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Transcript of SDA 3E Chapter 5
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2007 Pearson Education
Chapter 5: Hypothesis Testingand Statistical Inference
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Hypothesis Testing Hypothesis testing involves drawing
inferences about two contrasting propositions
(hypotheses) relating to the value of apopulation parameter, one of which isassumed to be true in the absence ofcontradictory data.
We seek evidence to determine if thehypothesis can be rejected; if not, we canonly assume it to be true but have notstatistically proven it true.
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Hypothesis Testing Procedure1. Formulate the hypothesis
2. Select a level of significance, which defines
the risk of drawing an incorrect conclusionthat a true hypothesis is false
3. Determine a decision rule
4. Collect data and calculate a test statistic5. Apply the decision rule and draw a
conclusion
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Hypothesis Formulation Null hypothesis, H0 a statement that is
accepted as correct
Alternative hypothesis, H1 a proposition thatmust be true if H0 is false
Formulating the correct set of hypothesesdepends on burden of proof what you
wish to prove statistically should be H1 Tests involving a single population parameter
are called one-sample tests; tests involvingtwo populations are called two-sample tests.
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Types of Hypothesis Tests One Sample Tests
H0: population parameter constant vs.
H1: population parameter < constant
H0: population parameter constant vs.H
1: population parameter > constant
H0: population parameter = constant vs.
H1: population parameter constant
Two Sample Tests H
0: population parameter (1) - population parameter (2) 0 vs.
H1: population parameter (1) - population parameter (2) < 0
H0: population parameter (1) - population parameter (2) 0 vs.
H1: population parameter (1) - population parameter (2) > 0
H0: population parameter (1) - population parameter (2) = 0 vs.H
1: population parameter (1) - population parameter (2) 0
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Four Outcomes1. The null hypothesis is actually true, and the
test correctly fails to reject it.
2. The null hypothesis is actually false, and thehypothesis test correctly reaches thisconclusion.
3. The null hypothesis is actually true, but the
hypothesis test incorrectly rejects it (Type Ierror).
4. The null hypothesis is actually false, but thehypothesis test incorrectly fails to reject it
(Type II error).
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Quantifying Outcomes Probability of Type I error (rejecting H0 when
it is true) = = level of significance Probability ofcorrectly failingto reject H0 = 1
= confidence coefficient Probability of Type II error (failing to reject H0
when it is false) = Probability ofcorrectly rejectingH0 when it is
false = 1 = power of the test
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Decision Rules Compute a test statistic from sample data and
compare it to the hypothesized sampling
distribution of the test statistic Divide the sampling distribution into a
rejection region and non-rejection region.
If the test statistic falls in the rejection region,reject H0 (concluding that H1 is true);
otherwise, fail to reject H0
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Rejection Regions
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Hypothesis Tests and
Spreadsheet SupportType of Test Excel/PHStatProcedure
One sample test for mean, unknown PHStat: One Sample Test Z-test for the
Mean, Sigma KnownOne sample test for mean, unknown PHStat: One Sample Test t-test for the
Mean, Sigma Unknown
One sample test for proportion PHStat: One Sample Test Z-test for theProportion
Two sample test for means, known Excel z-test: Two-Sample for Means
PHStat: Two Sample Tests Z-Test forDifferences in Two Means
Two sample test for means, unknown,unequal
Excel t-test: Two-Sample AssumingUnequal Variances
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Hypothesis Tests and
Spreadsheet Support (contd)Type of Test Excel/PHStatProcedure
Two sample test for means, unknown,assumed equal
Excel t-test: Two-Sample Assuming EqualVariances
PHStat: Two Sample Tests t-Test forDifferences in Two Means
Paired two sample test for means Excel t-test: Paired Two-Sample for Means
Two sample test for proportions PHStat: Two Sample Tests Z-Test forDifferences in Two Proportions
Equality of variances Excel F-test Two-Sample for Variances
PHStat: Two Sample Tests F-Test forDifferences in Two Variances
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One Sample Tests for Means
Standard Deviation Unknown Example hypothesis
H0: 0 versus H1: < 0 Test statistic:
Reject H0if t < -t
n-1,
ns
xt
/
0=
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Example For the Customer Support Survey.xls data, test thehypotheses
H0: mean response time 30 minutes H1: mean response time< 30 minutes
Sample mean = 21.91; sample standard deviation =
19.49; n = 44 observations
Reject H0 because t = 2.75 < -t43,0.05 = -1.6811
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PHStatTool: t-Test for Mean PHStatmenu > One Sample
Tests > t-Test for the Mean,
Sigma Unknown
Enter null hypothesis and alpha
Enter sample statistics or datarange
Choose type of test
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Results
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One Sample Tests for
Proportions Example hypothesis
H0:
0versusH
1:
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Example For the Customer Support Survey.xls data, test the hypothesis that the
proportion of overall quality responses in the top two boxes is at least0.75
H0: .75 H
0: < .75
Sample proportion = 0.682; n = 44
For a level of significance of 0.05, the critical value ofzis -1.645;therefore, we cannot reject the null hypothesis
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PHStat Tool: One Sample z-
Test for Proportions PHStat> One Sample Tests> z-Tests
for the Proportion
Enter null hypothesis,significance level, numberof successes, and sample
sizeEnter type of test
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Results
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Type II Errors and the Power
of a Test The probability of a Type II error, , and the
power of the test (1 ) cannot be chosen by
the experimenter. The power of the test depends on the true
value of the population mean, the level ofconfidence used, and the sample size.
A power curve shows (1 ) as a function of
1.
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Example Power Curve
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Two Sample Tests for Means
Standard Deviation Known Example hypothesis
H0:1 20 versusH1:1 -2 < 0
Test Statistic:
Reject if z < -z
2
2
21
2
1
21
// nn
xxz
+
=
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Two Sample Tests for Means
Sigma Unknown and Equal Example hypothesis
H0:1 20 versusH1:1 -2 > 0
Test Statistic:
Reject if z > z
21
21
21
2
22
2
11
21
2
)1()1(
nn
nn
nn
snsn
xxz
+
+
=
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Two Sample Tests for Means
Sigma Unknown and Unequal Example hypothesis
H0:1 2=0 versusH1:1 -2 0
Test Statistic:
Reject if z > z/2 or z < - z/2
t = (x1
-
x2) / 2
2
2
1
2
1
n
s
n
s+
+
+
1
)/(
1
)/(
2
2
2
2
2
1
2
1
2
1
2
2
2
2
1
2
1
n
ns
n
ns
ns
ns
with df =
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Excel Data Analysis Tool: Two
Sample t-Tests Tools> Data Analysis> t-test: Two Sample
Assuming Unequal Variances, or t-test: TwoSample Assuming Equal Variances
Enter range of data, hypothesized meandifference, and level of significance
Tool allows you to test H0:
1-
2= d
Output is provided for upper-tail test only For lower-tail test, change the sign on t
Critical one-tail, and subtract P(T
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PHStatTool: Two Sample
t-Tests PHStat> Two Sample Tests> t-Test
for Differences in Two Means
Test assumes equal variances Must compute and enter the sample
mean, sample standard deviation, and
sample size
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Comparison of Excel and PHStat
Results Lower-Tail Test
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Two Sample Test for Means
With Paired Samples Example hypothesis
H0: average difference=0 versus
H1: average difference0
Test Statistic:
Reject if t > tn-1,/2 or t < - tn-1,/2
ns
Dt
D
D
/
=
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Two Sample Tests for
Proportions Example hypothesis
H0: 1 2=0 versusH1: 1 -2 0
Test Statistic:
Reject if z > z/2 or z < - z/2
+
=
21
21
11)1(
nnpp
ppz
where21 nn
samplesbothinsuccessesofnumberp+
=
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Hypothesis Tests and
Confidence Intervals If a 100(1 )% confidence interval contains
the hypothesized value, then we would not
reject the null hypothesis based on this valuewith a level of significance . Example hypothesis
H0: 0 versus H1: < 0 If a 100(1-)% confidence interval does not
contain 0, then we can reject H0
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F-Test for Differences in Two
Variances Hypothesis
H0: 1
2 2
2=0 versusH
1: 1
2 -22 0
Test Statistic:
Assume s12 > s2
2
Reject if F > F/2,n1-1,n2-1 (see Appendix A.4)
Assumes both samples drawn from normaldistributions
2
2
21
s
sF =
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Excel Data Analysis Tool: F-
Test for Equality of Variances Tools> Data Analysis> F-test for
Equality of Variances
Specify data ranges Use /2 for the significance level! If the variance of Variable 1 is greater
than the variance of variable 2, theoutput will specify the upper tail;otherwise, you obtain the lower tailinformation.
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PHStat Tool: F-Test for
Differences in Variances PHStatmenu > Two Sample Tests> F-
test for Differences in Two Variances
Compute and enter sample standarddeviations
Enter the significance level , not /2as in Excel
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Excel and PHStatResults
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Analysis of Variance (ANOVA) Compare the means of m different
groups (factors) to determine if all are
equal H
0: 1 = 1 = ... m
H1: at least one mean is different from the
others
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ANOVA Theory nj = number of observations in sample j
SST = total variation in the data
SSB = variation between groups
SSW = variation within groups
SST = SSB + SSW
= =
=n
j
n
i
ij
j
XXSST
1 1
2)(
=
=n
j
jj XXnSSB1
2)(
= = =n
j
n
i
jij
j
XXSSW1 1
2)(
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ANOVA Test Statistic MSB = SSB/(m 1)
MSW = SSW/(n m)
Test statistic: F = MSB/MSW Has an F-distribution with m-1 and n-m
degrees of freedom
Reject H0 if F > F/2,m-1,n-m
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Excel Data Analysis Tool for
ANOVA Tools> Data Analysis>ANOVA: Single
Factor
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ANOVA Results
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ANOVA Assumptions The mgroups or factor levels being studied
represent populations whose outcome
measures are Randomly and independently obtained Are normally distributed Have equal variances
Violation of these assumptions can affect thetrue level of significance and power of thetest.
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Nonparametric Tests Used when assumptions (usually
normality) are violated. Examples: Wilcoxon rank sum test for testing
difference between two medians Kurskal-Wallis rank test for determining
whether multiple populations have equalmedians. Both supported by PHStat
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Tukey-Kramer Multiple
Comparison ProcedureANOVA cannot identify which means
may differ from the rest
PHStatmenu > Multiple Sample Tests> Tukey-Kramer Multiple ComparisonProcedure
Enter Q Statistic from Table A.5
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Chi-Square Test for
Independence Test whether two categorical variables
are independent H0: the two categorical variables are
independent
H1: the two categorical variables are
dependent
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Example Is gender independent of holding a CPA
in an accounting firm?
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Chi-Square Test for
Independence Test statistic
Reject H0 if2 > 2, (r-1)(c-1) PHStattool available in Multiple Sample
Testsmenu
e
eo
f
ff 22 )( =
where f0= observed frequency
fe= expected frequency if H0 true
in the cells of the contingency table
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ExampleExpected No CPA CPA Total
Female 6.74 7.26 14
Male 6.26 6.74 13Total 13 14 27
Critical value with = 0.05 and (2 - 1)(2 - 1) - 1 df =3.841; therefore, we cannot reject the null hypothesisthat the two categorical variables are independent.
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PHStat Procedure Results
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Design of ExperimentsA test or series of tests that enables the
experimenter to compare two or more
methods to determine which is better,or determine levels of controllablefactors to optimize the yield of a
process or minimize the variability of aresponse variable.
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Factorial Experiments All combinations of levels of each factor are considered.
With m factors at k levels, there are km experiments. Example: Suppose that temperature and reaction time
are thought to be important factors in the percent yield ofa chemical process. Currently, the process operates at atemperature of 100 degrees and a 60 minute reactiontime. In an effort to reduce costs and improve yield, theplant manager wants to determine if changing the
temperature and reaction time will have any significanteffect on the percent yield, and if so, to identify the bestlevels of these factors to optimize the yield.
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Designed ExperimentAnalyze the effect of two levels of each
factor (for instance, temperature at 100
and 125 degrees, and time at 60 and90 minutes)
The different combinations of levels of
each factor are commonly calledtreatments.
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Treatment Combinations
Low
High
Low High
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Experimental Results
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Main Effects Measures the difference in the response that
results from different factor levels Calculations
Temperature effect = (Average yield at high level) (Average yieldat low level)
= (B + D)/2 (A + C)/2
= (90.5 + 81)/2 (84 + 88.5)/2
= 85.75 86.25 = 0.5 percent. Reaction effect = (Average yield at high level) (Average yield at
low level)
= (C + D)/2 (A + B)/2
= (88.5 + 81)/2 (84 + 90.5)/2
= 84.75 87.25 = 2.5 percent.
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Interactions When the effect of changing one factor
dependson the level of other factors.
When interactions are present, wecannotestimate response changes bysimply adding main effects; the effect
of one factor must be interpretedrelative to levels of the other factor.
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Interaction Calculations Take the average difference in response
when the factors are both at the high or lowlevels and subtracting the average difference
in response when the factors are at oppositelevels. Temperature Time Interaction
= (Average yield, both factors at same level)
(Average yield, both factors at opposite levels)= (A + D)/2 (B + C)/2
= (84 + 81)/2 (90.5 + 88.5)/2 = -7.0 percent
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Graphical Illustration ofInteractions
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Two-Way ANOVA
Method for analyzing variation in a 2-factorexperiment
SST = SSA + SSB + SSAB + SSWwhere
SST = total sum of squares
SSA = sum of squares due to factor A
SSB = sum of squares due to factor B
SSAB = sum of squares due to interaction
SSW = sum of squares due to random variation (error)
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Mean Squares
MSA = SSA/(r 1)
MSB = SSB/(c 1)
MSAB = SSAB/(r-1)(c-1)
MSW = SSW/rc(k-1),
where k = number of replications ofeach treatment combination.
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Hypothesis Tests
Compute F statistics by dividing each mean squareby MSW. F = MSA/MSW tests the null hypothesis that means for
each treatment level of factor A are the same against thealternative hypothesis that not all means are equal.
F = MSB/MSW tests the null hypothesis that means foreach treatment level of factor A are the same against the
alternative hypothesis that not all means are equal. F = MSAB/MSW tests the null hypothesis that the
interaction between factors A and B is zero against thealternative hypothesis that the interaction is not zero.
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ExcelAnova: Two-Factor withReplication
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Results
Examine p-values forsignificance