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John LoucksSt. Edward’sUniversity

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Chapter 10, Part A Comparisons Involving Means, Experimental

Design, and Analysis of Variance

Inferences About the Difference Between Two Population Means: s 1 and s 2 Known

Inferences About the Difference Between Two Population Means: Matched Samples

Inferences About the Difference Between Two Population Means: s 1 and s 2 Unknown

Inferences About the Difference BetweenTwo Population Means: s 1 and s 2 Known

Interval Estimation of m 1 – m 2

Hypothesis Tests About m 1 – m 2

Estimating the Difference BetweenTwo Population Means

Let 1 equal the mean of population 1 and 2 equal

the mean of population 2. The difference between the two population means is 1 - 2. To estimate 1 - 2, we will select a simple random

sample of size n1 from population 1 and a simple

random sample of size n2 from population 2. Let equal the mean of sample 1 and

equal the mean of sample 2.

The point estimator of the difference between the

means of the populations 1 and 2 is .x x1 2

Expected Value

Sampling Distribution of x x1 2

E x x( )1 2 1 2

Standard Deviation (Standard Error)

x x n n1 2

where: 1 = standard deviation of population 1 2 = standard deviation of population 2

n1 = sample size from population 1 n2 = sample size from population 2

Interval Estimate

Interval Estimation of 1 - 2: s 1 and s 2 Known

2 21 2

1 2 / 21 2

x x zn n

where: 1 - is the confidence coefficient

In a test of driving distance using a mechanicaldriving device, a sample of Par golf balls wascompared with a sample of golf balls made by Rap,Ltd., a competitor. The sample statistics appear onthe next slide.

Par, Inc. is a manufacturer of golf equipment andhas developed a new golf ball that has been designed to provide “extra distance.”

Example: Par, Inc.

Sample SizeSample Mean

Sample #1Par, Inc.

Sample #2Rap, Ltd.

120 balls 80 balls275 yards 258 yards

Based on data from previous driving distancetests, the two population standard deviations areknown with s 1 = 15 yards and s 2 = 20 yards.

Example: Par, Inc.

Let us develop a 95% confidence interval estimateof the difference between the mean driving distances ofthe two brands of golf ball.

Estimating the Difference BetweenTwo Population Means

m1 – m2 = difference between the mean distances

x1 - x2 = Point Estimate of m1 – m2

Population 1Par, Inc. Golf Balls

m1 = mean driving distance of Par

golf balls

Population 2Rap, Ltd. Golf Balls

m2 = mean driving distance of Rap

golf balls

Simple random sample of n2 Rap golf balls

x2 = sample mean distance for the Rap golf balls

Simple random sample of n1 Par golf balls

x1 = sample mean distance for the Par golf balls

Point Estimate of 1 - 2

Point estimate of 1 - 2 =x x1 2

where:1 = mean distance for the population of Par, Inc. golf balls2 = mean distance for the population of Rap, Ltd. golf balls

= 275 - 258

= 17 yards

x x zn n1 2 2

17 1 9615120

/ .( ) ( )

Interval Estimation of 1 - 2: 1 and 2 Known

We are 95% confident that the difference betweenthe mean driving distances of Par, Inc. balls and Rap,Ltd. balls is 11.86 to 22.14 yards.

17 + 5.14 or 11.86 yards to 22.14 yards

Hypothesis Tests About m 1 - m 2:s 1 and s 2 Known

Hypotheses

2 21 2

( )x x Dz

1 2 0: aH D 0 1 2 0: H D 0 1 2 0: H D

1 2 0: aH D 0 1 2 0: H D 1 2 0: aH D

Left-tailed Right-tailed Two-tailed

Test Statistic

Example: Par, Inc.

Can we conclude, using a = .01, that themean driving distance of Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls?

H0: 1 - 2 < 0

Ha: 1 - 2 > 0where: 1 = mean distance for the population of Par, Inc. golf balls 2 = mean distance for the population of Rap, Ltd. golf balls

1. Develop the hypotheses.

p –Value and Critical Value Approaches

2. Specify the level of significance. a = .01

3. Compute the value of the test statistic.

2 21 2

( )x x Dz

(235 218) 0 17 6.49

2.62(15) (20)120 80

p –Value Approach

4. Compute the p–value.

For z = 6.49, the p –value < .0001.

5. Determine whether to reject H0.

Because p–value < a = .01, we reject H0.

At the .01 level of significance, the sample evidenceindicates the mean driving distance of Par, Inc. golfballs is greater than the mean driving distance of Rap,Ltd. golf balls.

Because z = 6.49 > 2.33, we reject H0.

Critical Value Approach

For a = .01, z.01 = 2.33

4. Determine the critical value and rejection rule.

Reject H0 if z > 2.33

The sample evidence indicates the mean drivingdistance of Par, Inc. golf balls is greater than the meandriving distance of Rap, Ltd. golf balls.

Inferences About the Difference BetweenTwo Population Means: s 1 and s 2

Unknown Interval Estimation of m1 – m2

Hypothesis Tests About m1 – m2

Interval Estimation of 1 - 2:s 1 and s 2 Unknown

When s 1 and s 2 are unknown, we will:

• replace za/2 with ta/2.

• use the sample standard deviations s1 and s2

as estimates of s 1 and s 2 , and

2 21 2

1 2 / 21 2

s sx x t

Where the degrees of freedom for ta/2 are:

Interval Estimation of 1 - 2:s 1 and s 2 Unknown

Interval Estimate

22 21 2

1 22 22 2

1 1 2 2

1 11 1

s sn n

n n n n

Example: Specific Motors

Difference Between Two Population Means:

s 1 and s 2 Unknown

Specific Motors of Detroit has developed a newAutomobile known as the M car. 24 M cars and 28 Jcars (from Japan) were road tested to compare miles-per-gallon (mpg) performance. The sample statisticsare shown on the next slide.

s 1 and s 2 Unknown Example: Specific Motors

Sample SizeSample MeanSample Std. Dev.

Sample #1M Cars

Sample #2J Cars

24 cars 28 cars29.8 mpg 27.3 mpg2.56 mpg 1.81 mpg

s 1 and s 2 Unknown

Let us develop a 90% confidence interval estimate of the difference between the mpg performances of the two models of automobile.

Point estimate of 1 - 2 =x x1 2

Point Estimate of m 1 - m 2

where:1 = mean miles-per-gallon for the population of M cars2 = mean miles-per-gallon for the population of J cars

= 29.8 - 27.3

= 2.5 mpg

Interval Estimation of m 1 - m 2:s 1 and s 2 Unknown

The degrees of freedom for ta/2 are:22 2

2 22 2

(2.56) (1.81)24 28

24.07 241 (2.56) 1 (1.81)

24 1 24 28 1 28

With a/2 = .05 and df = 24, ta/2 = 1.711

Interval Estimation of m 1 - m 2:s 1 and s 2 Unknown

2 2 2 21 2

1 2 / 21 2

(2.56) (1.81) 29.8 27.3 1.711

s sx x t

We are 90% confident that the difference betweenthe miles-per-gallon performances of M cars and J carsis 1.431 to 3.569 mpg.

2.5 + 1.069 or 1.431 to 3.569 mpg

Hypothesis Tests About m 1 - m 2:s 1 and s 2 Unknown

Hypotheses

2 21 2

( )x x Dt

s sn n

1 2 0: aH D 0 1 2 0: H D 0 1 2 0: H D

1 2 0: aH D 0 1 2 0: H D 1 2 0: aH D

Left-tailed Right-tailed Two-tailed Test Statistic

Can we conclude, using a .05 level of significance, that the miles-per-gallon (mpg) performance of M cars is greater than the miles-per-gallon performance of J cars?

H0: 1 - 2 < 0

Ha: 1 - 2 > 0where: 1 = mean mpg for the population of M cars 2 = mean mpg for the population of J cars

2. Specify the level of significance.

a = .05

2 2 2 21 2

( ) (29.8 27.3) 0 4.003

(2.56) (1.81)24 28

x x Dt

s sn n

p –Value Approach

4. Compute the p –value.

The degrees of freedom for ta are:

Because t = 4.003 > t.005 = 1.683, the p–value < .005.

2 22 2

(2.56) (1.81)24 28

40.566 411 (2.56) 1 (1.81)

24 1 24 28 1 28

We are at least 95% confident that the miles-per-gallon (mpg) performance of M cars is greater than the miles-per-gallon performance of J cars.

p –Value Approach

For a = .05 and df = 41, t.05 = 1.683

Reject H0 if t > 1.683

Because 4.003 > 1.683, we reject H0.

We are at least 95% confident that the miles-per-gallon (mpg) performance of M cars is greater than the miles-per-gallon performance of J cars.

With a matched-sample design each sampled item provides a pair of data values.

This design often leads to a smaller sampling error

than the independent-sample design because variation between sampled items is eliminated as a source of sampling error.

Inferences About the Difference BetweenTwo Population Means: Matched Samples

Example: Express Deliveries

A Chicago-based firm has documents that mustbe quickly distributed to district offices throughout the U.S. The firm must decide between two deliveryservices, UPX (United Parcel Express) and INTEX(International Express), to transport its documents.

Example: Express Deliveries

In testing the delivery times of the two services,the firm sent two reports to a random sample of its district offices with one report carried by UPX and the other report carried by INTEX. Do the data onthe next slide indicate a difference in mean deliverytimes for the two services? Use a .05 level ofsignificance.

3230191615181410 716

25241515131515 8 911

UPX INTEX DifferenceDistrict OfficeSeattleLos AngelesBostonClevelandNew YorkHoustonAtlantaSt. LouisMilwaukeeDenver

Delivery Time (Hours)

7 6 4 1 2 3 -1 2 -2 5

H0: d = 0

Ha: d Let d = the mean of the difference values for the two delivery services for the population of district offices

2. Specify the level of significance. a = .05

ddni ( ... )

.7 6 5

sd dndi

( ) ..

2.7 0 2.94

2.9 10d

We are at least 95% confident that there is a difference in mean delivery times for the two services.

4. Compute the p –value.

For t = 2.94 and df = 9, the p–value is between.02 and .01. (This is a two-tailed test, so we double the upper-tail areas of .01 and .005.)

p –Value Approach

For a = .05 and df = 9, t.025 = 2.262.

Reject H0 if t > 2.262

Because t = 2.94 > 2.262, we reject H0.

We are at least 95% confident that there is a difference in mean delivery times for the two services.

End of Chapter 10Part A

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