business statistics: a decision-making approach, …business statistics department of quantitative...

Business Statistics

Department of Quantitative Methods & Information Systems

Dr. Mohammad Zainal QMIS 220

Chapter 10

Estimation and Hypothesis Testing

for Two Population Parameters

Chapter Goals

After completing this chapter, you should be

able to:

Test hypotheses or form interval estimates for

two independent population means

Standard deviations known

Standard deviations unknown

two means from paired samples

the difference between two population proportions

Comparing two population variances QMIS 220, by Dr. M. Zainal Chap 10-2

Estimation for Two Populations

Estimating two

population values

Population

means,

independent

samples

Paired

samples

Population

proportions

Group 1 vs. independent

Group 2

Same group before vs. after

treatment

Proportion 1 vs. Proportion 2

Examples:

QMIS 220, by Dr. M. Zainal Chap 10-3

Difference Between Two Means

Population means,

independent

samples

σ1 and σ2 known

σ1 and σ2 unknown but assumed equal

σ1 and σ2 unknown, not assumed equal

Goal: Form a confidence

interval for the difference

between two population

means, μ1 – μ2

The point estimate for the

difference is

x1 – x2

*


Independent Samples

Population means,

independent

samples

σ1 and σ2 known

Different data sources

Unrelated

Independent

Sample selected from

one population has no

effect on the sample

selected from the other

population

Use the difference between

2 sample means

*




Population means,

independent

samples

σ1 and σ2 known

σ1 and σ2 known

Assumptions:

Samples are randomly and

independently drawn

population distributions are

normal or both sample sizes

are 30

Population standard

deviations are known

* σ1 and σ2 unknown but assumed equal



Population means,

independent

samples

σ1 and σ2 known …and the standard error of

x1 – x2 is

When σ1 and σ2 are known and

both populations are normal or

both sample sizes are at least 30,

the test statistic is a z value…

2

2

2

1

2

1

xx n

σ

n

σσ

21

(continued)

σ1 and σ2 known




Population means,

independent

samples

σ1 and σ2 known

2

2

2

1

2

1/221

n

σ

n

σxx z

The confidence interval for

μ1 – μ2 is:

σ1 and σ2 known (continued)




Population means,

independent

samples

σ1 and σ2 known

σ1 and σ2 unknown, large samples

Assumptions: Samples are randomly and independently drawn

Population standard

deviations are unknown

The two standard deviations

are equal




Population means,

independent

samples

σ1 and σ2 known


(continued)


Forming interval estimates:

The population standard

deviations are assumed equal,

so use the two sample

standard deviations and pool

them to estimate σ

the test statistic is a t value

with (n1 + n2 – 2) degrees

of freedom σ1 and σ2 unknown, not assumed equal


Population means,

independent

samples

σ1 and σ2 known


(continued)


The pooled standard

deviation is

2nn

s1ns1ns

21

2

22

2

11p



Population means,

independent

samples

σ1 and σ2 known


(continued)


21

p/221

n

1

n

1stxx


μ1 – μ2 is:

Where t/2 has (n1 + n2 – 2) d.f.,

and

2nn

s1ns1ns

21

2

22

2

11p



Population means,

independent

samples

σ1 and σ2 known

σ1 and σ2 unknown, small samples

Assumptions: populations are normally distributed

there is a reason to believe

that the populations do not

have equal variances

samples are independent

*




Population means,

independent

samples

σ1 and σ2 known


Forming interval

estimates:

The population variances

are not assumed equal, so

we do not pool them

the test statistic is a t value

with degrees of freedom

given by:

(continued)

*


1n

/ns

1n

/ns

)/ns/n(sdf

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2

1σ1 and σ2 unknown, not assumed equal

QMIS 220, by Dr. M. Zainal

Population means,

independent

samples

σ1 and σ2 known 2

2

2

1

2

1α/221

n

s

n

stxx


μ1 – μ2 is:


(continued)

Where t/2 has d.f. given by

*


1n

/ns

1n

/ns

)/ns/n(sdf

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2



Hypothesis Tests for the Difference Between Two Means

Testing Hypotheses about μ1 – μ2

Use the same situations discussed already:

Standard deviations known

Standard deviations unknown

Assumed equal

Assumed not equal


Hypothesis Tests for Two Population Proportions

Lower tail test:

H0: μ1 μ2

HA: μ1 < μ2

i.e.,

H0: μ1 – μ2 0

HA: μ1 – μ2 < 0

Upper tail test:

H0: μ1 ≤ μ2

HA: μ1 > μ2

i.e.,

H0: μ1 – μ2 ≤ 0

HA: μ1 – μ2 > 0

Two-tailed test:

H0: μ1 = μ2

HA: μ1 ≠ μ2

i.e.,

H0: μ1 – μ2 = 0

HA: μ1 – μ2 ≠ 0

Two Population Means, Independent Samples


Hypothesis tests for μ1 – μ2

Population means, independent samples

σ1 and σ2 known Use a z test statistic

Use sp to estimate unknown

σ , use a t test statistic with

n1 + n2 – 2 d.f.

Use s1 and s2 to estimate

unknown σ1 and σ2 , use a t

test statistic and calculate the

required degrees of freedom




Population means,

independent

samples

σ1 and σ2 known

2

2

2

1

2

1

2121

n

σ

n

σ

μμxxz

The test statistic for

μ1 – μ2 is:

σ1 and σ2 known




Population means,

independent

samples

σ1 and σ2 known



Where t has (n1 + n2 – 2) d.f.,

and

2nn

s1ns1ns

21

2

22

2

11p

21

p

2121

n

1

n

1s

μμxxt


μ1 – μ2 is:



Population means,

independent

samples

σ1 and σ2 known



μ1 – μ2 is:

*


2

2

2

1

2

1

2121

n

s

n

s

μμxxt

Where t has d.f. given by

1n

/ns

1n

/ns

)/ns/n(sdf

2

2

2

2

2

1

2

1

2

1

2

2

2

21

2



Two Population Means, Independent Samples

Lower tail test:

H0: μ1 – μ2 0

HA: μ1 – μ2 < 0

Upper tail test:

H0: μ1 – μ2 ≤ 0

HA: μ1 – μ2 > 0

Two-tailed test:

H0: μ1 – μ2 = 0

HA: μ1 – μ2 ≠ 0

/2 /2

-z -z/2 z z/2

Reject H0 if z < -z Reject H0 if z > z Reject H0 if z < -z/2

or z > z/2

Hypothesis tests for μ1 – μ2

Example: σ1 and σ2 known:


Pooled sp t Test Example σ1 and σ2 unknown, assumed equal

You’re a financial analyst for a brokerage firm. Is there a

difference in dividend yield between stocks listed on the

NYSE & NASDAQ? You collect the following data:

NYSE NASDAQ

Number 21 25

Sample mean 3.27 2.53

Sample std dev 1.30 1.16

Assuming equal variances, is

there a difference in average

yield ( = 0.05)?


Calculating the Test Statistic

1.2256

22521

1.161251.30121

2nn

s1ns1ns

22

21

2

22

2

11p

2.040

25

1

21

11.2256

02.533.27

n

1

n

1s

μμxxt

21

p

2121

The test statistic is:

Where:


Solution

H0: μ1 - μ2 = 0 i.e. (μ1 = μ2)

HA: μ1 - μ2 ≠ 0 i.e. (μ1 ≠ μ2)

= 0.05

df = 21 + 25 - 2 = 44

Critical Values: t = ± 2.0154

Test Statistic: Decision:

Conclusion:

Reject H0 at = 0.05

There is evidence that

the means are different.

t 0 2.0154 -2.0154

.025

Reject H0 Reject H0

.025

2.040

2.040

25

1

21

11.2256

2.533.27t


Paired Samples

Tests Means of 2 Related Populations

Paired or matched samples

Repeated measures (before/after)

Use difference between paired values:

Eliminates Variation Among Subjects

Assumptions:

Both Populations Are Normally Distributed

Or, if Not Normal, use large samples

Paired

samples

d = x1 - x2


Paired Differences

The ith paired difference is di , where

Paired

samples di = x1i - x2i

The point estimate for

the population mean

paired difference is d :

1n

)d(d

s

n

1i

2

i

d

n

d

d

n

1i

i

The sample standard

deviation is

n is the number of pairs in the paired sample


Paired Differences

The confidence interval for d is Paired

samples

1n

)d(d

s

n

1i

2

i

d

n

std d

Where t has n - 1 d.f. and sd is:

(continued)

n is the number of pairs in the paired sample


The test statistic for d is Paired

samples

1n

)d(d

s

n

1i

2

i

d

n

s

μdt

d

d

Where t has n - 1 d.f.

and sd is:

n is the

number

of pairs

in the

paired

sample

Hypothesis Testing for Paired Samples


Lower tail test:

H0: μd 0

HA: μd < 0

Upper tail test:

H0: μd ≤ 0

HA: μd > 0

Two-tailed test:

H0: μd = 0

HA: μd ≠ 0

Paired Samples

Hypothesis Testing for Paired Samples

/2 /2

-t -t/2 t t/2

Reject H0 if t < -t Reject H0 if t > t Reject H0 if t < -t/2

or t > t/2 Where t has n - 1 d.f.

(continued)


Assume you send your salespeople to a “customer

service” training workshop. Is the training effective?

You collect the following data:

Paired Samples Example

Number of Complaints: (2) - (1)

Salesperson Before (1) After (2) Difference, di

C.B. 6 4 - 2

T.F. 20 6 -14

M.H. 3 2 - 1

R.K. 0 0 0

M.O. 4 0 - 4

-21

d = di

n

5.67

1n

)d(ds

2

i

d

= -4.2


Has the training made a difference in the number of

complaints (at the 0.05 level)?

- 4.2 d =

1.6655.67/

04.2

n/s

μdt

d

d

H0: μd = 0

HA: μd 0

Test Statistic:

Critical Value = ± 2.7765 d.f. = n - 1 = 4

Reject

/2

- 2.7765 2.7765

Decision: Do not reject H0

(t stat is not in the reject region)

Conclusion: There is not a

significant change in the

number of complaints.

Paired Samples: Solution

Reject

/2

- 1.66 = .05


Two Population Proportions

Goal: Form a confidence interval for

or test a hypothesis about the

difference between two population

proportions, P1 – P2

The point estimate for

the difference is 𝑝 1 – 𝑝 2

Population

proportions

Assumptions:

n1P1 5 , n1(1-P1) 5

n2P2 5 , n2(1-P2) 5


Confidence Interval for Two Population Proportions

Population

proportions

2

22

1

1121

n

)p(1p

n

)p(1pzpp


P1 – P2 is:



Population proportions

Lower tail test:

H0: P1 P2

HA: P1 < P2

i.e.,

H0: P1 – P2 0

HA: P1 – P2 < 0

Upper tail test:

H0: P1 ≤ P2

HA: P1 > P2

i.e.,

H0: P1 – P2 ≤ 0

HA: P1 – P2 > 0

Two-tailed test:

H0: P1 = P2

HA: P1 ≠ P2

i.e.,

H0: P1 – P2 = 0

HA: P1 – P2 ≠ 0



Population

proportions

21

21

21

2211

nn

xx

nn

pnpnp

The pooled estimate for the

overall proportion is:

where x1 and x2 are the numbers from

samples 1 and 2 with the characteristic of interest

Since we begin by assuming the null

hypothesis is true, we assume P1 = P2

and pool the two p estimates



Population

proportions

21

2121

n

1

n

1)p(1p

PPppz


P1 – P2 is:

(continued)



Population proportions

Lower tail test:

H0: P1 – P2 0

HA: P1 – P2 < 0

Upper tail test:

H0: P1 – P2 ≤ 0

HA: P1 – P2 > 0

Two-tailed test:

H0: P1 – P2 = 0

HA: P1 – P2 ≠ 0

/2 /2

-z -z/2 z z/2

Reject H0 if z < -z Reject H0 if z > z Reject H0 if z < -z/2

or z > z/2


Example: Two population Proportions

Is there a significant difference between the

proportion of men and the proportion of

women who will vote Yes on Proposition A?

In a random sample, 36 of 72 men and 31 of

50 women indicated they would vote Yes

Test at the .05 level of significance


The hypothesis test is:

H0: P1 – P2 = 0 (the two proportions are equal)

HA: P1 – P2 ≠ 0 (there is a significant difference between proportions)

The sample proportions are:

Men: p1 = 36/72 = .50

Women: p2 = 31/50 = .62

.549122

67

5072

3136

nn

xxp

21

21

The pooled estimate for the overall proportion is:


(continued)


The test statistic for P1 – P2 is:


(continued)

.025

-1.96 1.96

.025

-1.31

Decision: Do not reject H0

Conclusion: There is not

significant evidence of a

difference in the proportion

who will vote yes between

men and women.

1.31

50

1

72

1.549)(1.549

0.62.50

n

1

n

1)p(1p

PPppz

21

2121

Reject H0 Reject H0

Critical Values = ±1.96 For = .05


Two Sample Tests in EXCEL

For independent samples:

Independent sample z test with variances known:

Data | data analysis | z-Test: Two Sample for Means

Independent sample t test with variance unknown:

Data | data analysis | t-Test: Two Sample Assuming Equal

Variances

Data | data analysis | t-Test: Two Sample Assuming Unequal

Variances

For paired samples (t test):

Data | data analysis… | t-Test: Paired Two Sample for Means


Comparing Two Population

Variances (F – Test)


Hypothesis Tests for Variances

Tests for Two

Population Variances

F test statistic

*

F Test for Difference in Two Population Variances

H0: σ12 = σ2

2

HA: σ12 ≠ σ2

2 Two tailed test

Lower tail test

Upper tail test

H0: σ12 σ2

2

HA: σ12 < σ2

2

H0: σ12 ≤ σ2

2

HA: σ12 > σ2

2


Hypothesis Tests for Variances

F test statistic *

F Test for Difference in Two Population Variances

Tests for Two

Population Variances 2

2

2

1

s

sF

The F test statistic is:

= Variance of Sample 1

D1 = n1 - 1 = numerator degrees of freedom

D2 = n2 - 1 = denominator degrees of freedom

= Variance of Sample 2

2

1s

2

2s

Where F has D1

numerator and D2

denominator

degrees of freedom


The F critical value is found from the F table

The are two appropriate degrees of freedom:

D1 (numerator) and D2 (denominator)

In the F table,

numerator degrees of freedom determine the column

denominator degrees of freedom determine the row

The F Distribution

where D1 = n1 – 1 ; D2 = n2 – 1 2

2

2

1

s

sF


The F distribution is continuous and skewed to the right.

The units of an F distribution, denoted by F ,D1,D2 are

nonnegative.

The F Distribution


Finding the F Value


Example: Find the F value for 4 degrees of freedom for the

numerator, 10 degrees of freedom for the denominator, and

0.1 area in the right tail of F distribution curve. (F .1, 4,10)

For a two-tailed test, always place the larger

sample variance in the numerator

For a one-tailed test, consider the alternative

hypothesis: place in the numerator the sample

variance for the population that is predicted

(based on HA) to have the larger variance

Formulating the F Ratio

where df1 = the size of the sample having the larger variance - 1

2

2

2

1

2

2

2

1

and ofsmaller the

and oflarger the

ss

ssF


df2 = the size of the sample having the smaller variance - 1

F 0

rejection region for a one-tail test is

Finding the Critical Value

F 0

2/2

2

2

1 Fs

sF F

s

sF

2

2

2

1

(where the larger sample variance in the numerator)

rejection region for a two-tailed test is

/2

F F/2 Reject H0 Do not

reject H0

Reject H0 Do not reject H0

H0: σ12 = σ2

2

HA: σ12 ≠ σ2

2

H0: σ12 σ2

2

HA: σ12 < σ2

2

H0: σ12 ≤ σ2

2

HA: σ12 > σ2

2


F Test: An Example

You are a financial analyst for a brokerage firm. You

want to compare dividend yields between stocks listed

on the NYSE & NASDAQ. You collect the following data:

NYSE NASDAQ

Number 21 25

Mean 3.27 2.53

Std dev 1.30 1.16

Is there a difference in the

variances between the NYSE

& NASDAQ at the = 0.05 level?


F Test: Example Solution

Form the hypothesis test:

H0: σ21 = σ2

2 (there is no difference between variances)

HA: σ21 ≠ σ2

2 (there is a difference between variances)

Find the F critical value for = .05:

Numerator:

D1 = n1 – 1 = 21 – 1 = 20

Denominator:

D2 = n2 – 1 = 25 – 1 = 24

F.05/2, 20, 24 = 2.327


The test statistic is:

0

256.116.1

30.1

s

sF

2

2

2

2

2

1

/2 = .025

F/2 =2.327

Reject H0 Do not reject H0

H0: σ12 = σ2

2

HA: σ12 ≠ σ2

2

F Test: Example Solution

F = 1.256 is not greater than

the critical F value of 2.327, so

we do not reject H0

(continued)

Conclusion: There is no evidence of a

difference in variances at = .05


Using EXCEL

EXCEL

F test for two variances:

Data | Data Analysis | F-test: Two Sample for Variances


Chapter Summary

Compared two independent samples

Formed confidence intervals for the differences between two

means

Performed z test for the differences in two means

Performed t test for the differences in two means

Compared two related samples (paired samples)

Formed confidence intervals for the paired difference

Performed paired sample t tests for the mean difference

Compared two population proportions

Formed confidence intervals for the difference between two

population proportions

Performed z test for two population proportions


Problems


A business analyst took a random sample of 32 advertising

managers from across the United States and a similar random

sample is taken of 34 auditing managers. They were asked

what their annual salary is. The analyst wants to test whether

there is a difference in the average wage of an advertising

manager and the auditing manager. The resulting salary data

are listed in the table below with the sample means, population

standard deviations and variances.

164.264

253.16

700.70

32

2

1

1

1

1

x

n

411.166

900.12

187.62

34

2

2

2

2

2

x

n

Advertising Auditing

Chap 10-56

Chap 10-57 QMIS 220, by Dr. M. Zainal

z-Test: Two Sample for Means

Adv Mgr Auditing Mgr

Mean 70.7001 62.187

Known Variance 264.164 166.411

Observations 32 34

Hypothesized Mean Difference 0

z 2.35

P(Z<=z) one-tail 0.0094

z Critical one-tail 1.64

P(Z<=z) two-tail 0.0189

z Critical two-tail 1.960


Problems


A consumer test group wants to determine the difference in

gasoline mileage of cars using regular gas and cars using

premium gas. Their researchers divided a fleet of 100 cars of

the same make in half and tested each car on one tank of gas.

50 of the cars were filled with regular gas and the rest filled

with premium gas. The sample average for the regular gasoline

group was 21.45 mpg, and the sample average for the

premium gasoline was 24.60. Assume the population standard

deviation of the regular gasoline is 3.46 mpg, and the

population standard deviation of the premium gas is 2.99 mpg.

Construct a 95% confidence interval to estimate the difference

in the mean gas mileage between the cars using regular and

premium gasoline.

Chap 10-60

Problems


At a certain company, new employees are expected to attend a

3 days seminar (Method A) to learn about the company. At the

end of the seminar, they are tested to measure their knowledge

about the company. The management decided to experiment

another training procedure (Method B) which ends with the

same knowledge test but saves a lot of time and money. To test

the effectiveness of the new method (B), managers randomly

selected two groups and the results were recorded in the

following tables. Using = 5%, the managers want to

determine whether there is a significant difference in the mean

score of the two groups assuming the score of the test are

normally distributed and the population variances are

approximately equal.

Chap 10-62

Training Method A

56 51 45

47 52 43

42 53 52

50 42 48

47 44 44

Training Method B

59

52

53

54

57

56

55

64

53

65

53

57

495.19

73.47

15

2

1

1

1

s

x

n

273.18

5.56

12

2

2

2

2

s

x

n


t-Test: Two-Sample Assuming Equal Variances

A B

Mean 4 7.73 56.5

Variance 19.495 18.27

Observations 15 12

Pooled Variance 18.957


df 25

t Stat - 5.20

P(T<=t) one-tail 1.12E-05

t Critical one-tail 1.71

P(T<=t) two-tail 2.23E-05

t Critical two-tail 2.06


Problems


A coffee manufacturer is interested in estimating the difference

in the average daily coffee consumption of a regular coffee

drinker and decaffeinated coffee drinker. Its researcher

randomly selects 13 regular coffee drinkers and asks how

many cups of coffee per day they drink. He randomly locates

15 decaffeinated coffee drinkers and how many cups of coffee

per day they drink. The average for the regular coffee drinkers

is 4.35 cups, with a standard deviation of 1.2 cups. The

average of the decaffeinated-coffee drinkers is 6.84 cups, with

a standard deviation of 1.42 cups. The researcher assumes, for

each population, that the daily consumption is normally

distributed, and their variances are approximately equal. He

wants to construct a 95% confidence interval to estimate the

difference in the averages of the two populations.

Chap 10-67

Problems


Suppose a stock market investor is interested in determining

whether there is a significant difference in P/E (price to

earning) ratio for companies from one year to the next. He

randomly samples nine companies and records the P/E ratios

for each of these companies at the end of the year 1 and 2.

Assume there is no prior information to indicate whether P/E

ratios have gone up or down. Also, assume that the P/E ratios

are normally distributed in the population. The data are shown

in the next table. (Use = 1%)

Chap 10-70

Company Year1 P/E Ratio Year2 P/E Ratio

1 8.9 12.7

2 38.1 45.4

3 43.0 10.0

4 34.0 27.2

5 34.5 22.8

6 15.2 24.1

7 20.3 32.3

8 19.9 40.1

9 61.9 106.5


Company Year1 P/E Ratio Year2 P/E Ratio d

1 8.9 12.7 -3.8

2 38.1 45.4 -7.3

3 43.0 10.0 33.0

4 34.0 27.2 6.8

5 34.5 22.8 11.7

6 15.2 24.1 -8.9

7 20.3 32.3 -12.0

8 19.9 40.1 -20.2

9 61.9 106.5 -44.6


t-Test: Paired Two Sample for Means

Year1 P/E

Ratio

Year2 P/E

Ratio

Mean 30.64 35.68

Variance 268.1 837.5

Observations 9 9

Pearson Correlation 0.674


df 8

t Stat -0.7

P(T<=t) one-tail 0.252

t Critical one-tail 1.86

P(T<=t) two-tail 0.504

t Critical two-tail 2.306


Problems


Time magazine reported the result of a telephone poll of 800

adult Americans. The question posed of the Americans who

were surveyed was: "Should the federal tax on cigarettes be

raised to pay for health care reform?" The results of the survey

were:

Chap 10-75

Non-Smokers Smokers

n1 = 605 N2 = 195

Said yes = 351 Said yes = 41

Problems


A random sample of 10 hot drinks from Dispenser A had a

mean volume of 203 ml and a standard deviation of 3 ml.

Another random sample of 15 hot drinks from Dispenser B

gave corresponding values of 206 ml and 5 ml. The amount

dispensed by each machine may be assumed to be normally

distributed. Test, at the 5% significance level, the hypothesis

that there is no difference in the variability of the volume

dispensed by the two machines.

Chap 11-77

Copyright

The materials of this presentation were mostly

taken from the PowerPoint files accompanied

Business Statistics: A Decision-Making Approach,

7e © 2008 Prentice-Hall, Inc.


business statistics: a decision-making approach, …business statistics department of quantitative...

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