chapter 14

14.1Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.

Chapter 14

Statistical Inference:Review of Chapters 12 & 13


Which technique to use?


Identifying the Correct Technique…

The two most important factors in determining the correct statistical technique to use are: the problem objective,

(i.e. describe one population or compare two populations) and the data type.

(i.e. interval data or nominal data)

Once these factors are determined, our analysis extends to other factors (e.g. type of descriptive measure [central location? variability?], etc.)


Figure 14.1…

The flowchart in your textbook (Figure 14.1) describes the logical process that allows us to identify the appropriate method to use for the problem.

Start at the top and work

your way down the chart…

The following slides are

an interactive version of

this flowchart…


Figure 14.1: Flowchart of Techniques…

Problem objective?

Describe a population Compare two populations

Click on the mouse icon to follow the branch of the flowchart to the

next level…

Skip flowchart,go to examples…



Problem objective?

Describe a population

Data type?

Interval Nominal



Problem objective?


Data type?

Interval Type of descriptivemeasurement?

Central location Variability


Slide 12.19 : Identifying Factors…

Factors that identify the t-test and estimator of :Top of

Flowchart



Factors that identify the chi-squared test and

estimator of :

Top ofFlowchart



Problem objective?


Data type?

Nominal



Factors that identify the z-test and

interval estimator of p:

Top ofFlowchart



Problem objective?

Compare two populations

Data type?

Interval Nominal



Problem objective?


Data type?

Nominal



Factors that identify the z-test

and estimator for p1–p2

Top ofFlowchart



Problem objective?


Data type?

Interval

Type of descriptivemeasurement?




Problem objective?


Data type?

Interval

Type of descriptivemeasurement?

Variability



Factors that identify the F-test and

estimator of :

Top ofFlowchart



Problem objective? Compare two populations Data type?

IntervalType of descriptive

measurement?Central Location

Independent samples

Matched pairsExperimental design?






Matched pairsExperimental design?



Factors that identify the t-test and estimator of :Top of

Flowchart






Independent samples

Experimental design?

Population variances?

UnequalEqual






Independent samples



Equal



Factors that identify the equal-variances t-test

and estimator of :

Top ofFlowchart






Independent samples



Unequal



Factors that identify the unequal-variances t-test and estimator of :

Top ofFlowchart


Example 14.1…

Is anti-lock braking system (ABS) in cars really effective?

We would expect if it were effective that:

The number of accidents would decrease, and

The cost of accident repairs would be less.

Data were collected on 500 cars with ABS and 500 cars without. The number of cars involved in accidents was recorded, as was the cost of repairs. What can we conclude?


Example 14.1 (a)…

Is there sufficient evidence to infer that the accident rate is lower in ABS-equipped cars than in cars without ABS?(If ABS is effective, we would expect a lower accident rate in ABS-equipped cars.)

Accident rate = number of cars in accidents

total number of cars

This is nominal (i.e. categorical) data; either a car had an accident or it didn’t. The accident rate is a proportion. We want to compare cars with ABS against cars without.

IDENTIFY


Example 14.1 (a)…

Identify the correct technique…

IDENTIFY

Problem objective?

Describing a single population Compare two populations

Data type?

Interval Nominal


Example 14.1 (a)…

The correct technique:

has been identified. The next step is to translate the rest of the problem into the symbols and language of statistics:

p1 = proportion of cars without ABS involved in an accident

p2 = proportion of cars with ABS involved in an accident

We want to test if ABS is effective, that is, we want to research if: p1 > p2 , that is if H1: (p1 – p2 ) > 0

IDENTIFY


Example 14.1 (a)…

Since H1: (p1 – p2 ) > 0

we have our null hypothesis: H0: (p1 – p2 ) = 0

Hence this is a Case 1 type problem.

Upon calculating our sample proportions…

…we can use Excel to complete our analysis…

COMPUTE


Example 14.1 (a)…

Noting that z = .4663 is not greater than zCritical = 1.6449 (or alternatively, by looking at the p-value of .3205), we cannot reject H0 in favor of H1, that is, there is not enough evidence to infer that ABS equipped cars have fewer accidents than non-ABS equipped cars…

INTERPRET


Example 14.1 (b)…

When accidents do occur, we would expect the severity of accidents to be lower in ABS-equipped cars (assuming that ABS is effective), thus we are interested in this question:

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?

The cost of repairs is interval data. We need a measure to compare the two populations of cars in a meaningful way…

IDENTIFY


Example 14.1 (b)…

Identify the correct technique…

IDENTIFY

Problem objective?

Describing a single population Compare two populations

Data type?

Interval Nominal

Type of descriptivemeasurements?


…continues…


Example 14.1 (b)…

Identify the correct technique (continued)…

IDENTIFY

Equal??

Population variancesequal?

Independent samples Matched pairs

Unequal??


Central location

e.g. we are not comparing a head-on

collision of an ABS equipped car with a

head-on collision of a non-ABS car…

Which one is it?


Example 14.1 (b)…


IDENTIFY

Equal??


Unequal??

Apply theF-test of

there is not enough evidence to infer

that the variances differ…


Example 14.1 (b)…


…we have the right technique! Let’s proceed with our hypotheses set-up…

IDENTIFY

Equal


Unequal


Example 14.1 (b)…

We want to research whether or not the mean cost of repairing cars without ABS brakes (population 1) is greater than the mean cost of repair of cars equipped with ABS brakes (population 2), i.e.:

IDENTIFY


Example 14.1 (b)…

Applying the Data Analysis tools in Excel to our data…

Indeed, there is sufficient evidence to support the belief that non-ABS equipped cars do indeed have higher accident repair costs than ABS equipped cars.

INTERPRET


Example 14.1 (c)…

In part (b) we’ve shown that ABS-equipped cars suffer less damage in accidents (as measured by repair costs); can we estimate how much cheaper they are to repair on average compared to cars without ABS brakes.

Our path through the flowchart is the same, that is, we are comparing the measure of central location of independent samples of interval data from two populations who’s variances are equal…

IDENTIFY


Example 14.1 (c)…

The estimator of interest is:

Assuming a 95% confidence interval…

…we estimate the cost of repair for a non-ABS equipped car of between $71 and $651 over an ABS equipped car.

CALCULATE

chapter 14

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