10. Bus Stat - Chapter 10 - April 4th

8/12/2019 10. Bus Stat - Chapter 10 - April 4th

1/51

Copyright 2011 Pearson Education, Inc. publishing as Prentice Hall 10-1

Business Statistics:

A Decision-Making Approach8thEdition

Chapter 10

Estimation and Hypothesis Testing

for Two Population Parameters


2/51


Estimation for Two Populations

Estimating two

population values

Population

means,

independent

samples

Paired

samples

Population

proportions

Group 1 vs.Group 2

Same groupbefore vs. after

Proportion 1 vs.Proportion 2


3/51


Difference Between Two Means

Population means,

independent

samples

1and 2known

1and 2unknownbut assumed equal

1and 2unknown,not assumed equal

Goal: Form a confidence

interval for the difference

between two populationmeans, 12

The point estimate for the

difference is

x1x2

*


4/51


1and 2known

Assumptions:

Samples are randomly and

independently drawn

population distributions are

normal or both sample sizes

are 30

Population standard

deviations are known


5/51


When 1and 2are known and both populations are normal or

both sample sizes are at least 30, the test statistic is a z

valueand the standard error of x1x2 is

2

22

1

21

xx n

n

21

(continued)

1and 2known


6/51


Confidence Interval:

1and 2known

The confidence interval for

12 is:

2

2

2

1

2

1/221

n

n

xx z


7/51Copyright 2011 Pearson Education, Inc. publishing as Prentice Hall 10-7

General Steps

1. Define the population parameter of interest

and select independent samples from each

population2. Specify the confidence interval

3. Compute the point estimate

4. Determine the standard error

5. Determine the critical value

6. Develop the confidence interval estimate


8/51

Example 1 with std dev known

RS Royal Jemursari mengadakan survey untuk

mengetahui berapa lama pasien

dikonsultasikan dengan dokter. Rata-rata untuk

100 pria adalah 35 menit dengan std dev 11menit dan 100 wanita 42 menit dengan std dev

16 menit.

Cari 95% Confidence Interval



9/51

Example 2 with std dev known

Perbandingan on time attendance di kelas

bisnis statistik antara kelas reguler dan IBM

telah disurvey. Rata-rata 150 kelas reguler

adalah 45 menit dengan std dev 15 menit dan100 siswa IBM 30 menit dng std dev 10 menit.

Cari 90 Confidence Interval



10/51

Example 3 Std Dev known

Cari 95% Confidence interval estimate utuk

kedua populasi yg disediakan.

Populasi 1

Mean1 : 345

Std Dev1 : 34

N1 : 50

Populasi 2

Mean1 : 320

Std Dev1 : 40

N1 : 80




Population means,

independent

samples

1and 2known

1and 2unknown,large samples

Assumptions: Samples are randomly and

independently drawn

Populations follow the normal

distribution

The two standard deviationsare equal*

1and

2unknown

but assumed equal





(continued)

Forming interval estimates:

The population standard deviations are

assumed equal, so use the two sample standarddeviations and pool them to estimate (called

pooled standard deviation)

the test statistic is a t value with (n1+ n22)degrees of freedom

d k




(continued)

The pooled standard deviation is:

2nn

s1ns1ns

21

2

22

2

11p

The confidence interval for 12 is:

21p/221

n

1

n

1stxx


14/51

Contoh 1, std dev unknown

Kalbe mengadakan penelitian utk obat terbaru

berapa lama obat itu akan larut dalam darah.

Rata rata dari 6 orang yang berumur diatas

50th adalah 13.6 menit, S1: 3.1 menit dan 8orang yang berumur lebih dari 50th adalah 11.2

menit, S2: 5 menit.

Cari 95% Confidence Interval



15/51

Contoh 2, std dev unknown

Cari 90% Confidence Interval dari

Sample 1

N1: 15

X1: 50

S1: 5

Sample 2

N1: 13 X1: 53

S1: 6




Population means,

independent

samples

1and 2known

1and 2unknown,small samples

Assumptions:

populations are normally

distributed

there is a reason to believe

that the populations do not

have equal variances

samples are independent

*

1and

2unknown

but assumed equal





Forming interval estimates:

The population variances

are not assumed equal, sowe 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

1



2

2

2

1

2

1/221

n

s

n

stxx



(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

1


19/51

Example 1

Visa



20/51




Hypothesis Tests for theDifference Between Two Means

Testing Hypotheses about 12

Use the same situations discussed already: Standard deviations known

Standard deviations unknown

Assumed equal

Assumed not equal (small samples)



Hypothesis Tests forTwo Population Means

Lower tail test:

H0: 12HA: 1< 2

i.e.,

H0: 120HA: 12< 0

Upper tail test:

H0: 12HA: 1> 2

i.e.,

H0: 120HA: 12> 0

Two-tailed test:

H0: 1= 2HA: 12

i.e.,

H0: 12= 0HA: 120

Two Population Means, Independent Samples



Hypothesis tests for 12

Population means, independent samples

1and 2knownUse a z test statistic

Use sp to estimate unknown

, use a t test statisticwith

n1+ n22 d.f.

Use s1and s2 to estimateunknown 1and 2, use a t

test statisticand calculate the

required degrees of freedom

1and 2unknownbut assumed equal



24/51


2

2

2

1

2

1

2121

n

n

xxz

The test statistic for 12 is:

1and 2known


25/51

Example 1

Pengeboran pertamina mempunyai 2 mesin

bor. Mesin Alpha mempunyai diameter 0.025

inch dan mesin Beta mempunyai 0.034 inch.

Apakah mesin Beta mempunyai rata2 yg lebihbesar drpd rata2 mesin Alpha. Jika 100 sample

dari kedua mesin diambil dan rata2 mesin aplha

0.501 inch dan mesin beta 0.509 inch



26/51

Contoh 2

Page 444 no 10-21


St H th i T t f T


27/51


Steps: Hypothesis Test for Two

Population Means

1. Specify parameter of interest

2. Formulate hypotheses

3. Specify the significance level ()4. Construct the rejection region and develop the

decision rule

5. Compute the test statistics and/or the p-value6. Reach a decision

7. Draw a conclusionSee Chapter 9


28/51



Where t has (n1+ n22) d.f.,and

2nn

s1ns1ns

21

2

22

2

11p

21

p

2121

n

1

n

1s

xx

t



29/51




2

22

1

21

2121

n

s

n

s

xx

t

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

1


30/51


Two Population Means, Independent Samples

Lower tail test:

H0: 120HA: 12< 0

Upper tail test:

H0: 120

HA: 12> 0

Two-tailed test:

H0: 12= 0

HA: 120

/2 /2

-z

-z/2z z/2

Reject H0if z < -z Reject H0if z > z Reject H0if z < -z/2or z > z

/2

Hypothesis tests for 12

Example: 1and 2known:


31/51


Example1and 2unknown, assumed equal

Youre 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)?


32/51


Calculating the Test Statistic

1.2256

22521

1.161251.30121

2nn

s1ns1ns

22

21

2

22

2

11p

2.040

251

2111.2256

02.533.27

n1

n1s

xxt

21

p

2121

The test statistic is:

Where:


33/51


Solution

H0: 1 - 2= 0 i.e. (1 = 2)

HA: 1 - 20 i.e. (1 2)

= 0.05

df = 21 + 25 - 2 = 44Critical Values: t = 2.0154

Test Statistic: Decision:

Conclusion:Reject H0at = 0.05

There is evidence that

the means are different.

t0 2.0154-2.0154

0.025

Reject H0 Reject H0

0.025

2.040

2.040

25

1

21

11.2256

2.533.27t


34/51

10.3 Paired Sample



35/51


Paired Samples

Tests Means of 2 RelatedPopulations

Paired or matched samples e.g., two different value appraisals on the same home

Repeated measures (before/after) e.g., weight before and after a diet program

Use difference between paired values:

Eliminates Variation Among Subjects Assumptions:

Both Populations Are Normally Distributed

Or, if Not Normal, use large samples

d = x1- x2


36/51


Paired Differences

The ith paired difference is di, where

di= x1i- x2i

The point estimate forthe population mean

paired difference is d :

1n

)d(d

s

n

1i

2

i

d

n

d

d

n

1i

i

The sample standarddeviation is

n is the number of pairs in the paired sample


37/51


Paired Differences

The confidence interval for d is

1n

)d(d

s

n

1i

2

i

d

nstd d

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

(continued)

n is the number of pairs in the paired sample

H th i T ti f


38/51


The test statistic for d is

1n

)d(d

s

n

1i

2

i

d

n

s dt dd

Where t has n - 1 d.f.

and sd is:

n is the

number

of pairs

in the

paired

sample

Hypothesis Testing forPaired Samples

H th i T ti f


39/51


Lower tail test:

H0: d0HA: d< 0

Upper tail test:

H0: d0HA: d> 0

Two-tailed test:

H0: d= 0HA: d0

Paired Samples

Hypothesis Testing forPaired Samples

/2 /2

-t

-t/2t t/2

Reject H0if t < -t Reject H0if t > t Reject H0if t < -t/2or t > t

/2

Where t has n - 1 d.f.

(continued)


40/51


Assume you send your salespeople to a customerservice training workshop. Is the training effective?

You collect the following data:

Paired Sample Example

Number of Complaints: (2) - (1)Salesperson Before (1) After (2) Difference,d

i

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 = din

5.67

1n

)d(ds

2

i

d

= -4.2


41/51


Has the training made a difference in the number of complaints

(at the 0.05 level)?

- 4.2d =

1.6655.67/

04.2

n/s

dt

d

d

H0: d = 0

HA:

d0

Test Statistic:

Critical Value = 2.7765d.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 Sample: Solution

Reject

/2

- 1.66= 0.05


42/51

10.4 Proportion



43/51


Two Population Proportions

Goal: Form a confidence interval for

or test a hypothesis about the

difference between two population

proportions, 12

The point estimate for

the difference isp1p2

Assumptions:

n115 , n1(1-1) 5

n225 , n2(1-2) 5

Confidence Interval for


44/51


Confidence Interval forTwo Population Proportions

222

1

11

21 n

)p(1p

n

)p(1p

zpp


Hypothesis Tests for


45/51


Hypothesis Tests forTwo Population Proportions

Population proportions

Lower tail test:

H0: 12HA: 1< 2

i.e.,

H0: 120HA: 12< 0

Upper tail test:

H0: 12HA: 1> 2

i.e.,

H0: 120HA: 12> 0

Two-tailed test:

H0: 1= 2HA: 12

i.e.,

H0: 12= 0HA: 120


46/51



21

21

21

2211

nn

xx

nn

pnpn

p

The pooled estimate for the

overall proportion is:

Where x1and x2are the numbers from

samples 1 and 2 with the characteristic of interest

Since we begin by assuming the null

hypothesis is true, we assume 1= 2

and pool the two p estimates


47/51



21

2121

n

1

n

1)p(1p

ppz


(continued)

Hypothesis Tests for


48/51


Hypothesis Tests forTwo Population Proportions

Population proportions

Lower tail test:

H0: 120HA: 12< 0

Upper tail test:

H0: 120

HA: 12> 0

Two-tailed test:

H0: 12= 0

HA: 120

/2 /2

-z

-z/2z z/2

Reject H0if z < -z Reject H0if z > z Reject H0if z < -z/2or z > z

/2

Example:


49/51


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 0.05 level of significance

Example:


50/51


The hypothesis test is:

H0: 12= 0 (the two proportions are equal)

HA: 120 (there is a significant difference between proportions)

The sample proportions are:

Men: p1= 36/72 = 0.50

Women: p2= 31/50 = 0.62

.5490122

67

5072

3136

nn

xxp

21

21

The pooled estimate for the overall proportion is:


(continued)

Example:


51/51



(continued)

0.025

-1.96

1.96

0.025

-1.31

Decision:Do not reject H0

Conclusion:There is not

significant evidence of a

difference in the proportion

who will vote yes between

1.31

50

1

72

1.549)0(1.5490

0.620.500

n1

n1)p(1p

ppz

21

2121

Reject H0 Reject H0

Critical Values = 1.96For = 0.05

10. Bus Stat - Chapter 10 - April 4th

Documents

Transcript of 10. Bus Stat - Chapter 10 - April 4th