Sampling Distribution of

Sampling Distribution of 1 2x x

1 2-

2 21 2

1 2n n

2x1xIf and are normally distributed and samples 1 and 2 are independent, their difference is

Sampling Distribution of 1 2x x

1 2-

2 21 2

1 2

s sn n

2x1xIf and are normally distributed and samples 1 and 2 are independent, their difference is

22 21 2

1 22 22 2

1 2

1 1 2 2

1 11 1

s sn n

dfs s

n n n n

The normal distribution is not used if 1 and 2 are unknown. The t-distribution is used instead with

Sampling Distribution of

n1p1 > 5 n1(1 p1) > 5n2p2 > 5 n2(1 p2) > 5

1 2p p

21 1 1(1 )p p 2

2 2 2(1 )p p

1 2-p p

2 21 2

1 2n n

If and are normally distributed and samples 1 and 2 are independent, their difference is

1p 2p

The sample proportions are pooled (averaged) when testing the equality of proportions.

Sampling Distribution of 1 2p p

21 (1 )p p 2

2 (1 )p p

1 2-p p

2 21 2

1 2n n

If and are normally distributed and samples 1 and 2 are independent, their difference is

1p 2p

The sample proportions are pooled (averaged) when testing whether the proportions are equal or not.

1 21 2

1 2

n nnp p

np

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 manufacturer of golf equipment and has developed a new golf ball that has been designed to provide “extra distance.”

Example 1

Interval estimate of 1 2

Sample SizeSample Mean

Sample #1Par, Inc.

Sample #2Rap, Ltd.

120 balls 80 balls275 yards 258 yards

Based on data from previous driving distance tests, the two population standard deviations are 15 and 20 yards, respectively. Develop a 95% confidence interval estimate of the difference between the mean driving distances of the two brands of golf balls.

x1 x2 = 275 – 258 = 17

Example 1


/

2 21 2

1 21 2

2x xn n

z

17 + 5.14 yards11.86 yards to 22.14 yards

.

2 2

0250(15) (20)275 258120 80

z

1.9617 6.875

We are 95% confident that the difference between the mean driving distances of Par and Rap golf balls is

between 11.86 to 22.14 yards.

Example 1


Specific Motors of Detroit has developed the M car. 24 M cars and 28 J cars (from Japan) were road tested to compare miles-per-gallon (MPG) performance. The sample statistics are given below. Develop a 90% confidence interval estimate of the difference between the MPG performances of the two models of automobile.

Example 2

nM Cars J Cars

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

Point estimate of 1 2= 29.8 – 27.3= 2.5 MPG

= x1 – x2

xs


22 2

2 22 2

2.56 1.81

2.5

( ) ( )24 28

1 ( ) 1 ( )24 1 24 28 1 2

6 1.8

81

The degrees of freedom are:

df

With 1 = .9000

t.0500 = 1.684

/2 =df =

= .1000

40.585

2

2 2

.2731 .1170

.2731 .117023 27

(the row of t-table).0500(the column of t-table)

40

40

-t.0500 = -1.684

Example 2


2.5 + 1.051

2 22.529.8 6( ) ( ) 1.6842

27.4 28

1.813

2 2

/ 21

1 21

1 2

( ) tns s

nx x

1.448 to 3.552 mpg

We are 90% confident that the difference between the MPG performances of M cars and J cars is

1.448 to 3.552 mpg

Example 2


Market Research Associates is conducting research to evaluate the effectiveness of a client’s new advertising campaign. Before the new campaign began, a telephone survey of 150 households in the test market area showed 60 households “aware” of the client’s product.

The new campaign has been initiated with TV and newspaper advertisements running for three weeks. A surveyconducted immediately after the new campaign showed 120of 250 households “aware” of the client’s product. Develop a 95% confidence interval estimate of the difference betweenthe proportion of households that are aware of the client’s product.

2p 60150.40

Example 3

1p 120250.48 1 2p p .08

Interval estimate of 1 2p p

.48(1 .48) .40(1 .40).48 .40 1.96250 150

z.0250 = 1.96

.08 + .10

1 1 2 21 2 / 2

1 2

(1 ) (1 )a

p p p pp p zn n

1 = .9500 /2 = = .0500 .0250Example 3

Interval estimate of 1 2p p

We are 95% confident that the “change in awareness”due to the campaign is between -0.02 and 0.18…

Can we conclude, using a 5% level of significance, that the mean driving distance of Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls?

1 2 1 2 2 20

1 2 0

Hypothesis Tests About 1 2

Ha: 1 – 2 > 0

1. Develop the hypotheses.

H0: 1 – 2 < 0

Example 1 (continued)

2. Determine the critical value = .0500

3. Compute the value of the test statistic.

-statz 1 2 02 21 2

1 2

( )

n n

Dx x

6.49

2 2

( )(15) (20)120

275 258 0

80

/ 1 .0500 z.0500 = 1.645


Do Not Reject H0 Reject H0

z 1.645 6.49

z-stat0

At the 5% level of significance, the sample evidence indicates the mean driving distance of

Par, Inc. golf balls is greater than the mean driving distance of Rap, Ltd. golf balls.

.050

4. Reject or do not reject the null hypothesis


Can we conclude, using a 5% level of significance, that the miles-per-gallon (MPG) performance of M cars is greater than the MPG performance of J cars?Recall


nM Cars J Cars

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

xs

1. Develop the hypotheses. H0: 1 - 2 < 0Ha: 1 - 2 > 0


/1 = .050 (column) t.050 = 1.684df = 40 (row)


3. Compute the value of the test statistic.1 2

22

2

1 2

0

1

( )-stat

x

s

x Dt

n ns

2 22.56

( )

( ) ( )24

029.8 27

1.828

.3

1

4.003

2. Determine the critical value.


t 1.684

t050

4.003 t-stat

0

At 5% significance, fuel economy of M cars is greater than

the mean fuel economy of

J cars.

.050



http://www.halsnarr.com/ppt/data_pelican.xlsx

Can we conclude, using a 5% level of significance, that the proportion of households aware of the client’s product increased after the new advertising campaign?

1120250 .48p 2

60150 .40p

Hypothesis Tests About p1 p2


1. Develop the hypotheses. H0: p1 p2 < 0Ha: p1 p2 > 0

2. Determine the critical value. = .0500 z.05 = 1.645


1.57..080510

.48 .( )-s 0t .05104at 0z

z


0

We cannot conclude that the proportion of households aware of the client’s product increased after the new campaign.

1.57 z-stat

1.645z.050

.050



John is a political candidate who wants to know if female support for his candidacy is the same as his male support using a 5% level of significance. A survey conducted shows 310 of 620 females surveyed said they will vote for John while the same survey said 362 of 680 males said they will vote for John.

1310620 .5p 2

362680 .532p

1 1 22

1 2

p pn nn n

0.5 0.532620 680620 680

310 362620 680 .5169p


Example 4

1. Develop the hypotheses.Ha: p1 p2 ≠ 0H0: p1 p2 = 0

2. Determine the critical value.


= .0500 / 2 .0250 ± z.0250 = ± 1.96

1.15.0

.0322775

.5 .5( )- 0st .032

5at 277z


1 2p ps .5169(1 .5169) .5169(1 .5169)

620 680

.02775

-1.96 1.96

1 – a = .9700

.0250.0250

0


-1.15z-stat

Reject H0 Do Not Reject H0 Reject H0

We cannot conclude that support from female voters is different from support from male voters.


Example: Express Deliveries

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

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 on the next slide indicate a difference in mean delivery times for the two services? Use a 5% level of significance.

Matched Samples

3230191615181410 716

25241515131515 8 911

UPX INTEX Difference (d)District OfficeSeattleLos AngelesBostonClevelandNew YorkHoustonAtlantaSt. LouisMilwaukeeDenver

Delivery Time (Hours)

7

6

4

1 2 3 -1 2 -2

5 Sum = 27

2.7d

Matched Samples

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

7– 2.76 – 2.74 – 2.71 – 2.72 – 2.73 – 2.7

-1 – 2.72 – 2.7

-2 – 2.75 – 2.7

4.33.31.3

-1.7-0.70.3

-3.7-0.7-4.72.3

18.4910.891.692.890.490.0913.690.4922.095.29

id id d 2( )id did d

Sum = 76.1 sd = 8.5

sd = 2.9

22( )

1di d

sdn

Matched Samples

H0: d = 0

with d = 1 – 21. Develop the hypotheses

A difference existsNo difference exists

2. Determine the critical values with = .050.

/2 = .025 (the column of the t-table)df = 10 – 1 = 9

Since this is atwo-tailed test

t.025 = 2.262 t.025 = 2.262

(the row of the t-table)

Ha: d

02.72.9 10-statt 2.94

0d

ds n


Matched Samples

2.262 2.262 t.025 t.025

2.94 t-stat

0

At the 5% level of significance, the sample evidence indicates there is a difference in mean delivery times for the two services.

Reject H0 Do Not Reject Reject H0

.0250.0250

H0: d = 0


Matched Samples

http://www.halsnarr.com/ppt/data_unemp.xlsx

Documents

Sampling Distribution of