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Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

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Page 1: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-1

Chapter 8Introduction to

Hypothesis Testing

Page 2: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-2.

Chapter Goals

After completing this chapter, you should be able to:

Formulate null and alternative hypotheses for applications involving a single population mean or proportion

Formulate a decision rule for testing a hypothesis Know how to use the test statistic, critical value, and

p-value approaches to test the null hypothesis Know what Type I and Type II errors are Compute the probability of a Type II error

Page 3: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-3.

What is a Hypothesis?

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

population mean

population proportion

Example: The mean monthly cell phone bill of this city is = $42

Example: The proportion of adults in this city with cell phones is p = .68

Page 4: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-4.

The Null Hypothesis, H0

States the assumption (numerical) to be tested

Example: The average number of TV sets in

U.S. Homes is at least three ( )

Is always about a population parameter, not about a sample statistic

3μ:H0

3μ:H0 3x:H0

Page 5: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-5.

The Null Hypothesis, H0

Begin with the assumption that the null hypothesis is true Similar to the notion of innocent until

proven guilty Refers to the status quo Always contains “=” , “≤” or “” sign May or may not be rejected

(continued)

Page 6: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-6.

The Alternative Hypothesis, HA

Is the opposite of the null hypothesis e.g.: The average number of TV sets in U.S.

homes is less than 3 ( HA: < 3 )

Challenges the status quo Never contains the “=” , “≤” or “” sign May or may not be accepted Is generally the hypothesis that is believed

(or needs to be supported) by the researcher

Page 7: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

.

Population

Claim: thepopulationmean age is 50.(Null Hypothesis:

REJECT

Supposethe samplemean age is 20: x = 20

SampleNull Hypothesis

20 likely if = 50?Is

Hypothesis Testing Process

If not likely,

Now select a random sample

H0: = 50 )

x

Page 8: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-8.

Sampling Distribution of x

= 50If H0 is true

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

... then we reject the null

hypothesis that = 50.

Reason for Rejecting H0

20

... if in fact this were the population mean…

x

Page 9: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-9.

Level of Significance,

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

distribution

Is designated by , (level of significance) Typical values are .01, .05, or .10

Is selected by the researcher at the beginning

Provides the critical value(s) of the test

Page 10: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-10.

Level of Significance and the Rejection Region

H0: μ ≥ 3

HA: μ < 3 0

H0: μ ≤ 3

HA: μ > 3

H0: μ = 3

HA: μ ≠ 3

/2

Represents critical value

Lower tail test

Level of significance =

0

0

/2

Upper tail test

Two tailed test

Rejection region is shaded

Page 11: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-11.

Errors in Making Decisions

Type I Error Reject a true null hypothesis Considered a serious type of error

The probability of Type I Error is Called level of significance of the test Set by researcher in advance

Page 12: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-12.

Errors in Making Decisions

Type II Error Fail to reject a false null hypothesis

The probability of Type II Error is β

(continued)

Page 13: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-13.

Outcomes and Probabilities

State of Nature

Decision

Do NotReject

H0

No error (1 - )

Type II Error ( β )

RejectH0

Type I Error( )

Possible Hypothesis Test Outcomes

H0 False H0 True

Key:Outcome

(Probability) No Error ( 1 - β )

Page 14: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-14.

Type I & II Error Relationship

Type I and Type II errors can not happen at the same time

Type I error can only occur if H0 is true

Type II error can only occur if H0 is false

If Type I error probability ( ) , then

Type II error probability ( β )

Page 15: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-15.

Factors Affecting Type II Error

All else equal, β when the difference between

hypothesized parameter and its true value

β when

β when σ

β when n

Page 16: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-16.

Critical Value Approach to Testing

Convert sample statistic (e.g.: ) to test statistic ( Z or t statistic )

Determine the critical value(s) for a specifiedlevel of significance from a table or computer

If the test statistic falls in the rejection region,

reject H0 ; otherwise do not reject H0

x

Page 17: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-17.

Reject H0 Do not reject H0

The cutoff value,

or , is called a

critical value

Lower Tail Tests

-zα

-zα xα

0

μ

H0: μ ≥ 3

HA: μ < 3

n

σzμx

Page 18: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-18.

Reject H0Do not reject H0

The cutoff value,

or , is called a

critical value

Upper Tail Tests

zα xα

H0: μ ≤ 3

HA: μ > 3

n

σzμx

Page 19: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-19.

Do not reject H0 Reject H0Reject H0

There are two cutoff values

(critical values):

or

Two Tailed Tests

/2

-zα/2

xα/2

± zα/2

xα/2

0μ0

H0: μ = 3

HA: μ

3

zα/2

xα/2

n

σzμx /2/2

Lower

Upperxα/2

Lower Upper

/2

Page 20: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-20.

Critical Value Approach to Testing

Convert sample statistic ( ) to a test statistic ( Z or t statistic )

x

Known Unknown

Hypothesis Tests for

Page 21: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-21.

Known Unknown

Hypothesis Tests for μ

The test statistic is:

Calculating the Test Statistic

n

σμx

z

Page 22: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-22.

Known

Working WithLarge Samples

Unknown

Hypothesis Tests for

The test statistic is:

Calculating the Test Statistic

n

sμx

t 1n

But is sometimes approximated using a z:

n

σμx

z

(continued)

Page 23: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-23.

Known Unknown

Hypothesis Tests for

Using Small

Samples

The test statistic is:

Calculating the Test Statistic

n

sμx

t 1n

(The population must be approximately normal)

(continued)

Page 24: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-24.

Review: Steps in Hypothesis Testing

1. Specify the population value of interest

2. Formulate the appropriate null and alternative hypotheses

3. Specify the desired level of significance

4. Determine the rejection region

5. Obtain sample evidence and compute the test statistic

6. Reach a decision and interpret the result

Page 25: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-25.

Hypothesis Testing Example

Test the claim that the true mean # of TV sets in US homes is at least 3.

(Assume σ = 0.8)

1. Specify the population value of interest The mean number of TVs in US homes

2. Formulate the appropriate null and alternative hypotheses H0: μ 3 HA: μ < 3 (This is a lower tail test)

3. Specify the desired level of significance Suppose that = .05 is chosen for this test

Page 26: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-26.

Reject H0 Do not reject H0

4. Determine the rejection region

= .05

-zα= -1.645 0

This is a one-tailed test with = .05. Since σ is known, the cutoff value is a z value:

Reject H0 if z < z = -1.645 ; otherwise do not reject H0

Hypothesis Testing Example(continued)

Page 27: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-27.

5. Obtain sample evidence and compute the test statistic

Suppose a sample is taken with the following results: n = 100, x = 2.84 ( = 0.8 is assumed known)

Then the test statistic is:

2.0.08

.16

100

0.832.84

n

σμx

z

Hypothesis Testing Example

Page 28: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-28.

Reject H0 Do not reject H0

= .05

-1.645 0

6. Reach a decision and interpret the result

-2.0

Since z = -2.0 < -1.645, we reject the null hypothesis that the mean number of TVs in US homes is at least 3

Hypothesis Testing Example(continued)

z

Page 29: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-29.

Reject H0

= .05

2.8684

Do not reject H0

3

An alternate way of constructing rejection region:

2.84

Since x = 2.84 < 2.8684, we reject the null hypothesis

Hypothesis Testing Example(continued)

x

Now expressed in x, not z units

2.8684100

0.81.6453

n

σzμx αα

Page 30: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-30.

p-Value Approach to Testing

Convert Sample Statistic (e.g. ) to Test Statistic ( z or t statistic )

Obtain the p-value from a table or computer

Compare the p-value with

If p-value < , reject H0

If p-value , do not reject H0

x

Page 31: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-31.

p-Value Approach to Testing

p-value: Probability of obtaining a test statistic more extreme ( ≤ or ) than the observed sample value given H0 is

true

Also called observed level of significance

Smallest value of for which H0 can be

rejected

(continued)

Page 32: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-32.

p-value =.0228

= .05

p-value example

Example: How likely is it to see a sample mean of 2.84 (or something further below the mean) if the true mean is = 3.0? n=100, sigma=0.8

2.8684 3

2.84

x

.02282.0)P(z

1000.8

3.02.84zP

3.0)μ|2.84xP(

Page 33: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-33.

Computing p value in R

R has a function called pnorm which computes the area under the standard normal distribution z~N(0,1). If we give it the z value, the function pnrom in R computes the entire area from negative infinity to that z. For examples:

> pnorm(-2) [1] 0.02275013 pnorm(0) # is 0.5

Page 34: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-34.

Compare the p-value with

If p-value < , reject H0

If p-value , do not reject H0

Here: p-value = .0228 = .05

Since .0228 < .05, we reject the null hypothesis

(continued)

p-value example

p-value =.0228

= .05

2.8684 3

2.84

Page 35: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-35.

Example: Upper Tail z Test for Mean ( Known)

A phone industry manager thinks that customer monthly cell phone bill have increased, and now average over $52 per month. The company wishes to test this claim. (Assume = 10 is known)

H0: μ ≤ 52 the average is not over $52 per month

HA: μ > 52 the average is greater than $52 per month(i.e., sufficient evidence exists to support the manager’s claim)

Form hypothesis test:

Page 36: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-36.

Reject H0Do not reject H0

Suppose that = .10 is chosen for this test

Find the rejection region:

= .10

zα=1.280

Reject H0

Reject H0 if z > 1.28

Example: Find Rejection Region

(continued)

Page 37: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-37.

Finding rejection region in R

Rejection region is defined from a dividing line between accept and reject. Given the tail area or the probability that z random variable is inside the tail, we want the z value which defines the dividing line. The z can fall inside the lower (left) tail or the right tail. This will mean looking up the N(0,1) table backwards. The R command:

qnorm(0.10, lower.tail=FALSE) [1] 1.281552

Page 38: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-38.

Review:Finding Critical Value - One

Tail

Z .07 .09

1.1 .3790 .3810 .3830

1.2 .3980 .4015

1.3 .4147 .4162 .4177z 0 1.28

.08

Standard Normal Distribution Table (Portion)What is z given =

0.10?

= .10

Critical Value = 1.28

.90

.3997

.10

.40.50

Page 39: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-39.

Obtain sample evidence and compute the test statistic

Suppose a sample is taken with the following results: n = 64, x = 53.1 (=10 was assumed known)

Then the test statistic is:

0.88

64

105253.1

n

σμx

z

Example: Test Statistic(continued)

Page 40: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-40.

Reject H0Do not reject H0

Example: Decision

= .10

1.280

Reject H0

Do not reject H0 since z = 0.88 ≤ 1.28

i.e.: there is not sufficient evidence that the mean bill is over $52

z = .88

Reach a decision and interpret the result:(continued)

Page 41: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-41.

Reject H0

= .10

Do not reject H0 1.28

0

Reject H0

z = .88

Calculate the p-value and compare to

(continued)

.1894

.3106.50.88)P(z

6410

52.053.1zP

52.0)μ|53.1xP(

p-value = .1894

p -Value Solution

Do not reject H0 since p-value = .1894 > = .10

Page 42: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-42.

Using R to find p-value

Given the test statistic (z value) finding the p value means finding a probability or area under the N(0,1) curve. This means we use the R command: pnorm(0.88, lower.tail=FALSE)

[1] 0.1894297 This is the p-value from the previous slide! If the z is negative, do not use the option

lower.tail=FALSE. For example, pnorm(-0.88) [1] 0.1894297

Page 43: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-43.

Example: Two-Tail Test( Unknown)

The average cost of a hotel room in New York is said to be $168 per night. A random sample of 25 hotels resulted in x = $172.50 and

s = $15.40. Test at the

= 0.05 level.(Assume the population distribution is normal)

H0: μ= 168

HA: μ

168

Page 44: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-44.

= 0.05

n = 25

is unknown, so use a t statistic

Critical Value:

t24 = ± 2.0639

Example Solution: Two-Tail Test

Do not reject H0: not sufficient evidence that true mean cost is different than $168

Reject H0Reject H0

/2=.025

-tα/2

Do not reject H0

0 tα/2

/2=.025

-2.0639 2.0639

1.46

25

15.40168172.50

n

sμx

t 1n

1.46

H0: μ= 168

HA: μ

168

Page 45: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-45.

Critical value of t in R

By analogy with pnorm and qnorm R has functions pt to find the probability under the t density and qt to look at t tables backward and get the critical value of t from knowing the tail probability. Here we need to specify the degrees of freedom (df).

qt(0.025, lower.tail=FALSE, df=24) [1] 2.063899 pt(1.46, lower.tail=FALSE, df=24) [1] 0.07862868

Page 46: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-46.

Hypothesis Tests for Proportions

Involves categorical values

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

“Failure” (does not possesses that characteristic)

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

Page 47: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-47.

Proportions

Sample proportion in the success category is denoted by p

When both np and n(1-p) are at least 5, p can be approximated by a normal distribution with mean and standard deviation

(continued)

sizesample

sampleinsuccessesofnumber

n

xp

pμP

n

p)p(1σ

p

Page 48: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-48.

The sampling distribution of p is normal, so the test statistic is a z value:

Hypothesis Tests for Proportions

n)p(p

ppz

1

np 5and

n(1-p) 5

Hypothesis Tests for p

np < 5or

n(1-p) < 5

Not discussed in this chapter

Page 49: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-49.

Example: z Test for Proportion

A marketing company claims that it receives 8% responses from its mailing. To test this claim, a random sample of 500 were surveyed with 25 responses. Test at the = .05 significance level.

Check:

n p = (500)(.08) = 40

n(1-p) = (500)(.92) = 460

Page 50: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-50.

Z Test for Proportion: Solution

= .05

n = 500, p = .05

Reject H0 at = .05

H0: p = .08

HA: p

.08

Critical Values: ± 1.96

Test Statistic:

Decision:

Conclusion:

z0

Reject Reject

.025.025

1.96

-2.47

There is sufficient evidence to reject the company’s claim of 8% response rate.

2.47

500.08).08(1

.08.05

np)p(1

ppz

-1.96

Page 51: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-51.

Do not reject H0

Reject H0Reject H0

/2 = .025

1.960

z = -2.47

Calculate the p-value and compare to (For a two sided test the p-value is always two sided)

(continued)

0.01362(.0068)

.4932)2(.5

2.47)P(x2.47)P(z

p-value = .0136:

p -Value Solution

Reject H0 since p-value = .0136 < = .05

z = 2.47

-1.96

/2 = .025

.0068.0068

Page 52: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-52.

Two-sided p value in R

As before, finding p value means finding a probability and it involves the function pnorm or pt as the case may be. Here we want two-sided p value given the z value of + or -2.47.

pnorm(-2.47) [1] 0.006755653 We have to double this as there are 2 tail

probabilities as 2*0.0068=0.0136

Page 53: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-53.

Reject H0: μ 52

Do not reject H0 : μ 52

Type II Error

Type II error is the probability of

failing to reject a false H0

5250

Suppose we fail to reject H0: μ 52 when in fact the true mean is μ = 50

Page 54: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-54.

Reject H0: 52

Do not reject H0 : 52

Type II Error

Suppose we do not reject H0: 52 when in fact the true mean is = 50

5250

This is the true distribution of x if = 50

This is the range of x where H0 is not rejected

(continued)

Page 55: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-55.

Reject H0: μ 52

Do not reject H0 : μ 52

Type II Error

Suppose we do not reject H0: μ 52 when in fact the true mean is μ = 50

5250

β

Here, β = P( x cutoff ) if μ = 50

(continued)

Page 56: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-56.

Reject H0: μ 52

Do not reject H0 : μ 52

Suppose n = 64 , σ = 6 , and = .05

5250

So β = P( x 50.766 ) if μ = 50

Calculating β

50.76664

61.64552

n

σzμxcutoff

(for H0 : μ 52)

50.766

Page 57: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-57.

Reject H0: μ 52

Do not reject H0 : μ 52

.1539.3461.51.02)P(z

646

5050.766zP50)μ|50.766xP(

Suppose n = 64 , σ = 6 , and = .05

5250

Calculating β(continued)

Probability of type II error:

β = .1539

Page 58: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-58.

Chapter Summary

Addressed hypothesis testing methodology

Performed z Test for the mean (σ known)

Discussed p–value approach to hypothesis testing

Performed one-tail and two-tail tests . . .

Page 59: Chap 8-1 Chapter 8 Introduction to Hypothesis Testing

Chap 8-59.

Chapter Summary

Performed t test for the mean (σ unknown)

Performed z test for the proportion

Discussed type II error and computed its probability

(continued)