Statistics for Business and Economics Chapter 6 Inferences Based on a Single Sample: Tests of Hypothesis

Statistics for Business and Economics

Chapter 6

Inferences Based on a Single Sample: Tests of Hypothesis

Learning Objectives

1. Distinguish Types of Hypotheses

2. Describe Hypothesis Testing Process

3. Explain p-Value Concept

4. Solve Hypothesis Testing Problems Based on a Single Sample

5. Explain Power of a Test

Statistical Methods

StatisticalMethods

EstimationHypothesis

Testing

InferentialStatistics

DescriptiveStatistics

Hypothesis Testing Concepts

Hypothesis Testing

Population

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

Mean X = 20

Random sample

Reject hypothesis! Not close.

Reject hypothesis! Not close.

What’s a Hypothesis?

A belief about a population parameter

• Parameter is population mean, proportion, variance

• Must be stated before analysis

I believe the mean GPA of this class is 3.5!

© 1984-1994 T/Maker Co.

Null Hypothesis

1. What is tested

2. Has serious outcome if incorrect decision made

3. Always has equality sign: , , or 4. Designated H0 (pronounced H-oh)

5. Specified as H0: some numeric value • Specified with = sign even if or • Example, H0: 3

Alternative Hypothesis

1. Opposite of null hypothesis

2. Always has inequality sign: ,, or

3. Designated Ha

4. Specified Ha: ,, or some value

• Example, Ha: < 3

Identifying HypothesesSteps

Example 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 alternative hypothesis ( 3)— Has the , <, or > sign

• State the null hypothesis ( = 3)

• State the question statistically: = 12

• State the opposite statistically: 12

• Select the alternative hypothesis: Ha: 12

• State the null hypothesis: H0: = 12

Is the population average amount of TV viewing 12 hours?

What Are the Hypotheses?

• State the question statistically: 12

• State the opposite statistically: = 12


• State the null hypothesis: H0: = 12

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





• State the null hypothesis: H0: 20

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





• State the null hypothesis: H0: 25

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


Basic Idea

Sample Means = 50H0

Sampling Distribution

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

20

... if in fact this were the population mean

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

Level of Significance

1. Probability

2. Defines unlikely values of sample statistic if null hypothesis is true

• Called rejection region of sampling distribution

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

4. Selected by researcher at start

Rejection Region (One-Tail Test)

Ho

ValueCriticalValue

Sample Statistic

RejectionRegion

NonrejectionRegion


1 –

Level of Confidence

Observed sample statistic

Rejection Region (One-Tail Test)

Sampling DistributionLevel of Confidence

Ho

ValueCriticalValue

Sample Statistic

RejectionRegion

NonrejectionRegion


1 –

Level of Confidence


Rejection Regions (Two-Tailed Test)

Ho

Value CriticalValue

CriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 –

Level of Confidence


Ho

Value CriticalValue

CriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 –

Level of Confidence




Ho

Value CriticalValue

CriticalValue

1/2 1/2

Sample Statistic

RejectionRegion

RejectionRegion

NonrejectionRegion


1 –

Level of Confidence




Decision Making Risks

Errors in Making Decision

1. Type I Error• Reject true null hypothesis• Has serious consequences• Probability of Type I Error is (alpha)

— Called level of significance

2. Type II Error• Do not reject false null hypothesis• Probability of Type II Error is (beta)

Decision Results

H0: Innocent

Jury Trial

Actual Situation

Verdict Innocent Guilty

Innocent Correct Error

Guilty Error Correct

H0 Test

Actual Situation

Decision H0 True H0

False

AcceptH0

1 – Type IIError

()

RejectH0

Type IError ()

Power(1 – )

& Have an Inverse Relationship

You can’t reduce both errors simultaneously!

Factors Affecting 1. True value of population parameter

• Increases when difference with hypothesizedparameter decreases

2. Significance level, • Increases when decreases

3. Population standard deviation, • Increases when increases

4. Sample size, n• Increases when n decreases

Hypothesis Testing Steps

H0 Testing Steps

• State H0

• State Ha

• Choose

• Choose n

• Choose test

• Set up critical values

• Collect data

• Compute test statistic

• Make statistical decision

• Express decision

One Population Tests

OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)

Mean Proportion Variance

2 Test(1 & 2tail)

Two-Tailed Z Test of Mean ( Known)


OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

Two-Tailed Z Test for Mean ( Known)

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

normal distribution (n 30)

2. Alternative hypothesis has sign

x

x

X XZ

n

3. Z-Test Statistic

Two-Tailed Z Test for Mean Hypotheses

H0:=0 Ha: ≠ 0

Z0

Reject H0

Reject H

.500 - .025

.475

Z0

= 1

Two-Tailed Z Test Finding Critical Z

What is Z given = .05?

= .025

Z .05 .07

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

.06

1.9 .4750

Standardized Normal Probability Table (Portion)

1.96-1.96

Two-Tailed Z Test Example

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

368 gm.368 gm.

Two-Tailed Z Test Solution

• H0:

• Ha:

• • n

• Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 368

368

.0525

Z0 1.96-1.96

.025

Reject H 0 Reject H0

.025

372.5 3681.50

15

25

XZ

n

Do not reject at = .05

No evidence average is not 368

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 of significance, is there evidence that the machine is not meeting the average breaking strength?

Two-Tailed Z Test Solution*

• H0:

• Ha:

• = • n = • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 70

70

.05

36

Z0 1.96-1.96

.025


.025

69.7 70.51

3.5

36

XZ

n


No evidence average is not 70

One-Tailed Z Test of Mean ( Known)

One-Tailed Z Test for Mean ( Known)

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

normal distribution (n 30)

2. Alternative hypothesis has < or > sign

3. Z-test Statistic

x

x

X XZ

n

One-Tailed Z Test for Mean Hypotheses

H0:=0 Ha: < 0

Z0

Reject H0

Must be significantly below

Z0

Reject H0

H0:=0 Ha: > 0

Small values satisfy H0 . Don’t reject!

.500 - .025

.475

Z0

= 1

One-Tailed Z Test Finding Critical Z

What Is Z given = .025?

= .025

1.96

Z .05 .07

1.6 .4505 .4515 .4525

1.7 .4599 .4608 .4616

1.8 .4678 .4686 .4693

.4744 .4756

.06

1.9 .4750

Standardized Normal Probability Table (Portion)

One-Tailed Z Test Example

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

368 gm.368 gm.

One-Tailed Z Test Solution

• H0: • Ha: • = • n = • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 368

> 368

.05

25

Z0 1.645

.05

Reject

372.5 3681.50

15

25

XZ

n


No evidence average is more than 368

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 of significance, is there evidence that the miles per gallon is at least 32?

One-Tailed Z Test Solution*

• H0: • Ha: • = • n =• Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 32 < 32

.01

60

Z0-2.33

.01

Reject

30.7 322.65

3.8

60

XZ

n

Reject at = .01

There is evidence average is less than 32

Observed Significance Levels: p-Values

p-Value

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

2. Called observed level of significance• Smallest value of for which H0 can be rejected

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

• If p-value < , reject H0

Two-Tailed Z Test p-Value Example

Does an average box of cereal contain 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified to be 25 grams. Find the p-Value.

368 gm.368 gm.

Two-Tailed Z Test p-Value Solution

Z0 1.50

Z value of sample statistic (observed)

372.5 3681.50

15

25

XZ

n

1/2 p-Value1/2 p-Value


Z value of sample statistic (observed)

p-value is P(Z -1.50 or Z 1.50)

Z0 1.50-1.50

From Z table: lookup 1.50

.4332

.5000- .4332

.0668


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


.5000- .4332

.0668

Z0 1.50-1.50


0 1.50-1.50 Z

Reject H0Reject H0

1/2 p-Value = .06681/2 p-Value = .0668

1/2 = .0251/2 = .025

(p-Value = .1336) ( = .05). Do not reject H0.

Test statistic is in ‘Do not reject’ region

One-Tailed Z Test p-Value Example

Does an average box of cereal contain more than 368 grams of cereal? A random sample of 25 boxes showed x = 372.5. The company has specified to be 25 grams. Find the p-Value.

368 gm.368 gm.

One-Tailed Z Test p-Value Solution

Z0 1.50


372.5 3681.50

15

25

XZ

n


Use alternative hypothesis to find direction

p-Value is P(Z 1.50)


p-Value

Z0 1.50From Z table: lookup 1.50

.4332

.5000- .4332

.0668


p-Value.0668




.5000- .4332

.0668

p-Value is P(Z 1.50) = .0668

Z0 1.50

.4332

= .05


0 1.50 Z

Reject H0

p-Value = .0668

(p-Value = .0668) ( = .05). Do not reject H0.

Test statistic is in ‘Do not reject’ region

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)?


p-Value Solution*

Z0-2.65Z value of sample statistic From Z table:

lookup 2.65

.4960

p-Value.004

.5000- .4960

.0040

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

Two-Tailed t Test of Mean ( Unknown)


OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

t Test for Mean ( Unknown)

1. Assumptions• Population is normally distributed• If not normal, only slightly skewed & large

sample (n 30) taken

2. Parametric test procedure

3. t test statisticX

tS

n

t0

Two-Tailed t Test Finding Critical t Values

Given: n = 3; = .10

/2 = .05

/2 = .05

df = n - 1 = 2

v t .10 t .05 t .025

1 3.078 6.314 12.706

2 1.886 2.920 4.303

3 1.638 2.353 3.182

Critical Values of t Table (Portion)

2.920-2.920

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 and a standard deviation of 12 grams. Test at the .05 level of significance.

368 gm.368 gm.

Two-Tailed t Test Solution

• H0: • Ha: • = • df = • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 368 368

.05

36 - 1 = 35

t0 2.030-2.030

.025

Reject H0 Reject H0

.025

372.5 3682.25

12

36

Xt

S

n

Reject at = .05

There is evidence population average is not 368

Two-Tailed t TestThinking 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 of significance, is the manufacturer correct?

3.25 lb.3.25 lb.

Two-Tailed t Test Solution*

• H0: • Ha: • • df • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 3.25

3.25

.01

64 - 1 = 63

t0 2.656-2.656

.005

Reject H0 Reject H0

.005

3.238 3.25.82

.117

64

Xt

S

n


There is no evidence average is not 3.25

One-Tailed t Test of Mean ( Unknown)

One-Tailed t TestExample

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

One-Tailed t Test Solution

• H0: • Ha: • =• df =• Critical Value(s):

Test Statistic:

Decision:

Conclusion:

= 140

< 140

.0520 - 1 = 19

t0-1.729

.05

Reject H0

138.47 1402.57

2.66

20

Xt

S

n

Reject at = .05

There is evidence population average is less than 140

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 of significance, is there evidence that the average bear sales per store is more than 5 ($ 00)?

One-Tailed t Test Solution*


Test Statistic:

Decision:

Conclusion:

= 5 > 5

.05

10 - 1 = 9

t0 1.833

.05

Reject H0

6.4 51.31

3.373

10

Xt

S

n


There is no evidence average is more than 5

Z Test of Proportion

Data Types

Data

Quantitative Qualitative

ContinuousDiscrete

Qualitative Data

1. Qualitative random variables yield responses that classify

• e.g., Gender (male, female)

2. Measurement reflects number in category

3. Nominal or ordinal scale

4. Examples• Do you own savings bonds? • Do you live on-campus or off-campus?

Proportions1. Involve qualitative variables

2. Fraction or percentage of population in a category

3. If two qualitative outcomes, binomial distribution

• Possess or don’t possess characteristic

4. Sample Proportion (p)number of successes

ˆsample size

xp

n

^

Sampling Distribution of Proportion

1. Approximated by Normal Distribution

– Excludes 0 or n

2. Mean

3. Standard Error

P̂p


where p0 = Population Proportion

.0.0

.1.1

.2.2

.3.3

.0.0 .2.2 .4.4 .6.6 .8.8 1.01.0

PP^̂

P(PP(P^̂ )) ˆ1ˆ3ˆ ppnpn

n

ppp

)1( 00ˆ

Z = 0

zz= 1

Z

Standardizing Sampling Distribution of Proportion


Standardized Normal Distribution

P^P

P

^

^

Zp p p

p p

n

^p

p

^

^

( )1

^0

0 0


OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

One-Sample Z Test for Proportion

1. Assumptions

• Random sample selected from a binomial population

• Normal approximation can be used if

0 0ˆ ˆ15 and 15np nq

2. Z-test statistic for proportion0

0 0

p̂ pZ

p qn

Hypothesized

population proportion

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 of significance.

One-Proportion Z Test Solution

• H0: • Ha: • = • n =• Critical Value(s):

Test Statistic:

Decision:

Conclusion:

p = .10

p < .10

.05

200

Z0-1.645

.05

Reject H0

0

0 0

11.10ˆ 200 2.12

.10 .90200

p pZ

p qn

Reject at = .05

There is evidence new system < 10% defective

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 of significance?

One-Proportion Z Test Solution*

• H0: • Ha: • = • n = • Critical Value(s):

Test Statistic:

Decision:

Conclusion:

p = .04

p .04

.05

500

Z0 1.96-1.96

.025


.025

0

0 0

25.04ˆ 500 1.14

.04 .96500

p pZ

p qn


There is evidence proportion is not 4%

Calculating Type II Error Probabilities

Power of Test

1. Probability of rejecting false H0

• Correct decision

2. Designated 1 -

3. Used in determining test adequacy

4. Affected by• True value of population parameter

• Significance level • Standard deviation & sample size n

Finding PowerStep 1

X0 = 368

Reject H0

Do NotReject H0

Hypothesis:H0: 0 368Ha: 0 < 368 = .05

Draw15

25

n

Finding PowerSteps 2 & 3

Xa = 360

‘True’ Situation: a = 360 (Ha)

Draw

Specify

1-

X0 = 368

Reject H0

Do NotReject H0

Hypothesis:H0: 0 368Ha: 0 < 368 = .05

Draw15

25

n

Finding PowerStep 4

363.065363.065

065.363

25

1564.13680

n

ZX L

XXaa = 360= 360


Draw

Specify1-1-

X0 = 368

Reject H0

Do NotReject H0

Hypothesis:H0: 0 368Ha: 0 < 368 = .05

Draw15

25

n

Finding PowerStep 5

363.065363.065

065.363

25

1564.13680

n

ZX L

065.363

25

1564.13680

n

ZX L

XXaa = 360= 360


Draw

Specify

X0 = 368

Reject H0

Do NotReject H0

Hypothesis:H0: 0 368Ha: 0 < 368 = .05

Draw15

25

n

= .154

1- =.846

Z Table

Power Curves

Power Power

Power

Possible True Values for aPossible True Values for a

Possible True Values for a

H0: 0 H0: 0

H0: =0

= 368 in Example

Chi-Square (2) Test of Variance


OnePopulation

Z Test(1 & 2tail)

t Test(1 & 2tail)

Z Test(1 & 2tail)


2 Test(1 & 2tail)

Chi-Square (2) Testfor Variance

1. Tests one population variance or standard deviation

2. Assumes population is approximately normally distributed

3. Null hypothesis is H0: 2 = 02

4. Test statistic

Hypothesized pop. variance

Sample variance

2

2

2

1)

(n S

0

Chi-Square (2) Distribution

Select simple randomsample, size n.Compute s2

Compute 2 =(n-1)s 2 /2

Astronomical numberof 2 values

PopulationSampling Distributionsfor Different SampleSizes

21 2 30

What is the critical 2 value given:Ha: 2 > 0.7

n = 3 =.05?

Finding Critical Value Example

20

Upper Tail AreaDF .995 … .95 … .051 ... … 0.004 … 3.8412 0.010 … 0.103 … 5.991

2 Table (Portion)

df = n - 1 = 25.991

Reject

= .05


What is the critical 2 value given:Ha: 2 < 0.7

n = 3 =.05?

What do you do if the rejection region is on the left?

What is the critical 2 value given:Ha: 2 < 0.7

n = 3 =.05?


.103 20

Upper Tail AreaDF .995 … .95 … .051 ... … 0.004 … 3.8412 0.010 … 0.103 … 5.991

2 Table (Portion)

Upper Tail Area for Lower Critical Value = 1-.05 = .95 = .05

Reject H0

df = n - 1 = 2

Chi-Square (2) Test Example

Is the variation in boxes of cereal, measured by the variance, equal to 15 grams? A random sample of 25 boxes had a standard deviation of 17.7 grams. Test at the .05 level of significance.

Chi-Square (2) Test Solution


Test Statistic:

Decision:

Conclusion:

2 = 15

2 15

.0525 - 1 = 24

2200

/2 = .025

39.36412.401

= 33.42

2 22

2 20

( 1) (25 1) 17.7

15

n S


There is no evidence 2 is not 15

Conclusion

1. Distinguished Types of Hypotheses

2. Described Hypothesis Testing Process

3. Explained p-Value Concept

4. Solved Hypothesis Testing Problems Based on a Single Sample

5. Explained Power of a Test

Documents

Statistics for Business and Economics Chapter 6 Inferences Based on a Single Sample: Tests of Hypothesis