Statistical Inference: A Review of Chapters 12 and 13

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Statistical Inference:A Review of

Chapters 12 and 13

Statistical Inference:A Review of

Chapters 12 and 13

Chapter 14

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14.1 Introduction

• In this chapter 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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Problem objective?

Describe a population Compare two populations

Data type? Data type?

Interval Nominal Interval Nominal

Type of descriptivemeasurement?


Z test &estimator of p


Z test &estimator of p1-p2


Central location Variability Central location Variability

t- test &estimator of


- test &estimator of 2


F- test &estimator of

2/2


2/2Experimental design?

Continue Continue

Summary

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Continue Continue

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Problem objective?

Describe a population Compare two populations

Data type? Data type?

Interval Nominal Interval Nominal







Central location Variability Central location Variability






2/2


2/2Experimental design?

Continue Continue

t- test &estimator of 1-2

(Unequal variances)


(Unequal variances)

Population variances? t- test &estimator of D

t- test &estimator of D


(Equal variances)


(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?


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• Solution– Question 1: Is there sufficient evidence to infer that

the accident rate is lower in ABS equipped cars than in cars without ABS?

– Question 2: 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 3: How much cheaper is it to repair ABS equipped cars than cars without ABS?


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Question 1: Compare the accident rates

• Solution – continued

Problem objective?

Describing a single population Compare two populations

Data type?

Interval Nominal



A car had an accident: Yes / No

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• Solution – continued– p1 = proportion of cars without ABS involved with an

accidentp2 = proportion of cars with ABS involved with an accident

– The hypotheses testH0: p1 – p2 = 0H1: p1 – p2 > 0

Use case 1 test statistic

21

21

n1

n1

)p̂1(p̂

)p̂p̂(Z


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• Solution – continued – Use Test Statistics workbook: z-Test_2

Proportions(Case 1) worksheetz-Test of the Difference Between Two Populations (Case 1)

Sample Sample1 2 z Stat 0.47

Sample Proportion 0.084 0.076 P(Z<=z) 0.3205Sample size 500 500 z Critical one-tail 1.6449Alpha 0.05 P(Z<=z) two-tail 0.641

z Critical two-tail 1.96

42500 38500 Do not reject H0.


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Question 2: Compare the mean repair costs per accident

• Solution - continued

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


(Equal variances)


(Equal variances)

Run the F test for the ratio of two variances.

Equal


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• Solution – continued– 1 = mean cost of repairing cars without ABS

2 = mean cost of repairing cars with ABS– The hypotheses tested

H0: 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


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• Solution – continued– To determine whether the population variances differ

we apply the F test– From Excel Data Analysis we have (Xm14-01)

Do not reject H0.There is insufficientevidence to concludethat the two variances areunequal.


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.


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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.

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Statistical Inference: A Review of Chapters 12 and 13