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Statistics. Statistical Inference About Means and Proportions With Two Populations. STATISTICS in PRACTICE. Statistics plays a major role in pharmaceutical research. Statistical methods are used to test and develop new drugs.

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statistics

Statistics

Statistical Inference About Means and Proportions With Two Populations

statistics in practice
STATISTICSin PRACTICE
  • Statistics plays a major role in

pharmaceutical research.

Statistical methods are used to

test and develop new drugs.

  • In most studies, the statistical method involves hypothesis testing for the difference between the means of the new drug population and the standard drug population.
statistics in practice1
STATISTICSin PRACTICE
  • In this chapter you will learn how to construct interval estimates and make hypothesis tests about means and proportions with two populations.
contents
Contents
  • Inferences About the Difference Between Two Population Means: s 1 and s2 Known
  • Inferences About the Difference Between Two Population Means: s 1 and s2 Unknown
  • Inferences About the Difference Between Two Population Means: Matched Samples
  • Inferences About the Difference Between Two Population Proportions
inferences about the difference between two population means s 1 and s 2 known
Inferences About the Difference Between Two Population Means: s 1 and s 2 Known
  • Point and Interval Estimation of m1–m2

The Point Estimator of m1–m2 is

Interval Estimation of m1–m2 is

inferences about the difference between two population means s 1 and s 2 known1
Inferences About the Difference Between Two Population Means: s 1 and s 2 Known
  • Hypothesis Tests About m1–m2

D0: hypothesized difference between m1–m2

slide7

Estimating the Difference BetweenTwo Population Means

  • Let1equal the mean of population 1 and
  • 2equal 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 sizen1from population 1
  • and a simple random sample of sizen2from

population 2.

slide8

Estimating the Difference Between Two Population Means

  • Letequal 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 2is .
slide9

Sampling Distribution of

  • Expected Value
  • Standard Deviation (Standard Error)

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

slide10

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

  • Interval Estimate

where:

1 - is the confidence coefficient

slide11

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

  • Example: Par, Inc.

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

In a test of driving distance using a mechanical driving device, a sample of Par golf balls was compared with a sample of golf balls made by Rap, Ltd., a competitor. The sample statistics appear on the next slide.

slide12

Interval Estimation of1 - 2:s1 and s2 Known

  • Example: Par, Inc.

Sample #1

Par, Inc.

Sample #2

Rap, Ltd.

Sample Size

120 balls 80 balls

Sample Mean

275 yards 258 yards

Based on data from previous driving distance

tests, the two population standard deviations are

known with s 1 = 15 yards and s 2 = 20 yards.

slide13

Interval Estimation of1 - 2:s 1ands 2Known

  • Example: Par, Inc.

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

slide14

Population 2

Rap, Ltd. Golf Balls

m2= mean driving

distance of Rap

golf balls

Population 1

Par, Inc. Golf Balls

m1 = mean driving

distance of Par

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 μ1– μ2

Estimating the Difference BetweenTwo Population Means

μ1– μ2 = difference between

the mean distances

slide15

Point Estimate of1 - 2

Point estimate of1-2 =

= 275 - 258

= 17 yards

where:

1= mean distance for the population

of Par, Inc. golf balls

2 = mean distance for the population

of Rap, Ltd. golf balls

slide16

17 5.14 or 11.86 yards to 22.14 yards

Interval Estimation of 1 - 2:1 and 2 Known

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

slide17

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

  • Hypotheses

Left-tailed

Right-tailed

Two-tailed

  • Test Statistic
slide18

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

  • Example: Par, Inc.

Can we conclude, using

α = .01, that the mean driving

distance of Par, Inc. golf balls

is greater than the mean driving

distance of Rap, Ltd. golf balls?

slide19

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

  • p –Value and Critical Value Approaches

1. Develop the hypotheses.

H0: 1 - 2< 0

Ha: 1 - 2 > 0

where:

1 = mean distance for the population

of Par, Inc. golf balls

2 = mean distance for the population

of Rap, Ltd. golf balls

2. Specify the level of significance.

a = .01

slide20

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

  • p –Value and Critical Value Approaches

3. Compute the value of the test statistic.

slide21

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

  • 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

evidence indicates the mean driving distance

of Par, Inc. golf balls is greater than the mean

driving distance of Rap,Ltd. golf balls.

slide22

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

  • Critical Value Approach

4. Determine the critical value and rejection rule.

For a= .01,z.01= 2.33

Reject H0 if z> 2.33

5. Determine whether to reject H0.

Because z = 6.49 > 2.33, we reject H0.

slide23

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

  • Critical Value Approach

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.

slide24

Inferences About the Difference Between Two Population Means:

s 1 and s 2 Unknown

  • Interval Estimation of m1–m2
  • Hypothesis Tests About m1–m2
    • H0: m1–m2 =0
    • Ha: m1–m2 0
slide25

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

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

  • use the sample standard deviations s1
  • and s2 as estimates ofs1ands 2, and
  • replaceza/2withta/2
slide26

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

  • Interval Estimate

where the degrees of freedom forta/2are:

slide27

Difference Between Two Population Means : s1ands 2Unknown

  • Example: Specific Motors

Specific Motors of Detroit

has developed a new automobile

known as 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 shown

on the next slide.

slide28

Difference Between Two Population Means : s1and s 2Unknown

  • Example: Specific Motors

Sample #1

M Cars

Sample #2

J Cars

24 cars 28 cars

Sample Size

29.8 mpg 27.3 mpg

Sample Mean

2.56 mpg 1.81 mpg

Sample Std. Dev.

slide29

Difference Between Two Population Means : s1ands 2Unknown

  • Example: Specific Motors

Let us develop a 90% confidence

interval estimate of the difference

between the mpg performances of

the two models of automobile.

slide30

Point estimate of1-2 =

Point Estimate of m 1-m 2

= 29.8 - 27.3

= 2.5 mpg

where:

1=mean miles-per-gallon for the

population of M cars

2= mean miles-per-gallon for the

population of J cars

slide31

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

The degrees of freedom forta/2are:

Witha/2 = .05anddf = 24, ta/2= 1.711

slide32

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

2.5 + 1.069 or 1.431 to 3.569mpg

We are 90% confident that the difference

between the miles-per-gallon performances of

M cars and J cars is 1.431 to 3.569 mpg.

slide33

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

  • Hypotheses

Left-tailed

Right-tailed

Two-tailed

  • Test Statistic
slide34

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

  • Example: Specific Motors

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?

slide35

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

  • p –Value and Critical Value Approaches

1. Develop the hypotheses.

H0: 1 - 2< 0

Ha: 1 - 2 > 0

where:

1 = mean mpg for the population of M cars

2 = mean mpg for the population of J cars

slide36

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

  • p –Value and Critical Value Approaches

2. Specify the level of significance.

a= .05

3. Compute the value of the test statistic.

slide37

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

  • p –Value Approach

4. Compute the p –value.

The degrees of freedom forta/2are:

Becauset= 4.003 > t.005= 2.797,the p–value< .005.

slide38

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

  • p –Value Approach

5. Determine whether to reject H0.

Becausep–value<a = .05, we rejectH0.

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?.

slide39

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

  • Critical Value Approach

Determine the critical value and

rejection rule.

For a = .05 anddf= 24,t.05= 1.711

RejectH0ift>1.711

slide40

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

  • Critical Value Approach

5. Determine whether to rejectH0.

Because4.003>1.711, we rejectH0.

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?.

slide41

Inferences About the Difference Between Two Population Means: Matched Samples

  • With a matched-sample design each sampled itemprovides 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.
slide42

Inferences About the Difference Between Two Population Means: Matched Samples

  • 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.

slide43

Inferences About the Difference Between Two Population Means: Matched Samples

  • 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 on

thenext slide indicate a difference in mean

delivery times for the two services? Use a .05 level of significance.

slide44

Inferences About the Difference Between Two Population Means: Matched Samples

Delivery Time (Hours)

District Office

UPX

INTEX

Difference

32

30

19

16

15

18

14

10

7

16

25

24

15

15

13

15

15

8

9

11

7

6

4

1

2

3

-1

2

-2

5

Seattle

Los Angeles

Boston

Cleveland

New York

Houston

Atlanta

St. Louis

Milwaukee

Denver

slide45

Inferences About the Difference Between Two Population Means: Matched Samples

  • p –Value and Critical Value Approaches

1. Develop the hypotheses.

H0: d = 0

Ha: d

Let d = the mean of the difference values for the

two delivery services for the population

of district offices

slide46

Inferences About the Difference Between Two Population Means: Matched Samples

  • p –Value and Critical Value Approaches

a = .05

2. Specify the level of significance.

3. Compute the value of the test statistic.

slide47

Inferences About the Difference Between Two Population Means: Matched Samples

  • p –Value Approach

4. Compute the p –value.

For t = 2.94anddf= 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.)

slide48

Inferences About the Difference Between Two Population Means: Matched Samples

  • p –Value Approach

5. Determine whether to rejectH0.

Because p–value<a= .05, we rejectH0.

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

slide49

Inferences About the Difference Between Two Population Means: Matched Samples

  • Critical Value Approach

Determine the critical value and

rejection rule.

Fora= .05 anddf= 9, t.025= 2.262.

RejectH0if t>2.262

slide50

Inferences About the Difference Between Two Population Means: Matched Samples

  • Critical Value Approach

5. Determine whether to rejectH0.

Becauset= 2.94> 2.262, we rejectH0.

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

inferences about the difference between two population proportions
Inferences About the Difference Between Two Population Proportions
  • Interval Estimation of p1 - p2
  • Hypothesis Tests About p1 - p2
slide52

Sampling Distribution of

  • Expected Value
  • Standard Deviation (Standard Error)

where:n1= size of sample taken from population 1

n2= size of sample taken from population 2

slide53

If the sample sizes are large, the sampling

  • distribution of can be approximated
  • by a normal probability distribution.

Sampling Distribution of

The sample sizes are sufficiently large if all

of these conditions are met:

n1p1> 5

n1(1 - p1) > 5

n2p2> 5

n2(1 - p2) > 5

slide56

Interval Estimation of p1 - p2

  • Example: Market Research Associates

Market Research Associates is

conducting research to evaluate the

effectiveness of a client’s new adver-

tising 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.

slide57

Interval Estimation of p1 - p2

  • Example: Market Research Associates

The new campaign has been

initiated with TV and

newspaper advertisements

running for three weeks.

slide58

Interval Estimation of p1 - p2

  • Example: Market Research Associates

A survey conducted immediately

after the new campaign showed 120

of 250 households “aware” of the

client’s product.

Does the data support the position

that the advertising campaign has

provided an increased awareness of

the client’s product?

slide59

Point Estimator of the Difference Between Two Population Proportions

p1 = proportion of the population of households

“aware” of the product after the new campaign

p2 = proportion of the population of households

“aware” of the product before the new campaign

= sample proportion of households “aware”

of the product after the new campaign

= sample proportion of households “aware”

of the product before the new campaign

slide60

Interval Estimation of p1 - p2

For α= .05, z.025 = 1.96

.08 + 1.96(.0510)

.08 + .10

Hence, the 95% confidence interval for the

difference in before and after awareness of the

product is -.02 to +.18.

slide61

Hypothesis Tests about p1 - p2

  • Hypotheses

We focus on tests involving no difference between the two population proportions (i.e. p1 = p2)

Left-tailed

Right-tailed

Two-tailed

slide62

Pooled Estimate of Standard Error of

Hypothesis Tests about p1 - p2

where:

slide64

Hypothesis Tests about p1 - p2

  • Example: Market Research Associates

Can we conclude, using a .05 level

of significance, that the proportion of

households aware of the client’s product

increased after the new advertising

campaign?

slide65

Hypothesis Tests about p1 - p2

  • p -Value and Critical Value Approaches

1. Develop the hypotheses.

H0: p1 - p2< 0

Ha: p1 - p2 > 0

p1= proportion of the population of households

“aware” of the product after the new campaign

p2= proportion of the population of households

“aware” of the product before the new campaign

slide66

Hypothesis Tests about p1 - p2

  • p -Value and Critical Value Approaches

a = .05

2. Specify the level of significance.

3. Compute the value of the test statistic.

slide67

Hypothesis Tests about p1 - p2

  • p –Value Approach

4. Compute the p –value.

For z = 1.56, the p–value = .0594

5. Determine whether to reject H0.

Because p–value > a= .05, we cannot reject H0.

  • We cannot conclude that the proportion
  • of households aware of the client’s product
  • increased after the new campaign.
slide68

Hypothesis Tests about p1 - p2

  • Critical Value Approach

Determine the critical value and

rejection rule.

Fora= .05,z.05= 1.645

RejectH0if z>1.645

slide69

Hypothesis Tests about p1 - p2

  • Critical Value Approach

5. Determine whether to rejectH0.

Because 1.56 < 1.645, we cannot rejectH0.

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