7 - 1 © 1998 prentice-hall, inc. statistics for managers using microsoft excel, 1/e statistics for...

7 - 1

© 1998 Prentice-Hall, Inc.

Statistics for Managers Using Microsoft Excel, 1/e

Statistics for Managers Using Microsoft Excel

Fundamentals of Hypothesis TestingChapter 7

Learning Objectives Describe the hypothesis testing process Distinguish the types of hypotheses Explain hypothesis testing errors Solve hypothesis testing problems

One population mean One population proportion One & two-tailed tests

7 - 2



Statistical Methods

StatisticalMethods

DescriptiveStatistics

InferentialStatistics

EstimationHypothesis

Testing

7 - 3



Reject hypothesis! Not close.

Reject hypothesis! Not close.

Hypothesis Testing

Population

I believe the population mean age is 50 (hypothesis).

MeanMean X X = 20= 20

Random sample

7 - 4



What’s a Hypothesis?

A belief about a population parameter

Parameter is population mean, proportion, variance

Must be statedbefore analysis

I believe the mean GPA of this class is 3.5!

© 1984-1994 T/Maker Co.

7 - 5



Null Hypothesis

What is tested Has serious outcome if incorrect

decision made Always has equality sign: , or Designated H0

Pronounced ‘H sub-zero’ or ‘H oh’

Example H0: 3

7 - 6



Alternative Hypothesis

Opposite of null hypothesis Always has inequality sign: ,, or Designated H1

Example H1: < 3

7 - 7



Identifying Hypotheses in Problems

Problem: Test that the population mean is not 3

Steps State the question statistically: 3 State the opposite statistically: = 3

Must be mutually exclusive & exhaustive Select the null hypothesis: = 3

Has the =, , or sign

7 - 8



Identifying Hypotheses Thinking Challenge

Is the population average amount of TV viewing 12 hours?

What are the hypotheses?

7 - 9





Is the population average amount of TV viewing different from 12 hours?

What are the hypotheses?What are the hypotheses?

7 - 10






Is the average cost per hat less than or equal to $20?


7 - 11






Is the average cost per hat less than or equal to $20?

Is the average amount spent in the bookstore greater than $25?


7 - 12



Basic Idea

Sample Mean = 50

Sampling DistributionSampling Distribution

It is unlikely that we would get a sample mean of this value ...

... if in fact this were the population mean

... therefore, we reject the hypothesis that = 50.

20

H0

7 - 13



Level of Significance

Defines unlikely values of sample statistic if null hypothesis is true Called rejection region of sampling

distribution

Designated (alpha) Typical values are .01, .05, .10

Selected by researcher at start

7 - 14



HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

Rejection Region (One-Tail Test)


1 -

Level of ConfidenceLevel of Confidence

Observed sample statisticObserved sample statistic

7 - 15



HoValueCritical

Value

Sample Statistic

RejectionRegion

NonrejectionRegion

Rejection Region (One-Tail Test)


1 -

Level of ConfidenceLevel of Confidence

ObservedObserved sample statistic sample statistic

7 - 16



Rejection Regions (Two-Tailed Test)

HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion

Sampling Distribution

1 - 1 -

Level of Confidence


7 - 17




HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 - 1 -

Level of Confidence


7 - 18




HoValue Critical

ValueCriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 - 1 -

Level of Confidence


7 - 19



Risk of Errors in Making Decision

Type I error Reject true null hypothesis Has serious consequences Probability of Type I error is

Called level of significance

Type II error Do not reject false null hypothesis Probability of Type II error is (Beta)

7 - 20



Decision Results

HH00: Innocent: Innocent

Jury Trial H0 Test

Actual Situation Actual Situation

Verdict Innocent Guilty Decision H0 True H0

False

Innocent Correct ErrorDo NotReject

H0

1 - Type IIError

()

Guilty Error Correct RejectH0

Type IError ()

Power(1 - )

7 - 21



& Have an Inverse Relationship

You can’t reduce both errors simultaneously!

7 - 22



Factors Affecting

True value of population parameter increases when difference with

hypothesized parameter decreases

Significance level, Increases when decreases

Population standard deviation, Increases when increases

Sample size, n Increases when n decreases

7 - 23



Hypothesis H0 Testing Steps

State H0

State H1

Choose

Choose n

Choose test

7 - 24



H0 Testing Steps

Set up critical values

Collect data

Compute test statistic

Make statistical decision

Express decision

State HState H00

State HState H11

Choose Choose

Choose Choose nn

Choose testChoose test

7 - 25



One Population Tests

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Known

Z Test(1 & 2tail)

Mean ProportionUnknown

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Known

Z Test(1 & 2tail)


7 - 26



Two-Tailed Z Test for Mean ( Known)

Assumptions Population is normally distributed If not normal, can be approximated by

normal distribution (n 30)

Null hypothesis has = sign only Z-test statistic

ZX X

n

x

x

Z

X X

n

x

x

7 - 27



Two-Tailed Z Test Example

Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showedX = 372.5. The company has specified to be 15 grams. Test at the .05 level. 368 gm.368 gm.

7 - 28



Two-Tailed Z Test Solution

H0: = 368

H1: 368

.05

n 25

Critical Value(s):

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

Z0 1.96-1.96

.025

Reject H 0 Reject H 0

.025

Z0 1.96-1.96

.025


.025

ZX

n

372 5 368

1525

150.

.

Do not reject at = .05

No evidence No evidence average is not 368average is not 368

7 - 29



p-Value

Probability of obtaining a test statistic more extreme (or than actual sample value given H0 is true

Called observed level of significance Smallest value of H0 can be rejected

Used to make rejection decision If p-value , do not reject H0

If p-value < , reject H0

7 - 30



Two-Tailed Z Test p-Value Solution

Z0 1.50-1.50

1/2 p-Value.0668

1/2 p-Value.0668

p-value is P(Z -1.50 or Z 1.50) = .1336

Z value of sample statistic

From Z table: lookup 1.50

.4332

.5000- .4332

.0668

7 - 31



Two-Tailed Z Test p-Value Solution

0 1.50-1.50 Z

RejectReject

(p-Value = .1336) ( = .05).

Do not reject.1/2 p-Value = .06681/2 p-Value = .0668

1/2 = .0251/2 = .025

Test statistic is in ‘Do not reject’ region

7 - 32



Two-Tailed Z Test Thinking Challenge

You’re a Q/C inspector. You want to find out if a new machine is making electrical cords to customer specification: average breaking strength of 70 lb. with = 3.5 lb. You take a sample of 36 cords & compute a sample mean of 69.7 lb. At the .05 level, is there evidence that the machine is not meeting the average breaking strength?

7 - 33



Two-Tailed Z Test Solution*

H0: = 70

H1: 70

= .05

n = 36Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Z0 1.96-1.96

.025


.025

ZX

n

69 7 70

3 536

51.

..


No evidence average is not 70

7 - 34



One-Tailed Z Test for Mean ( Known)



Null hypothesis has or sign only

7 - 35



One-Tailed Z Test for Mean ( Known)



Null hypothesis has or sign only Z-test statistic

ZX X

n

x

x

7 - 36



Z0

Reject H 0

Z0

Reject H 0

One-Tailed Z Test for Mean Hypotheses

H0:0 H1: < 0

H0:0 H1: > 0

Must be significantly below

Small values satisfy H0 . Don’t reject!

7 - 37



Z0

= 1

One-Tailed Z Test Finding Critical Z

.500 - .025

.475

What Is Z given = .025?

= .025

7 - 38



Z0

= 1

1.96

Z .05 .07

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

One-Tailed Z Test Finding Critical Z

.500 - .025

.475

.06

1.9 .4750

Standardized Normal Probability Table (Portion)

What Is Z given = .025?

= .025

7 - 39



One-Tailed Z Test Example

Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showedX = 372.5. The company has specified to be 15 grams. Test at the .05 level.

368 gm.368 gm.

7 - 40



One-Tailed Z Test Solution

H0: 368

H1: > 368

= .05

n = 25

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Z0 1.645

.05

Reject

ZX

n

372 5 368

1525

150.

.


No evidence average is more than 368

7 - 41



One-Tailed Z Test p-Value Solution

0 1.50 Z

Reject

0 1.50 Z

Reject

(p-Value = .0668) ( = .05).

Do not reject.p-Value = .0668

= .05

Test statistic is in ‘Do not reject’ region

7 - 42



One-Tailed Z Test Thinking Challenge

You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. At the .01 level, is there evidence that the miles per gallon is at least 32?

7 - 43



Solution Template

H0:

H1:

=

n =

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Z0

7 - 44



One-Tailed Z Test Solution*

H0:

H1:

=

n =

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

7 - 45




H0: 32

H1: < 32

=

n =

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

7 - 46




H0: 32

H1: < 32

= .01

n = 60

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

7 - 47




H0: 32

H1: < 32

= .01

n = 60

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Z0-2.33

.01

Reject

7 - 48




H0: 32

H1: < 32

= .01

n = 60

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Z0-2.33

.01

Reject

ZX

n

30 7 32

3 860

2 65.

..

7 - 49




H0: 32

H1: < 32

= .01

n = 60

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Z0-2.33

.01

Reject

ZX

n

30 7 32

3 860

2 65.

..

Reject at = .01

7 - 50




H0: 32

H1: < 32

= .01

n = 60

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Z0-2.33

.01

Reject

ZX

n

30 7 32

3 860

2 65.

..

Reject at = .01

There is evidence average is less than 32

7 - 51



p-Value Thinking Challenge

You’re an analyst for Ford. You want to find out if the average miles per gallon of Escorts is at least 32 mpg. Similar models have a standard deviation of 3.8 mpg. You take a sample of 60 Escorts & compute a sample mean of 30.7 mpg. What is the value of the observed level of significance (p-Value)?

7 - 52



p-Value Solution*

Z0-2.65

p-Value.004

Z value of sample statistic

From Z table: lookup 2.65

.4960.4960

Use alternative hypothesis to find direction

.5000- .4960

.0040

p-Value is P(Z -2.65) = .004. p-Value < ( = .01). Reject H0.

7 - 53




OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Known

Z Test(1 & 2tail)


7 - 54



t Test for Mean ( Unknown)

Assumptions Population is normally distributed If not normal, only slightly skewed &

large sample (n 30) taken

Parametric test procedure t test statistic

tX

Sn

7 - 55



t0 2.920-2.920

Two-Tailed t Test Finding Critical t Values

/2 = .05

/2 = .05

Given: n = 3; = .10

df = n - 1 = 2

7 - 56



Upper Tail Area

df .25 .10 .05

1 1.000 3.078 6.314

2 0.817 1.886

3 0.765 1.638 2.353

t0 2.920-2.920

Two-Tailed t Test Finding Critical t Values

Critical Values of t Table (Portion)

/2 = .05

/2 = .05

2.9202.920

Given: n = 3; = .10

df = n - 1 = 2

7 - 57



Two-Tailed t Test Example

Does an average box of cereal contain 368 grams of cereal? A random sample of 36 boxes had a mean of 372.5 & a standard deviation of 12 grams. Test at the .05 level.

368 gm.368 gm.

7 - 58



Two-Tailed t Test Solution

H0: = 368

H1: 368

= .05

df = 36 - 1 = 35Critical Value(s):

Test Statistic:

Decision:

Conclusion:

7 - 59



Two-Tailed t Test Solution

H0: = 368

H1: 368

= .05

df = 36 - 1 = 35Critical Value(s):

Test Statistic:

Decision:

Conclusion:

t0 2.0301-2.0301

.025


.025

tX

Sn

372 5 368

1236

2 25.

.

Reject at = .05

There is evidence pop. average is not 368

7 - 60



Two-Tailed t Test Thinking Challenge

You work for the FTC. A manufacturer of detergent claims that the mean weight of detergent is 3.25 lb. You take a random sample of 64 containers. You calculate the sample average to be 3.238 lb. with a standard deviation of .117 lb. At the .01 level, is the manufacturer correct?

3.25 lb.

7 - 61



Two-Tailed t Test Solution*

H0: = 3.25

H1: 3.25

.01

df 64 - 1 = 63

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

t0 2.6561-2.6561

.005


.005

tX

Sn

3 238 3 25

11764

82. .

..


There is no evidence average is not 3.25

7 - 62



One-Tailed t Test Example

Is the average capacity of batteries at least 140 ampere-hours? A random sample of 20 batteries had a mean of 138.47 & a standard deviation of 2.66. Assume a normal distribution. Test at the .05 level.

7 - 63



t0-1.7291

.05

Reject

One-Tailed t Test Solution

H0: 140

H1: < 140

= .05

df = 20 - 1 = 19

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

tX

Sn

138 47 140

2 6620

2 57.

..

Reject at = .05

There is evidence pop. average is less than 140

7 - 64



One-Tailed t Test Thinking Challenge

You’re a marketing analyst for Wal-Mart. Wal-Mart had teddy bears on sale last week. The weekly sales ($ 00) of bears sold in 10 stores was: 8 11 0 4 7 8 10 5 8 3. At the .05 level, is there evidence that the average bear sales per store is more than 5 ($ 00)?

7 - 65



t0 1.8331

.05

Reject

t0 1.8331

.05

Reject

One-Tailed t Test Solution*

H0: 5

H1: > 5

= .05

df = 10 - 1 = 9

Critical Value(s):

Test Statistic: Test Statistic:

Decision:Decision:

Conclusion:Conclusion:

tX

Sn

6 4 5

3 37310

131..

.

Do not reject at Do not reject at = .05 = .05

There is no evidence There is no evidence average is more than 5average is more than 5

7 - 66



Data Types

Data

Numerical(Quantitative)

Categorical(Qualitative)

Discrete Continuous

7 - 67



Categorical Data

Categorical random variables yield responses that classify e.g., gender (male, female)

Measurement reflects # in category Nominal or ordinal scale Examples

Do you own savings bonds? Do you live on-campus or off-campus?

7 - 68



Proportions

Involve categorical variables Fraction or % of population in a category If two categorical outcomes, binomial

distribution Possess or don’t possess characteristic

7 - 69



Proportions

Involve categorical variables Fraction or % of population in a category If two categorical outcomes, binomial

distribution Possess or don’t possess characteristic

Sample proportion (ps)

pXns

number of successessample size

7 - 70




OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Known

Z Test(1 & 2tail)


OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Known

Z Test(1 & 2tail)


7 - 71



One-Sample Z Test for Proportion

Assumptions Two categorical outcomes Population follows binomial distribution Normal approximation can be used

n·p 5 & n·(1 - p) 5

7 - 72



One-Sample Z Test for Proportion

Assumptions Two categorical outcomes Population follows binomial distribution Normal approximation can be used

n·p 5 & n·(1 - p) 5

Z-test statistic for proportion

n

PP

PpZ s

)(

1Hypothesized Hypothesized population proportion proportion

7 - 73



One-Proportion Z Test Example

The present packaging system produces 10% defective cereal boxes. Using a new system, a random sample of 200 boxes had11 defects. Does the new system produce fewer defects? Test at the .05 level.

n·p 5n·(1 - p) 5

7 - 74



One-Proportion Z Test Solution

H0: p .10

H1: p < .10

= .05

n = 200

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Z0-1.645

.05

Reject

7 - 75



One-Proportion Z Test Solution

H0: p .10

H1: p < .10

= .05

n = 200

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Z0-1.645

.05

Reject Reject at = .05

There is evidence new system < 10% defective

Zp pp p

n

s

( )

.

. ( . ).

1

11200

10

10 1 10200

2 12

7 - 76



One-Proportion Z Test Thinking Challenge

You’re an accounting manager. A year-end audit showed 4% of transactions had errors. You implement new procedures. A random sample of 500 transactions had 25 errors. Has the proportion of incorrect transactions changed at the .05 level?

n·p 5n·(1 - p) 5

7 - 77



One-Proportion Z Test Solution*

H0: p = .04

H1: p .04

= .05

n = 500

Critical Value(s):

Test Statistic:

Decision:

Conclusion:

Z0 1.96-1.96

.025


.025


There is evidence proportion is still 4%

Zp pp p

n

s

( )

.

. ( . ).

1

25500

04

04 1 04500

114

7 - 78



Conclusion

Described the hypothesis testing process

Distinguished the types of hypotheses Explained hypothesis testing errors Solved hypothesis testing problems

One population mean One population proportion One & two-tailed tests

7 - 1 © 1998 prentice-hall, inc. statistics for managers using microsoft excel, 1/e statistics for...

Documents

population average

challenge n

e statistics

e null hypothesis n

n steps

hypothesis testing population

population mean

n designated h