Business Statistics, 4e, by Ken Black. © 2003 John Wiley & Sons. 9-1 Business Statistics, 4e by Ken Black Chapter 9 Statistical Inference: Hypothesis Testing

Business Statistics, 4e, by Ken Black. © 2003 John Wiley & Sons. 9-1

Business Statistics, 4eby Ken Black

Chapter 9

Statistical Inference: Hypothesis Testing

for Single Populations

Discrete Distributions


Learning Objectives• Understand the logic of hypothesis testing, and know how

to establish null and alternate hypotheses.• Understand Type I and Type II errors, and know how to

solve for Type II errors.• Know how to implement the HTAB system to test

hypotheses.• Test hypotheses about a single population mean when is

known.• Test hypotheses about a single population mean when is

unknown.• Test hypotheses about a single population proportion.• Test hypotheses about a single population variance.


Types of Hypotheses

• Research Hypothesis– a statement of what the researcher believes will

be the outcome of an experiment or a study.• Statistical Hypotheses

– a more formal structure derived from the research hypothesis.

• Substantive Hypotheses– a statistically significant difference does not

imply or mean a material, substantive difference.


Example Research Hypotheses

• Older workers are more loyal to a company• Companies with more than $1 billion of

assets spend a higher percentage of their annual budget on advertising than do companies with less than $1 billion of assets.

• The price of scrap metal is a good indicator of the industrial production index six months later.


Statistical Hypotheses• Two Parts

– a null hypothesis– an alternative hypothesis

• Null Hypothesis – nothing new is happening

• Alternative Hypothesis – something new is happening

• Notation– null: H0

– alternative: Ha


Null and Alternative Hypotheses

• The Null and Alternative Hypotheses are mutually exclusive. Only one of them can be true.

• The Null and Alternative Hypotheses are collectively exhaustive. They are stated to include all possibilities. (An abbreviated form of the null hypothesis is often used.)

• The Null Hypothesis is assumed to be true.• The burden of proof falls on the Alternative

Hypothesis.


Null and Alternative Hypotheses: Example

• A manufacturer is filling 40 oz. packages with flour.

• The company wants the package contents to average 40 ounces.

ozH

ozH

a 40:

40:0


• One-tailed Tests

One-tailed and Two-tailed Tests

40:

40:0

aH

H

18.0:

18.0:0

pH

pH

a

12:

12:0

aH

H

• Two-tailed Test


HTAB System to Test Hypotheses

Task 1:HYPOTHESIZE

Task 2:TEST

Task 3:TAKE STATISTICAL ACTION

Task 4:DETERMINING THE

BUSINESS IMPLICATIONS


Steps in Testing Hypotheses

1. Establish hypotheses: state the null and alternative hypotheses.

2. Determine the appropriate statistical test and sampling distribution.

3. Specify the Type I error rate (4. State the decision rule.5. Gather sample data.6. Calculate the value of the test statistic.7. State the statistical conclusion.8. Make a managerial decision.


HTAB Paradigm – Task 1

Task 1: Hypotheses

Step 1. Establish hypotheses: state the null and alternative hypotheses.



Task 2: Test

Step 2. Determine the appropriate statistical test and sampling distribution.

Step 3. Specify the Type I error rate (Step 4. State the decision rule.Step 5. Gather sample data.Step 6. Calculate the value of the test

statistic.



Task 3: Take Statistical Action

Step 7. State the statistical conclusion.



Task 4: Determine the business implications

Step 8. Make a managerial decision.


Rejection and Non Rejection Regions

=40 oz

Non Rejection Region

Rejection Region

Critical Value

Rejection Region

Critical Value


Type I and Type II Errors

• Type I Error– Rejecting a true null hypothesis – The probability of committing a Type I error is

called , the level of significance.

• Type II Error– Failing to reject a false null hypothesis– The probability of committing a Type II error is

called .


Decision Table for Hypothesis Testing

(

( )

Null True Null False

Fail toreject null

CorrectDecision

Type II error)

Reject null Type I error

Correct Decision


One-tailed Tests

40:

40:0

aH

H

40:

40:0

aH

H

=40 oz

Rejection Region


Critical Value

=40 oz

Rejection Region


Critical Value


Two-tailed Tests

40:

40:

a

o

H

H

=12 oz

Rejection Region


Critical Values

Rejection Region

40:

40:0

aH

H


CPA Net Income Example: Two-tailed Test (Part 1)

914,74$:

914,74$:

a

0

H

HRejection Region


=0

Zc 196.

Rejection Region

Zc 196.

2

025.2

025.


CPA Net Income Example: Two-tailed Test (Part 2)

.Hreject not do ,96.1 If

.Hreject ,96.1 If

0

0

c

c

zz

zz

75.2

112

530,14914,74695,78

n

xz

.Hreject 1.96, = z 2.75 = z 0c


CPA Net Income Example:Critical Value Method (Part 1)

605,77112

530,1496.1914,74

n

zx

Upper

cc

914,74$:

914,74$:0

aH

H

223,72112

530,1496.1914,74

n

zx

Lower

cc

Rejection Region


=0 Zc 196.

Rejection Region

Zc 196.

2

025.2

025.

72,223 77,605

96.1cz 0z 96.1cz


CPA Net Income Example:Critical Value Method (Part 2)

.Hreject not do ,605,7777,223 If

.Hreject ,605,77or 223,77 If

0

0

x

x x

.Hreject ,605,77695,78 Since o cxx

Rejection Region


=0 Zc 196.

Rejection Region

Zc 196.

2

025.2

025.

72,223 77,605

96.1cz 0z 96.1cz


Demonstration Problem 9.1: z Test (Part 1)

30.4:

30.4:0

aH

H

Rejection Region


0

=.05

Zc 1645.645.1cz


Demonstration Problem 9.1: z Test (Part 2)

Rejection Region


0

=.05

Zc 1645.

.reject not do ,6451 If

.reject ,6451 If

0

0

H.z

H.z

42.1

32

574.030.4156.4

n

xxz

.reject not do

,645142.1

0H

.z

645.1cz


Demonstration Problem 9.1: Critical Value (Part 1)

30.4:

30.4:0

aH

HRejection Region


0

=.05

Zc 1645.

cx 4133. 4.30

133.432

574.0)645.1(30.4

n

zxc

645.1cz

133.4cx


Demonstration Problem 9.1: Critical Value (Part 2)

Rejection Region


0

=.05

Zc 1645.

cx 4133. 4.30

.reject not do ,133.4 If

.reject ,133.4 If

0

0

Hx

Hx

.reject not do ,133.4156.4 0

Hx

645.1cz

133.4cx


Rejection Region


0

=.05

Demonstration Problem 9.1: Using the p-Value

30.4:

30.4:0

aH

H

.reject not do , value- If

.reject , < value- If

0

0

Hp

Hp

.reject not do

.05, = > .0778 = value- Since

0H

p

0778.)42.1(

42.1

32

574.030.4156.4

zpn

xz


Demonstration Problem 9.1: MINITAB

Test of mu = 4.300 vs mu < 4.300The assumed sigma = 0.574

Variable N MEAN STDEV SE MEAN Z P VALUERatings 32 4.156 0.574 0.101 -1.42 0.078


Demonstration Problem 9.1: Excel (Part 1)


Demonstration Problem 9.1: Excel (Part 2)

H0: = 4.3

Ha: < 4.3

3 4 5 5 4 5 5 4

4 4 4 4 4 4 4 5

4 4 4 3 4 4 4 3

5 4 4 5 4 4 4 5

n = =COUNT(A4:H7)

= 0.05

Mean = =AVERAGE(A4:H7)

S = =STDEV(A4:H7)

Std Error = =B12/SQRT(B9)

Z = =(B11-B1)/B13

p-Value =NORMSDIST(B14)


Two-tailed Test: Unknown, = .05 (Part 1)

Weights in Pounds of a Sample of 20 Plates

22.622.2 23.2 27.4 24.527.026.6 28.1 26.9 24.926.225.3 23.1 24.2 26.125.830.4 28.6 23.5 23.6

20 = and 2.1933,= ,51.25 nsx


Two-tailed Test: Unknown, = .05 (part 2)

Critical Values


Rejection Regions

ct 2 093. ct 2 093.

2

025.2

025.

H

H

o

a

:

:

25

25

df n 1 19

25:

25:0

aH

H


Two-tailed Test: Unknown, = .05 (part 3)

tX

S

n

2551 25 0

2 1933

20

104. .

. .

Since t do not reject H .o 104 2 093. . ,Critical Values


Rejection Regions

ct 2 093. ct 2 093.

2

025.2

025.

If t reject H .

If t do not reject H .

o

o

2 093

2 093

. ,

. , .reject not do 2.093, If

.reject 2.093, If

0

0

Ht

Ht

04.1

20

1933.20.2551.25

n

sx

t

.reject not do ,093.204.1 Since0

Ht


MINITAB Computer Printout for the Machine Plate Example

Test of mu = 25.000 vs mu not = 25.000

Variable N MEAN STDEV SE MEAN T P VALUEPlatewt 20 25.510 2.193 0.490 1.04 0.31


Machine Plate Example: Excel(Part 1)


Machine Plate Example: Excel(Part 2)

A B C D E

1 H0: = 25

2 Ha: 253

4 22.6 22.2 23.2 27.4 24.5

5 27 26.6 28.1 26.9 24.9

6 26.2 25.3 23.1 24.2 26.1

7 25.8 30.4 28.6 23.5 23.68

9 n = =COUNT(A4:E7)

10 = 0.05

11 Mean = =AVERAGE(A4:E7)

12 S = =STDEV(A4:E7)

13 Std Error = =B12/SQRT(B9)

14 t = =(B11-B1)/B13

15 p-Value =TDIST(B14,B9-1,2)


Demonstration Problem 9.2 (Part 1)

Size in Acres of 23 Farms

445 489 474 505 553 477 545463 466 557 502 449 438 500466 477 557 433 545 511 590561 560

23 = and 46.94,= ,78.498 nsx



471:

471:0

aH

H

df n 1 22

Critical Value


Rejection Region

ct 1717.

.05




.reject ,717.1 If

0

0

Ht

Ht

84.2

23

94.4647178.498

n

sx

t

.reject ,717.184.2 Since0

Ht Critical Value


Rejection Region

ct 1717.

.05


z Test of Population Proportion

pq

p

pn

qp

ppz

-1

proportion population

proportion sampleˆ :where

ˆ

5

and ,5

qn

pn


Testing Hypotheses about a Proportion: Manufacturer Example

(Part 1)

08.:

08.:

a

0

pH

pH

cZ 1645.

Critical Values


Rejection Regions

cZ 1645.

2

05. 2

05.

cz cz


Testing Hypotheses about a Proportion: Manufacturer Example

(Part 2)

.

. .

(. )(. ).

p

Zp P

P Qn

33

200165

165 08

08 92200

4 43

If Z reject H .

If Z do not reject H .

o

o

1645

1645

. ,

. ,

Since Z reject H .o 4 43 1645. . ,

cZ 1645.

Critical Values


Rejection Regions

cZ 1645.

2

05. 2

05.

cz cz

.0

0

reject not do 1.645, If

.reject ,645.1 If

Hz

Hz

43.4

200)92)(.08(.

08.165.ˆ

165.200

33ˆ

nqp

ppz

p

.reject ,645.143.4 Since0

Hz



H P

H P

o

a

: .

: .

17

17

Critical Value


Rejection Region

cZ 1645.

.0517.:

17.:0

pH

pH

a

cz



.

. .

(. )(. ).

p

Zp P

P Qn

115

550209

209 17

17 83550

2 44

If reject H .

If do not reject H .

o

o

Z

Z

1645

1645

. ,

. ,

Since Z = 2.44 reject H .o1645. ,Critical Value


Rejection Region

cZ 1645.

.05

cz


.reject ,645.1 If

0

0

Hz

Hz

44.2

550)83)(.17(.

17.209.ˆ

209.550

115ˆ

nqp

ppz

p

.reject ,645.144.2 Since 0Hz


Hypothesis Test for 2:


0

df = 15

.05

.05

.95

7.26094 24.9958

H

H

o

a

:

:

2

2

25

25

25:

25:2

2

0

0

H

H


Hypothesis Test for 2:


0

df = 15

.05

.05

.95

7.26094 24.9958

If or reject H .

If 7.26094 do not reject H .

2 2o

2o

7 26094 24 9958

24 9958

. . ,

. ,

22

2

1 15 281

251686

n S .

.

Since

do not reject H .

2

o

1686 24 995805 15

2. . ,

. ,

86.1625

)1.28)(15()1(2

22

sn

.0

0

reject not do ,9958.2427.26094 If

.reject ,9958.242or 26094.72 If

H

H

.reject not do

,9958.2486.16 Since

0

2

15,05.

2

H


Solving for Type II Errors: The Beverage Example

H

H

o

a

:

:

12

12

Rejection Region

Non Rejection Region=0

=.05

Zc 1645.

c cX Z n

12 1645010

6011979

( . ).

.

If X reject H .

If X do not reject H .

o

o

11979

11979

. ,

. ,

cz

12:

12:0

aH

H

11.979 60

10.0)645.1(12

n

zx cc


.reject ,979.11 If

0

0

Hx

Hx


Type II Error for Beverage Example with =11.99 oz

=.05

Reject Ho Do Not Reject Ho

Ho is True

Ho is False

95%

=.8023

CorrectDecision

Type IError

Type IIError

CorrectDecision 19.77%

X

Z0

Z1

0z

1z

x


Type II Error for Demonstration Problem 9.5, with =11.96 oz

=.05

Ho is True

Ho is False

95%

Reject Ho Do Not Reject Ho

=.0708

CorrectDecision

Type IError

Type IIError

CorrectDecision 92.92%

X

Z0

Z1

0z

1z

x


Values and Power Values for the Soft-Drink Example

Power

11.999 .94 .06

11.995 .89 .11

11.990 .80 .20

11.980 .53 .47

11.970 .24 .76

11.960 .07 .93

11.950 .01 .99


Operating Characteristic Curve for the Soft-Drink Example

0

0.10.2

0.3

0.4

0.50.6

0.7

0.80.9

1

11.95 11.96 11.97 11.98 11.99 12

Pro

bab

ilit

y


Power Curve for the Soft-Drink Example

0

0.10.2

0.3

0.4

0.50.6

0.7

0.80.9

1

11.95 11.96 11.97 11.98 11.99 12

Pro

bab

ilit

y

Documents

Business Statistics, 4e, by Ken Black. © 2003 John Wiley & Sons. 9-1 Business Statistics, 4e by Ken Black Chapter 9 Statistical Inference: Hypothesis Testing