Lecture 8
description
Transcript of Lecture 8
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Lecture 8
• Matched Pairs
• Review– Summary– The Flow approach to problem solving– Example
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Announcements
Extra office hours:– Today, after class – 1:00.– Sunday, 2-5– Monday 9-11:30. I also should be in my office
most of the afternoon after 2:15
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• Example 13.04 – It was suspected that salary offers were affected by
students’ GPA, (which caused S12 and S2
2 to increase).
– To reduce this variability, the following procedure was used:
• 25 ranges of GPAs were predetermined.• Students from each major were randomly selected, one
from each GPA range.• The highest salary offer for each student was recorded.
– From the data presented can we conclude that Finance majors are offered higher salaries?
The Matched Pairs ExperimentThe Matched Pairs Experiment
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Adminstrative Info for Midterm• Location: Steinberg Hall-Dietrich Hall 351
• Time: Monday (Feb. 10th), 6:00-8:00 p.m.
• Closed book.
• Allowed one double-sided 8.5 x 11 note sheet
• Bring calculator.
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Matched Pairs => One-Sample Test• After taking differences of observations
within each pair, one can continue with a one-sample test with 0:H0 D
GPA Group Finance Marketing1 95171 893292 88009 927053 98089 992054 106322 990035 74566 748256 87089 770387 88664 782728 71200 594629 69367 5155510 82618 81591
. .
. .
. .
Difference5842
-4696-11167319-259
100511039211738178121027
.
.
.
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• Solution 13.04 (by hand)– The parameter tested is D (=1 – 2)
– The hypotheses:H0: D = 0H1: D > 0
– The t statistic:
Finance Marketing
The matched pairs hypothesis testThe matched pairs hypothesis test
Degrees of freedom = nD – 1
The rejection region is t > t.05,25-1 = 1.711
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• Solution 13.04
– Calculate t
647,6
065,5
D
D
s
x
81.325664705065
nsx
tD
DD
The matched pairs hypothesis testThe matched pairs hypothesis test
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Conclusion: There is sufficient evidence to infer at 5% significance level that the Finance MBAs’ highest salary offer is, on the average, higher than that ofthe Marketing MBAs.
The matched pairs hypothesis testThe matched pairs hypothesis test
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Matched Pairs vs. Independent Samples
• Potential advantage of matched pairs: The variance of can be much less than the variance of if the variable(s) on which the pairs are matched account for much of the variation within each group.
• However, if the variance of is close to the variance of , the independent samples design will produce a more powerful test because it has almost twice as many degrees of freedom.
DX
21 XX DX
21 XX
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Recognizing Matched Pairs Experiments
• Does there exist some natural relationship between the first pairs of observations that makes it more appropriate to compare the first pair than the first observation in group 1 and the second observation in group 2?
• Before and after designs• Example: A researcher for OSHA wants to see
whether cutbacks in enforcement of safety regulations coincided with an increase in work related accidents. For 20 industrial plants, she has number of accidents in 1980 and 1995.
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Examples
• Example: A scientist claims that tomatoes will grow large if they are played soft, soothing music. To test the claim, 10 tomatoes grown without music and 10 tomatoes grown with music are randomly selected. The tomatoes are weighed in ounces.
• Example: Many Americans use tax preparation companies to prepare their taxes. In order to investigate whether there are are any differences between companies, an experiment was conducted in which two of the largest companies were asked to prepare the tax returns of a sample of 55 taxpayers. The amounts of tax payable for each taxpayer at the two companies were recorded.
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Caveat
• Keep in mind the difference between observational and experimental data in terms of what can be inferred. – Does the study of finance/marketing majors in
Example 13.4 show that students educated in finance are more attractive to prospective employers?
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Additional Example-Problem 13.75
Tire example contd. (Problem 13.49) Suppose
now we redo the experiment: On 20 randomly
selected cars, one of each type of tire is
installed on the rear wheels and, as before, the
cars are driven until the tires wear out. The
number of miles(in 1000s) is stored in Xr13-
75. Can we conclude that the new tire is superior?
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A Review of Chapters 12 and 13
• Summary of techniques seen
• Here (Chapter 14) we build a framework that helps decide which technique (or techniques) should be used in solving a problem.
• Logical flow chart of techniques for Chapters 12 and 13 is presented next.
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Summary of statistical inference:Chapters 12 and 13
• Problem objective: Describe a population.– Data type: Interval
• Descriptive measurement: Central location– Parameter:
– Test statistic:
– Interval estimator:
– Required condition: Normal population
nsx
t
ns
tx 2
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Summary - continued• Problem objective: Describe a population.
– Data type: Interval• Descriptive measurement: Variability.
– Parameter: s2
– Test statistic:
– Interval estimator:
– Required condition: normal population.
2
22 s)1n(
21
2
2
2
2
2
)1(
)1(
snUCL
snLCL
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Summary - continued• Problem objective: Describe a population.
– Data type: Nominal
– Parameter: p
– Test statistic:
– Interval estimator:
– Required condition:
n)p1(ppp̂
z
n)p̂1(p̂
zp̂ 2
estimatefor(5)p̂1(nand5p̂n
)testfor(5)p1(nand5np
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Summary - continued
• Problem objective: Compare two populations.– Data type: Interval
• Descriptive measurement: Central location– Experimental design: Independent samples
» population variances:
» Parameter: m1 - m2
» Test statistic: Interval estimator:
» Required condition: Normal populations
22
21
)n1
n1
(s
)()xx(t
21
2p
2121
)
n1
n1
(stxx21
2p221
d.f. = n1 + n2 -2
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Summary - continued
• Problem objective: Compare two populations.– Data type: Interval.
• Descriptive measurement: Central location– Experimental design: Independent samples
» population variances:
» Parameter: m1 - m2
» Test statistic: Interval estimator:
» Required condition: Normal populations
22
21
2
22
1
21
2121 )()(
n
s
n
s
xxt
2
22
1
21
221 n
s
n
stxx
1
)(
1
)(
)/(d.f.
2
22
22
1
21
21
22
221
21
n
ns
n
ns
nsns
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Summary - continued
• Problem objective: Compare two populations.– Data type: Interval.
• Descriptive measurement: Central location– Experimental design: Matched pairs
» Parameter: mD
» Test statistic: Interval estimator:
» Required condition: Normal differences
DD
DD
ns
xt
D
DnD
n
stx 1,2/
d.f. = nD - 1
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Problem objective?
Describe a population Compare two populations
Data type? Data type?
Interval Nominal Interval Nominal
Type of descriptivemeasurement?
Type of descriptivemeasurement?
Z test &estimator of p
Z test &estimator of p
Z test &estimator of p1-p2
Z test &estimator of p1-p2
Central location Variability Central location Variability
t- test &estimator of
t- test &estimator of
- test &estimator of 2
- test &estimator of 2
F- test &estimator of
2/2
F- test &estimator of
2/2Experimental design?
Continue Continue
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Continue Continue
t- test &estimator of 1-2
(Unequal variances)
t- test &estimator of 1-2
(Unequal variances)
Population variances? t- test &estimator of D
t- test &estimator of D
t- test &estimator of 1-2
(Equal variances)
t- test &estimator of 1-2
(Equal variances)
Independent samples Matched pairs
Experimental design?
UnequalEqual
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Identifying the appropriate technique
• Example 14.1– Is the antilock braking system (ABS) really effective?– Two aspects of the effectiveness were examined:
• The number of accidents. • Cost of repair when accidents do occur.
– An experiment was conducted as follows:• 500 cars with ABS and 500 cars without ABS were randomly
selected.• For each car it was recorded whether the car was involved in an
accident.• If a car was involved with an accident, the cost of repair was
recorded.
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• Example – continued– Data
• 42 cars without ABS had an accident,
• 38 cars equipped with ABS had an accident
• The costs of repairs were recorded (see Xm14-01).
– Can we conclude that ABS is effective?
Identifying the appropriate technique
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• Solution– Question 1: Is there sufficient evidence to infer
that the cost of repairing accident damage in ABS equipped cars is less than that of cars without ABS?
– Question 2: How much cheaper is it to repair ABS equipped cars than cars without ABS?
Identifying the appropriate technique
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Question 2: Compare the mean repair costs per accident
• Solution
Problem objective?
Describing a single population Compare two populations
Data type?
Interval Nominal
Type of descriptivemeasurements?
Central location Variability
Cost of repair per accident
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Equal
• Solution - continued
Population variancesequal?
Independent samples Matched pairs
Unequal
Experimental design?
Central location
t- test &estimator of 1-2
(Equal variances)
t- test &estimator of 1-2
(Equal variances)
Run the F test for the ratio of two variances.
Equal
Question 2: Compare the mean repair costs per accident
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• Solution – continued– 1 = mean cost of repairing cars without ABS
2 = mean cost of repairing cars with ABS
– The hypotheses testedH0: 1 – 2 = 0H1: 1 – 2 > 0
– For the equal variance case we use
2nn.f.d
)n1
n1
(s
)()xx(t
21
21
2p
21
Question 2: Compare the mean repair costs per accident
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• Solution – continued– To determine whether the population variances
differ we apply the F test– From JMP we have (Xm14-01)
Do not reject H0.There is insufficientevidence to concludethat the two variances areunequal.
Question 2: Compare the mean repair costs per accident
F-Test Two-Sample for Variances
Cost 1991 Cost 1992Mean 2075 1714Variance 450343 390409Observations 42 38df 41 37F 1.15P(F<=f) one-tail 0.3313F Critical one-tail 1.7129
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Mean 2075 1714Variance 450343 390409Observations 42 38Pooled Variance 421913Hypothesized Mean Difference0df 78t Stat 2.48P(T<=t) one-tail 0.0077t Critical one-tail 1.6646P(T<=t) two-tail 0.0153t Critical two-tail 1.9908
• Solution – continued – Assuming the variances are really equal we run
the equal-variances t-test of the difference between two means
At 5% significance levelthere is sufficient evidenceto infer that the cost of repairsafter accidents for cars with ABS is smaller than the cost of repairs for cars without ABS.
Question 2: Compare the mean repair costs per accident
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Checking required conditions
• The two populations should be normal (or at least not extremely nonnormal)
Cost without ABS
0
5
10
15
900 1400 1900 2400 2900 3400 More
Frequency
0
5
10
15
20
900 1400 1900 2400 2900 3400 3900 More
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Question 3: Estimate the difference in repair costs
• Solution– Use Estimators Workbook: t-Test_2 Means
(Eq-Var) worksheett-Estimate of the Difference Between Two Means (Equal-Variances)
Sample 1 Sample 2 Confidence Interval EstimateMean 2075 1714 361 +/-' 290Variance 450343 390409 Lower confidence limit 71Sample size 42 38 Upper confidence limit 651Pooled Variance 421913Confidence level 0.95
We estimate that the cost of repairing a car not equipped with ABS is between $71 and $651 more expensive than to repair an ABS equipped car.