chapter 10 comparisons involving means

Chapter 10 Comparisons Involving

Means Inferences About the Difference Between

Two Population Means: when s 1 and s 2 Known

Inferences About the Difference Between Two Population Means: Matched Samples

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

Introduction to Analysis of Variance (Inference about the difference between more than two population means

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 m1 equal the mean of population 1 and m2 equal

the mean of population 2. The difference between the two population means is m1 - m2. To estimate m1 - m2, 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 m m

Standard Deviation (Standard Error)

s s sx x n n1 2

where: s1 = standard deviation of population 1 s2 = standard deviation of population 2

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

Interval Estimate

Interval Estimation of m1 - m2: s 1 and s 2 Known

2 21 2

1 2 / 21 2

x x zn ns s

where: 1 - is the confidence coefficient

Example:

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 on thenext slide.

Par, Inc. is a manufacturerof golf equipment and hasdeveloped a new golf ballthat has been designed toprovide “extra distance.”

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.

Develop a 95% confidence interval estimate of 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 Ballsm1 = mean driving

distance of Pargolf balls

Population 2Rap, Ltd. Golf Ballsm2 = mean driving

distance of Rapgolf balls

Simple random sample of n2 Rap golf ballsx2 = sample mean distance for the Rap golf balls

Simple random sample of n1 Par golf ballsx1 = sample mean distance for the Par golf balls

Point Estimate of m1 - m2

Point estimate of m1 m2 = x x1 2

where:m1 = mean distance for the population of Par, Inc. golf ballsm2 = mean distance for the population of Rap, Ltd. golf balls

= 275 258

= 17 yards

x x zn n1 2 2

2 217 1 96 15

1202080

/ . ( ) ( )

Interval Estimation of m1 - m2:s 1 and s 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

Hypothesis Testing

1 2 02 21 2

( )x x Dz

n ns s

m m 1 2 0: aH Dm m 0 1 2 0: H Dm m 0 1 2 0: H D

m m 1 2 0: aH Dm m 0 1 2 0: H Dm m 1 2 0: aH D

Left-tailed Right-tailed Two-tailed Test Statistic

Can we conclude, using = .01, that the mean drivingdistance of Par, Inc. golf ballsis greater than the mean drivingdistance of Rap, Ltd. golf balls?

H0: m1 - m2 < 0

Ha: m1 - m2 > 0

where: m1 = mean distance for the population of Par, Inc. golf balls m2 = mean distance for the population of Rap, Ltd. golf balls

1. Develop the hypotheses.

2. Specify the level of significance. = .01

3. Compute the value of the test statistic.

1 2 02 21 2

( )x x Dz

2 2(235 218) 0 17 6.492.62(15) (20)

120 80

5. Compare the Test Statistic with the Critical Value.Because z = 6.49 > 2.33, we reject H0.

Using the Critical Value Approach

For = .01, z.01 = 2.334. Determine the critical value and rejection rule.

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

Using the 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 < = .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.

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

Interval Estimation of m 1 – m 2

Hypothesis Tests About m 1 – m 2

Interval Estimation of m1 - m2:s 1 and s 2 Unknown

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

• replace z/2 with t/2.

• we 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 tn n

Where the degrees of freedom for t/2 are:

Interval Estimation of m1 - m2: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

Difference Between Two Population Means:

s 1 and s 2 Unknown

Specific Motors of Detroithas developed a new automobileknown as the M car. 24 M carsand 28 J cars (from Japan) were roadtested to compare miles-per-gallon (mpg) performance. The sample statistics are shown on the next slide.

s 1 and s 2 Unknown

Sample Size

Sample MeanSample Std. Dev.

Sample #1M Cars

Sample #2J Cars

24 cars 28 cars

29.8 mpg 27.3 mpg2.56 mpg 1.81 mpg

s 1 and s 2 Unknown

Develop a 90% confidenceinterval estimate of the differencebetween the mpg performances ofthe two models of automobile.

Point estimate of m1 m2 =x x1 2

Point Estimate of m 1 m 2

where:m1 = mean miles-per-gallon for the population of M carsm2 = 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 t/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 /2 = .05 and df = 24, t/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 1.71124 28

s sx x tn n

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

Hypothesis Testing

1 2 02 21 2

( )x x Dts sn n

m m 1 2 0: aH Dm m 0 1 2 0: H Dm m 0 1 2 0: H D

m m 1 2 0: aH Dm m 0 1 2 0: H Dm m 1 2 0: aH D

Left-tailed Right-tailed Two-tailed Test Statistic

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

H0: m1 - m2 = 0

Ha: m1 - m2 > 0where: m1 = mean mpg for the population of M cars m2 = mean mpg for the population of J cars

2. Specify the level of significance.

1 2 02 2 2 21 2

( ) (29.8 27.3) 0 4.003(2.56) (1.81)

x x Dts sn n

Using the p –Value Approach4. Compute the p –value.

Compute the the degrees of freedom: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

4. Determine the critical value and rejection rule.

For = .05 and df = 24, t.05 = 1.711

5. Compare the Test Statistic with the Critical Value.

As t he test statistic 4.003 is greater than the Critical value 1.711, 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?.

5. Compute the P-value and determine whether to 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?.

As the p–value < = .05, we reject H0.

Using the p-Value Approach

In 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: A Chicago-based firm hasdocuments that must be quicklydistributed to district officesthroughout the U.S. The firmmust decide between two deliveryservices, UPX (United Parcel Express) and INTEX(International Express), to transport its documents.

In testing the delivery timesof the two services, the firm senttwo reports to a random sampleof its district offices with onereport carried by UPX and theother report carried by INTEX. Do the data on thenext 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: md = 0

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

2. Specify the level of significance. = .05

d dni ( ... ) .7 6 5

s d dndi

( ) . .2

2.7 0 2.942.9 10d

4. Determine the critical value and rejection rule. Using the Critical Value Approach

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

5. Compare the Test Statistic with the Critical ValueBecause 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?

5. Determine whether to reject H0.

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.)

Because p–value < = .05, we reject H0.

Using the p –Value Approach

chapter 10 comparisons involving means

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