
22
Area = .1Area = .1
18.549318.5493
Figure 8.24Figure 8.24
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Chi-Square DistributionChi-Square Distribution
Figure 8.25Figure 8.25
22
Area = .025Area = .025
bb
Area = .95Area = .95
Area = .025Area = .025
aa
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Hypothesis Testing on 2Hypothesis Testing on 2
One tail testOne tail test
Ho: 2 = oHo: 2 = o
Ha: 2 < oHa: 2 < o
reject Ho if 2* < , n-1reject Ho if 2* < , n-1
22
22
22
Ho: 2 = oHo: 2 = o
Ha: 2 > oHa: 2 > o
reject Ho if 2* > , n-1reject Ho if 2* > , n-1
22
22
22
Two-tailed testTwo-tailed test
Ho: 2 = oHo: 2 = o
Ha: 2 ≠ oHa: 2 ≠ o
22
22
Test statistic: 2 =Test statistic: 2 =
(n - 1)s2(n - 1)s2
22
oo
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Chi-Square DistributionChi-Square Distribution
Figure 8.26Figure 8.26
22
Area = p valueArea = p value
152.6152.6
2 curve with 14 df2 curve with 14 df
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Chi-Square DistributionChi-Square Distribution
Figure 8.27Figure 8.27
