statistics for managers using microsoft® excel 7th edition chapter 9 fundamentals of hypothesis...

Statistics for ManagersUsing Microsoft® Excel

7th EditionChapter 9

Fundamentals of Hypothesis Testing:

One Sample Tests

Statistics for Managers Using Microsoft Excel, 7e © 2014 Pearson-Prentice-Hall, Inc. Philip A. Vaccaro , PhD

Learning Objectives

The Hypothesis

A hypothesis is a claim (assumption) about a population parameter: population mean ( μ )

pop. proportion ( π )

The Null Hypothesis, H0

3μ:H0

it is written in terms of the population

The Null Hypothesis, H0

The Alternative Hypothesis, H1

The Hypothesis Testing Process

Claim: The population mean age is 50. H0: μ = 50, H1: μ ≠ 50

Sample the population and find sample mean.Population

Sample


_


Sampling Distribution of X

μ = 50If H0 is true

20X

The Test Statistic and Critical Values

The Test Statistic and Critical Values

Critical Values

Distribution of the test statistic

Region of Rejection

Region of Rejection

Do Not Reject Ho

Thecriticalvalues

are statedin either‘z’ or ‘t’values

Errors in Decision Making

Type I Error Reject a true null hypothesis Considered a serious type of error The probability of a Type I Error is ‘’ Called level of significance of the test Set by researcher in advance:

Type II Error

Failure to reject false null hypothesis The probability of a Type II Error is ‘β’

within your control or dictated by management

Errors in Decision Making

Level of Significance, α

H0: μ ≥ 50

H1: μ < 500

H0: μ ≤ 50

H1: μ > 50

a

a

Lower-tail test

0

Upper-tail test

Two-tail test

0

H0: μ = 50

H1: μ ≠ 50

Claim: The population mean age is 50.

/2a

/2a

The probability of rejecting Ho when it is true

typical values

areα =.01α =.02α =.05α =.10

Hypothesis Testing: σ Known

_


Do not reject H0 Reject H0Reject H0

/2

-Z0

H0: μ = 3

H1: μ ≠

3

+Z

/2

Lower critical value

Upper critical value

3

Z

X


_


Is the test statistic in the rejection region?

Reject H0 Do not reject H0

a = .05/2

-Z= -1.96 0

= .05/2

Reject H0

+Z= +1.96

Here, Z = -2.0 < -1.96, so the test statistic is in the rejection region

a = .05level of

significance

Z 0.06

- 1.9 .0250

Z 0.06

+ 1.9 .9750

.025 .025

Hypothesis Testing: σ Known ,the ‘p’- Value Approach

* Because we can reject or not reject Ho “at a glance“ .

Hypothesis Testing: σ Knownp-Value Approach


.0228

/2 = .025

-1.96 0-2.0

Z1.962.0

.0228

/2 = .025

Z 0.00

- 2.00 .0228

Z 0.00

+ 2.00 .9772

Z = 2.84 - 3.00 .8 √100

= - .16 = - 2.00 .08

Z = 3.16 - 3.00 .8 √100

= +.16 = + 2.00 .08

Ho: μ = 3.00

2.84 3.16

(.9772)


Compare the p-value with If p-value < , reject H0

If p-value , do not reject H0

.0228

/2 = .025

-1.96 0

-2.0

Z1.96

2.0

.0228

/2 = .025

The probabilityof seeing a

sample meanof 2.84 or less,

or 3.16 or more , if

the populationmean is really 3.0

is only 4.56%

Hypothesis Testing: σ KnownConfidence Interval Connections

Simpler,faster

hypothesistest

Canonly be

used witha two-tailhypothesis

test !

Ifa = .05,

constructa 95%

confidenceinterval

Hypothesis Testing: σ KnownOne Tail Tests

H0: μ ≥ 3

H1: μ < 3

H0: μ ≤ 3

H1: μ > 3

Hypothesis Testing: σ KnownLower Tail Tests

There is only one critical value, since the rejection area is in

only one tail.

Reject H0 Do not reject H0

α

-Z

μ

Z

XCritical value

Hypothesis Testing: σ KnownUpper Tail Tests

There is only one critical value, since the rejection area is in

only one tail.

Reject H0

Do not reject H0

α

Zμ

Critical value

Z

X

Hypothesis Testing: σ KnownUpper Tail Test Example

Form the hypothesis test:

Hypothesis Testing: σ KnownUpper Tail Test Example Suppose that = .10 is chosen for this test Find the rejection region:

Reject H0Do not reject H0

= .10

Z0

Reject H0

1- = .90

Z 0.08

+ 1.2 .8997

Theentirea = .10goes to theupper

tail


What is Z given a = 0.10?

Z .07 .09

1.1 .8790 .8810 .8830

1.2 .8980 .9015

1.3 .9147 .9162 .9177z 0 1.28

.08a = .10

Critical Value = 1.28

.90

.8997

.10

.90


-


Reach a decision and interpret the result:

= .10

1.280

Reject H0

1- = .90

Z = .88

Do Not Reject Ho


Calculate the p-value and compare to ‘’

Reject H0

= .10

Do not reject H0

1.28

0

Reject H0

Z = .88 .1894

.810610.88)P(Z

6410/

52.053.1ZP

53.1)XP(

p-value = .1894Z 0.08

+ .8 .8106

Z = 53.1 - 52.0 = 1.1 10 / √ 64 1.25

= .88

1 - .8106 = .1894

“Talk”is notcheap,

especiallywhen

you arebeing

cheated !

Hypothesis Testing: σ Unknown

df =1df =2df =3df =4df =5df =6


Recall that the ‘t’ test statistic with n-1 degrees of freedom is:

Hypothesis Testing: σ Unknown Example

_

H0: μ =

168 H1: μ ¹ 168

_


H0: μ = 168

H1: μ ≠ 168

Reject H0Reject H0

α/2=.025

-t n-1,α/2

Do not reject H0

0

α/2=.025

-2.0639 2.0639

t n-1,α/2

Determine the regions of rejection

t .025

24 df 2.0639


a/2=.025

-t n-1,α/2 0

a/2=.025

-2.0639 2.0639

t n-1,α/21.46Since t df = 24, a=.025 = 1.46 < 2.0639,

Do Not Reject HoRejectHo

RejectHo

Hypothesis Testing: Connection to Confidence Intervals

For X = 172.5, S = 15.40 and n = 25, the 95% confidence interval is:

166.14 ≤ μ ≤ 178.86

Since this interval contains the hypothesized mean (168), you do not reject the null hypothesis at = .05 _

Hypothesis Testing:Proportions

Involves categorical variables ( yes/no , male/female )

Two possible outcomes “Success” (possesses a certain characteristic)

“Failure” (does not possesses that characteristic)

Fraction or proportion of the population in the “success” category is denoted by π


sizesample

sampleinsuccessesofnumber

n

Xp

pμn

)(1σ

p

Sample proportion in the success category is denoted by p

When both nπ and n(1-π) are at least 5, p can be approximated

by a normal distribution with mean and standard deviation


The sampling distribution of p is approximately normal, so the test statistic is a ‘Z’ value:

n

pZ

)1(

Hypothesis Testing:Proportions Example

A marketing company claims that it receives 8% responses from its mailings. To test this claim, a random sample of 500 were surveyed with 30 responses. Test at the = .05 significance level.

Hypothesis Testing:Proportions ExampleH0: π = .08 H1: π ≠ .08

α = .05

n = 500, p = .06

Critical Values: ± 1.96

z0

Reject Reject

.025.025

1.96-1.96

Determine region of rejection

Do Not RejectHo

30 / 500 = .06 = p (sample proportion)

Z 0.06

- 1.9 .0250

Z 0.06

+ 1.9 .9750

Hypothesis Testing:Proportions Example

Test Statistic: Decision:

Conclusion:

z0

.025.025

1.96-1.96-1.646

Do Not RejectHo

Potential Pitfalls and Ethical Considerations

Chapter Summary

In this chapter, we have

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