Ken Black QA ch09

8/3/2019 Ken Black QA ch09

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Business Statistics, 5th ed.

by Ken Black

Chapter 9

Statistical Inference:Hypothesis Testing

for SinglePopulations

Discrete Distributions

PowerPoint presentations prepared by Lloyd Jaisingh,Morehead State University


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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 testhypotheses.

Test hypotheses about a single population mean when s isknown.

Test hypotheses about a single population mean when s isunknown.

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


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Types of Hypotheses

Research Hypothesisa statement of what the researcher believes will

be the outcome of an experiment or a study

Statistical Hypothesesa formal structure used to scientifically test the

research hypothesis

Substantive Hypotheses

a statistically significant difference does notimply a material, or substantive difference


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Example Research Hypotheses

Older workers are more loyal to a company.

Companies with more than $1 billion ofassets spend a higher percentage of their

annual budget on advertising than docompanies with less than $1 billion ofassets.

The price of scrap metal is a good indicator

of the industrial production index sixmonths later.


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Statistical Hypotheses

Two Partsa null hypothesis

an alternative hypothesis

Null Hypothesisnothing new ishappening; the null condition exists

Alternative Hypothesissomething new ishappening

Notationnull: H0alternative: Ha


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Null and Alternative Hypotheses

The Null and Alternative Hypotheses aremutually exclusive. Only one of them canbe true.

The Null and Alternative Hypotheses arecollectively exhaustive. They are stated toinclude all possibilities. (An abbreviatedform of the null hypothesis is often used.)

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

Hypothesis.


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Null and Alternative Hypotheses:

Example

A manufacturer is filling 40 oz. packageswith flour.

The company wants the package contents toaverage 40 ounces.

ozH

ozH

a40:

40:0


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HTAB System to Test Hypotheses

Task 1:

HYPOTHESIZE

Task 2:TEST

Task 3:

TAKE STATISTICAL ACTION

Task 4:

DETERMINING THEBUSINESS IMPLICATIONS


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Steps in Testing Hypotheses

1. Establish hypotheses: state the null andalternative 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.


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HTAB ParadigmTask 1

Task 1: Hypotheses

Step 1. Establish hypotheses: state thenull and alternative hypotheses.


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HTAB ParadigmTask 2

Task 2: Test

Step 2. Determine the appropriate

statistical test and samplingdistribution.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.


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HTAB ParadigmTask 3

Task 3: Take Statistical Action

Step 7. State the statistical conclusion.


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HTAB ParadigmTask 4

Task 4: Determine the businessimplications

Step 8. Make a managerial decision.


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Rejection and Nonrejection Regions

Using the critical values established atStep 4 of the hypothesis testing process, thepossible statistical outcomes of a study can

be divided into two groups: Those that cause the rejection of the nullhypothesis

Those that do not cause the rejection of the

null hypothesis


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Rejection and Nonrejection Regions

Conceptually and graphically, statisticaloutcomes that result in the rejection of thenull hypothesis lie in what is termed the

rejection region. Statistical outcomes that fail to result in therejection of the null hypothesis lie in what istermed the nonrejection region.


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Possible Rejection and Nonrejection

Regions -

There are three possibilities which can bestipulated in the alternative hypothesis.

The three possibilities are: >,


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Regions -

Rejection region

for hypothesis

which involve the

standard normal

distribution and

the > symbol

(righttailed test)


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Regions -

Rejection region

for hypothesis

which involve the

standard normal

distribution and

the < symbol

(lefttailed test)


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Regions -

Rejection region

for hypothesis

which involve the

standard normal

distribution and

the symbol(twotailed test)


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Type I and Type II Errors

Type I ErrorRejecting a true null hypothesis

The probability of committing a Type I error is

called , the level of significance.

Type II ErrorFailing to reject a false null hypothesis

The probability of committing a Type II error iscalled .


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Decision Table

for Hypothesis Testing

(

( )

Null True Null False

Fail to

reject null

Correct

Decision

Type II error

)

Reject null Type I error Correct Decision


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Testing Hypotheses about a Population

Mean Using the z Statistic (s Known) Example: A survey, done 10 years ago, of

CPAs in the U.S. found that their averagesalary was $74,914. An accounting

researcher would like to test whether thisaverage has changed over the years. Asample of 112 CPAs produced a meansalary of $78,695. Assume that thepopulation standard deviation of salariess = $14,530.


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Mean Using the z Statistic (s Known) Step 1: Hypothesize

Step 2: Test

nXz/s

914,74$:

914,74$:0

aH

H


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Mean Using the z Statistic (s Known) Step 3: Specify the Type I error rate-

= 0.05 z/2 = 1.96

Step 4: Establish the decision rule- Reject H0 if the test statistic < -1.96 or it the

test statistic > 1.96.


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Mean Using the z Statistic (s Known) Step 5: Gather sample data-

x-bar = $78,695, n = 112, s = $14,530,hypothesized = $74,914.

Step 6: Compute the test statistic.

75.2112/530,14

914,74695,78

z


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Mean Using the z Statistic (s Known) Step 7: Reach a statistical conclusion-

Since z = 2.75 > 1.96, reject H0.

Step 8: Business decision-

Statistically, the researcher has enoughevidence to reject the figure of $74,914 asthe true average salary for CPAs. Inaddition, based on the evidence gathered, it

may suggest that the average has increasedover the 10-year period.


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Mean Using the z Statistic (s Known)from a Finite Population

Test statistic:

1

N

nN

n

Xzs


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Using the p-Value to Test Hypotheses

Another way to reach a statisticalconclusion in hypothesis testing problems isby using the p-value, sometimes referred to

as the observed significance level. p-value < reject H0

p-value do not reject H0


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One should be careful when using p-valuesfrom statistical software outputs.

Both MINITAB and EXCEL report theactual p-values for hypothesis tests.

MINITAB doubles the p-value for a two-tailed test so you can compare with .

EXCEL does not double the p-value for a

two-tailed test. So when using the p-valuefrom EXCEL, you may multiply the valueby 2 and then compare with .


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Demonstration Problem: MINITAB


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Critical Value Method to Test Hypotheses

The critical value method determines thecritical mean value required forz to be inthe rejection region and uses it to test the

hypotheses.

n

xz cc

s


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For the previous example,

605,77and223,72

691,2914,74

112

530,1496.1914,74

112530,14

914,74

96.1

cc

c

c

xupperxlower

xor

x


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Thus, a sample mean greater than $77,605or less than $72,223 will result in therejection of the null hypothesis.

This method is particularly attractive inindustrial settings where standards can beset ahead of time and then quality controltechnicians can gather data and compare

actual measurements of products tospecifications.



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Mean Using the t Statistic (s Unknown) In this case, the test statistic will be

1

/

ndf

nsXt


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Two-tailed Test: s Unknown, = .05

(Part 1)

Example: Weights in Pounds of a Sampleof 20 Plates

22.6 22.2 23.2 27.4 24.5

27.0 26.6 28.1 26.9 24.9

26.2 25.3 23.1 24.2 26.1

25.8 30.4 28.6 23.5 23.6


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MINITAB Computer Printout

for the Machine Plate ExampleH0: = 25Ha: 25

Do not reject the null hypothesis

since P-value = 0.311 > = 0.05.


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Machine Plate Example: Excel

(Part 1)

Do not reject the

null hypothesis

sinceP-value = 0.3114 > = 0.05.


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Machine Plate Example: Excel

(Part 2)A B C D E

1 H0: = 25

2 Ha: 25

3

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 = 0.0511 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)


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z Test of Population Proportion

pq

p

pn

qp

ppz

-1

proportionpopulation

proportionsample:where

5

and,5

qn

pn


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z Test of Population Proportion

A manufacturer believes exactly 8% of itsproducts contain at least one minor flaw.Suppose the company wants to test thisbelief. A sample of 200 products resulted in33 items have at least one minor flaw. Usea probability of a Type I error of 0.10.

H0:p = 0.08Ha:p 0.08

Testing Hypotheses about a


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Testing Hypotheses about aProportion: Manufacturer Example

(Part 2)

.

. .

(. )(. ).

p

Zp P

P Q

n

33200

165

165 08

08 92

200

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 cz cZ 1645.

Critical Values

Non Rejection Region

Rejection Regions

cZ 1645.

205.

205.

.0

0

rejectnotdo1.645,If

.reject,645.1If

Hz

Hz

43.4

200

)92)(.08(.

08.165.

165.20033

n

qp

ppz

p

.reject,645.143.4Since0

Hz


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MINITAB Computer Printout

for the Minor Flaw ExampleH0:p = 0.08

Ha:p 0.08Reject the null hypothesis

since P-value = 0.000 < = 0.1.


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Using the Critical Value Method

112.0and048.0

032.008.0200)92.0)(08.0(

645.108.0

200)92.0)(08.0(

08.0645.1

2/

cpuppercplower

cpor

cp

npq

pcpz

Since the sample

proportion of

0.165

falls outside theinterval, the null

hypothesis

is rejected

H0:p = 0.08

Ha:p 0.08


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Testing Hypotheses About a Variance

The test statistic for this test is

2

22 )1(

s sn


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Testing Hypotheses About a Variance:

Demonstration Problem 9.4

Step 1:

Step 2: Test statistic

H0: s2 = 25Ha: s2 25

2

2

2 )1( s sn


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Step 3: Because this is a two-tailed test, = 0.10 and /2 = 0.05.

Step 4: The degrees of freedom are 161 =15. The two critical chi-square values are 2(10.05), 15 = 2 0.95, 15 = 7.26093 and 2 0.05,15 = 24.9958.

Step 5: The data are listed in the text.

Step 6: The sample variance is s2 = 28.1. The

observed chi-square value is calculated as2 = 16.86.


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Step 7: The observed chi-square value is inthe nonrejection region because 2 0.95, 15 =7.26093 < 2observed = 16.86 <

20.05), 15 =

24.9958.

Step 8: This result indicates to the companymanagers that the variance of weeklyovertime hours is about what they expected.


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Solving for Type II Errors

When the null hypothesis is not rejected, theneither a correct decision is made or anincorrect decision is made.

If an incorrect decision is made, that is, if the

null hypothesis is not rejected when it is false,then a Type II error has occurred.

Finding the probability of a Type II error ismore complex than finding the probability of

Type I error. A Type II error, , varies with possiblevalues of the alternative parameter.


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Solving for Type II Errors (Soft Drink)

Suppose a test is conducted on the followinghypotheses: H0: = 12 ounces vs. Ha: < 12ounces when the sample size is 60 with meanof 11.985.

The first step in determining the probabilityof a Type II error is to calculate a criticalvalue for the sample mean (in this case).

For an =0.05, then the critical value for the

sample mean is (given on next slide).


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979.11

60/10.0

12645.1

/

cx

cx

n

cx

cz

s

In testing the null hypothesis

by the critical value method,

this value is used as the cutoff

for the nonrejection region.

For any sample mean

obtained that is less than

11.979, the null hypothesis is

rejected. Any sample mean

greater than 11.979, the nullhypothesis is not rejected.


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Since a Type II error, , varies with possiblevalues of the alternative parameter, then foran alternative mean of 11.99 (< 12) thecorresponding z-value is

85.0

60/10.0

99.11979.11

/1

1

1

1

z

z

n

cxz

s


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INSERT FIGURE 9.20


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The value of z yields an area of 0.3023. The probability of committing a Type II error

is equal to the area to the right of the criticalvalue of the sample mean of 11.979.

This area is = 0.3023 + 0.5000 = 0.8023. Thus, there is an 80.23% chance ofcommitting a Type II error if the alternativemean is 11.99.

Note: equivalent problems can be solved forsample proportions (See DemonstrationProblem 9.6).

O i Ch i i d P


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Operating Characteristic and Power

Curve

Because the probability of committing a TypeII error changes for each different value ofthe alternative parameter, it is best to examinea series of possible alternative values.

The power of a test is the probability ofrejecting the null hypothesis when it is false.

Power = 1 - .

O i Ch i i d P


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Curve (Soft Drink)

O ti Ch t i ti d P


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Curve (Soft Drink)

12.0011.9911.9811.9711.9611.95

0.9

0.8

0.7

0.6

0.5

0.4

0.3

0.2

0.1

0.0

Alternative Means

Type

IIError

O ti Ch t i ti d P


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Curve (Soft Drink)

12.0011.9911.9811.9711.9611.95

1.0

0.8

0.6

0.4

0.2

0.0

Alternative Means

Power


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Ken Black QA ch09

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