lecture 8 matched pairs review –summary –the flow approach to problem solving –example

Lecture 8

• Matched Pairs

• Review– Summary– The Flow approach to problem solving– Example

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

• 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

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.

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

.

.

.

• 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

• 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

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

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

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.

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.

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?

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?

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.

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

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

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

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

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

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

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

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

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.

• 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

• 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

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

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

• 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

• 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

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

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

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.