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Chapter 7

Hypothesis Testing

Lesson 7-1

Introduction

Introduction

Use a random sample to learn something

about a larger population.

Two ways to learn about a population

Confidence Intervals ( Chapter 6)

Hypothesis Testing (Chapter 7)

Confidence Intervals and

Hypothesis Testing

Allow us to use sample data to estimate a population

value, like the true mean or the true proportion.

Example: What is the true average amount

students spend weekly on gas?

Allows us to use sample data to test a claim about a

population, such as testing whether a population

proportion or population mean equals some number.

Example: Is the true average amount that

students spend weekly on gas $20.

Lesson 7-2

Basic of Hypothesis Testing

Illustrating Hypothesis Testing

Population of

Light Bulbs

Sample of 49

Light Bulbs

The manufacture claims

that the average lifetime

of a bulb is 500 hours?

A consumer advocate group

would like to know if the mean

lifetime of a bulb is less than 500

hours if the population average

was 500 hours.

Illustrate Hypothesis Testing

426

49 s

n

μ=

50

0

49

4

48

8

48

2

51

8

51

2

50

6

42

476

491

X

X

Illustrating Hypothesis Testing The sample mean of 476 hours is “too far” from the claimed

mean (500 hours) that we have reason to believe that the manufacture’s claim is probably not true.

Consumer advocate group would conclude that the population mean is likely some value less than 500 hours.

The sample mean of 491 hours is “close enough” to the claimed mean (500 hours) that we have reason to believe that the manufacture’s claim is probably true.

Consumer advocate group would conclude that the population mean is likely 500 hours.

Illustrating Hypothesis Testing

This raises an interesting question. How far

from the presumed mean of 500 hours can our

sample man be before we reject the idea that

the bulb lasts 500 hours?

To answer this question requires hypothesis

testing.

Definitions

In statistics, a hypothesis is a claim or

statement about a characteristic of a

population.

A hypothesis test (or test of significance) is

standard procedure for testing a claim about a

characteristic of a population.

The claims that we test regard the population

mean, population proportion and population

standard deviations.

Rare Event Rule for Inferential

Statistics If, under a given assumption, the probability

of a particular observed event is exceptionally

small, we conclude that the assumption if

probably not correct.

Example – Page 385, #2

What do you conclude? Use the rare event rule (Under a given assumption, the

probability of a particular observed event is exceptionally small, we conclude that

the assumption is probability not correct.

Claim: A gender selection method is effective in helping couples

have baby girls and among 50 babies, 49 are girls.

Gender selection appears to be effective. There is sufficient

evidence to support the claim. Under the assumption that the claim

is not true, having 49 girls among 50 babies would be a

rare event (unusual).

Two Types of Hypotheses

Null Hypothesis Ho

Includes the assumed value of the population parameter

Is the statement to be tested.

It is assume true until evidence we have indicated otherwise

Alternative Hypothesis H1 or Ha

The statement that the parameter has a value that somehow differs from the null hypothesis.

Is the claim to be tested.

Criminal Trial Analogy

First, State the 2 hypothesis

H0: Defendant is not guilty

H1: Defendant is guilty.

Then, collect evidence such finger prints, blood spots, hair samples, etc.

In statistics, the data is the evidence

Then, make initial assumption.

Defendant is innocent until proven guilty.

In statistics, we always assume the null hypothesis is true.

Criminal Trial Analogy

Then, make a decision based on the available evidence.

If there is sufficient evidence “beyond a reasonable doubt”

Behave as if defendant is guilty

Reject the null hypothesis

If there is not enough evidence

Behave as if defendant is not guilty

Do not reject the null hypothesis

Important Points

Neither decision entails proving the null

hypothesis or the alternative hypothesis.

We merely state there is enough evidence to

behave one way or the other.

This is also always true in statistics! No

matter what decision we make, there is

always a chance we made an error.

Right-Tailed Test

0

1

____

____

H

H

Critical Region

Area of α

Critical Value

Left-Tailed Test

0

1

____

____

H

H

Two-Tailed Test

0

1

____

____

H

H


Examine the given statements, than express the null hypothesis and alternative

hypothesis in symbolic form. Be sure to use the correct symbol (μ, σ, p) for

the indicated parameter.

The mean IQ of statistic students is at least 110.

0 : 110H

1 : 110H


Examine the given statements, than express the null hypothesis and alternative

hypothesis in symbolic form. Be sure to use the correct symbol (μ, σ, p) for

the indicated parameter.

Salaries among women business analysts have a standard

deviation greater than $3000.

0 : 3000H

1 : 3000H

Test StatisticThe test statistic is a value computed from the sample data, and

it is used in making the decision about the rejection of the null

hypothesis.

Proportion Mean

Standard Deviation

p̂ pz

pq

n

2

2

2

1n s

xz

xt

s

n


Find the test statistics. The claim is that the proportion of drivers stopped by

police in a year is different from the 10.3% reported by the Department of

Justice. Sample statistics include n = 800 randomly selected drivers with

12% of them stopped in the past year.

p̂ pz

pq

n

0.103

800

0.12ˆ

p

n

p

0.12 0.1031.5819 1.58

0.103 0.897

800

Definitions

The critical region or rejection region is the set of all values

of the test statistics that cause us to reject the null hypothesis.

The significance level () is the probability that the test

statistic will fall in the critical region when the null hypothesis

is actually true.

If the test statistic falls in the critical region, we reject the null hypothesis, so is the probability of making a mistake

of rejecting the null hypothesis when it is true.

Definitions

A critical value is any value that separates the

critical region (where reject the null

hypotheses) from the values of the test statistics

that do not lead to rejection of the null

hypothesis.


Find the critical values. Two-tailed test; = 0.01

0

0.0050.005

(1 .005,0,1) 2.575invnorm

2.575

(0.005,0,1) 2.575invnorm

-2.575

.010.005

2A


Find the critical values. Left-tailed test; = 0.05

0

0.05

(0.05,0,1) 1.645invnorm

-1.645

0.05A

P-Values or Probability Values

The p-value is the probability of getting a value of the test statistic that is at least as extreme as the one representing the sample data, assuming that the null hypothesis is true.

The p-value represents how likely we would be to observe such an extreme sample if the null hypothesis were true.

If the p-value is “small” (Typically less than 0.05), we reject the null hypothesis.


Find the p-value. The test statistic in a left-tailed test is z = -1.72

0-1.72

( 1.72)P z

( 99, 1.72,0,1)normalcdf E 0.0427

0.0427

0.0427p value


Find the p-value. With H1 : p = 0.30, test statistic is z = 2.44.

0-2.44

( 2.44)P z

( 99, 2.44,0,1) 0.00734normalcdf E

2.44( 2.44)P z

(2.44, 99,0,1) 0.00734normalcdf E

2(0.00734) 0.01469p value

0.007340.00734

Conclusions in Hypothesis Testing

We always test the null hypothesis

Reject the HO

Fail to reject the HO

Traditional Method

Reject Ho if the test statistics falls within the

critical region.

Fail to reject Ho if the test statistics does not fall

within the critical region.

Conclusions in Hypothesis Testing

P-value Method

Reject the HO if P-value α (where α is the

significance level, such as 0.05).

Fail to reject the HO if P-value > α

Confidence Intervals

Because the confidence interval estimate of a

population parameter contains the likely values of

that parameter, reject a claim that the population

parameter has value that is not included in the

confidence interval.


State the conclusion. Original claim: The proportion of college graduates who

smoke is less than 0.27. Initial conclusions: Reject the null hypothesis

There is sufficient evidence to support the claim that the

proportion of college graduates who smoke is less than 0.27.

: 0.27

: 0.27

o

a

H p

H p

Reject H0


State the conclusion. Original claim: The proportion of M&Ms that are blue

is equal to 0.10. Initial conclusions: Reject the null hypothesis.

There is sufficient evidence to reject the claim that the

M&M’s are blue is equal to 0.10.

: 0.10

: 0.10

o

a

H p

H p

Reject H0

Type I and Type II Errors

A Type I error is the mistake of rejecting the

null hypothesis when it is true.

The symbol α (alpha) is used to represent the

probability of making a type I error.

A Type II error is the mistake of failing to

reject the null hypothesis when it is false.

The symbol β (beta) is used to represent the

probability of a type II error

Errors in Criminal Trials

Truth

Jury Decision Not Guilty (Ho) Guilty (Ha)

Guilty Type I Error

Putting an innocent person in jail.

Reject Ho when Ho is true

Not Guilty Type II Error

Letting a guilty person go free

Fail to reject Ho when Ha is true

Controlling Type I and Type II Errors

For any fixed , an increase in the sample size n will

cause a decrease in β.

For any fixed sample size n, a decrease in will

cause an increase in β.

For any fixed sample size n, an increase in will

cause a decrease in β.

To decrease both and β, increase the sample size.

Example – 7, #38

Identify Type I and Type II errors. The proportion of college graduates who

smoke is less than 0.27.

1

: 0.27

: 0.27

oH p

H p

Type I Error: Supports the alternative hypothesis when the null

hypothesis is true.

Type II Error: Supports the null hypothesis when the alternative

hypothesis is true.

Lesson 7-3

Testing a Claim about a Proportion

Assumptions

The sample observations are simple random sample.

Population size > 10n

The conditions np ≥ 5 and nq ≥ 5 are both satisfied.

Binomial distribution of sample proportions can

be approximated by a normal distribution with

np npq

Test Statistic

n = sample size of number of trials

p = population proportion

(used in the null hypothesis)

ˆx

pn

1q p

p̂ pz

pq

n

sample proportion


In a Gallup survey, 1087 randomly selected adults were asked

“Do you have occasion to use alcoholic beverages such as liquor,

wine, or beer or are you a total abstainer?” Sixty-two percent of

the subjects said that they used alcoholic beverages. Consider a

hypothesis test that uses a 0.05 significance level to test the claim

that the majority (more than 50%) of adults used alcoholic

beverages.

1087

0.62ˆ

0.50

0.05

n

p

p

1

: 0.50

: 0.50

oH p

H p


A). What is the test statistic? 1087

0.62ˆ

0.50

0.05

n

p

p

1

: 0.50

: 0.50

oH p

H p

p̂ pz

pq

n

0.62 0.507.91

0.50 0.50

1087


1087

0.62ˆ

0.50

0.05

n

p

p

1

: 0.50

: 0.50

oH p

H p

B). What is the critical value?

ˆ7.91pz

0.05

(1 .05,0,1) 1.645invnorm

1.645 7.91


1087

0.62ˆ

0.50

0.05

n

p

p

1

: 0.50

: 0.50

oH p

H p

C). What is the P-value?

ˆ7.91pz

15(7.91, 99,0,1) 1.29 0.0001normalcdf E E

( 7.91)P z


1

: 0.50

: 0.50

oH p

H p

D). What is the conclusion?

ˆ7.91

0.0001

pz

P value

0.05

1.645 7.91

If P-value ≤ , reject Ho

There is sufficient evidence to reject Ho since (p-value = 0.0001

α = 0.05) and conclude that the majority of adult use alcoholic

beverages.


E). Based on the preceding results, can we conclude that 62% is

significantly greater than 50% for all such hypothesis tests?

No, 62% might not be significantly greater than 50% for some

samples, such as one with n = 50


Stats/Tests/5:1-PropZTest


In a recent year, of the 109,857 arrests for Federal offenses, 29.1%

were for drug offenses (based on the data from the U.S. Department

of Justice). Use a 0.01 significance level to test the claim that

the drug offense is equal 30%. How can the result be explained,

given that 29.1% appears to be so close to 30%.

1

: 0.30

: 0.30

oH p

H p

109,857

0.30

109857 0.291 31969

0.01

n

p

x


109,857

0.30

109857 0.291 31969

0.01

n

p

x

There is sufficient evidence to reject Ho since p-value = 0.0001 ≤ = 0.01 and conclude that arrests for drugs offenses is less than

30%. The sample size was so large that we were able to distinguish

between 29.1% and 30%.

Example – Page 397, #10In 1990, 5.8% of job applicants who were test for drugs failed

the test. At the 0.01 significance level, test the claim that the failure

rate is not lower if a simple random sample of 1520 current job

applicants results in 58 failures (based on the data from the American

Management Association). Does the result suggest that fewer

job applicants now use drugs?

1520

0.058

58

0.01

n

p

x

1

: 0.058

: 0.058

oH p

H p


1520

0.058

58

0.01

n

p

x

1

: 0.058

: 0.058

oH p

H p

There is sufficient evidence to reject Ho since p-value = 0.0004≤ = .01 and conclude that fewer job applicants now use drugs.

Example – Page 397, #16A simple random sample of households with TV sets in use show that 1024 of

them were tuned to 60 Minutes while 3836 were tuned to some other show. Use a

0.025 significance level to test the claim of a CBS executive that “60 Minutes gets

more than a 20 share,” which means that more than 20% of the sets in use are

turned to 60 Minutes. If you are a commercial advertiser and you are trying to

negotiate lower costs, what would you argue?

4860

1024

0.20

0.025

n

x

p

1

: 0.20

: 0.20

oH p

H p

p = proportion of households that were tuned to 60 Minutes.

Step 1 – Identify the population of interest and the parameter you want to draw

conclusions about.


4860

1024

0.20

0.025

n

x

p

1

: 0.20

: 0.20

oH p

H p

Use a one proportion z-test

Question stated SRS

Population size > 10(4860) = 48600

np = (4860)(0.20) = 972 ≥ 5

and nq = (4860)(.80) = 3888 ≥ 5

Step 2 – Choose the appropriate inference procedure. Verify

conditions for using the selected procedure.


4860

1024

0.20

0.025

n

x

p

1

: 0.20

: 0.20

oH p

H p

Step 3 – Carry out the inference procedure.


4860

1024

0.20

0.025

n

x

p

1

: 0.20

: 0.20

oH p

H p

p̂ pz

pq

n

0.025

0.2107 0.20

0.20(.80)

4860

= 1.865 p-value = 0.0311

1.96


There is sufficient evidence to fail to reject Ho since P-value = 0.0311 > = 0.025 and are unable to conclude

that more than 20% of households are watching 60

Minutes.

Even though happens to be greater than 20%,

that is not enough evidence to 97.5% (1 – 0.025 = 0.975)

certain that the true population proportion is greater

than 20%.

0.211p̂

Step 4 – Interpret your results in the context of the problem.

Lesson 7-4

Testing a Claim About a Mean: σ Known

Assumptions

A simple random sample is obtained.

The population standard deviation σ is known

The population from which the sample is

drawn is normally distribution or n > 30.

Mean

Standard Deviation

X

Xn

Test Statistic

Xxz

n


Determine whether the given conditions justify using the methods

of this section when testing a claim about the population mean μ.

The sample size n = 7, σ is not know, and the original population

is normally distributed.

No


Determine whether the given conditions justify using the methods

of this section when testing a claim about the population mean μ.

The sample size n = 47, σ = 12.6 and the original population

is not normally distributed.

yes


Find the test statistic, P-value, critical values and state the final

conclusion.

Claim: The mean body temperature of healthy adults is less

than 98.6° F.

Sample Data:

106

98.20

0.62

0.01

n

x

σ

α

1

: 98.60

: 98.60

oH μ

H μ


106

98.20

0.62

0.01

n

x

σ

α

1

: 98.60

: 98.60

oH μ

H μ

Xxz

n

98.20 98.60

6.640.62

106

11

( 99, 6.64,98.20,0.62)

1.577 0.0001

p value normalcdf E

E


106

98.20

0.62

0.01

n

x

σ

α

1

: 98.60

: 98.60

oH μ

H μ

0.01

2.33

(0.01) 2.33invnorm

Critical Value


Final Conclusion

Reject Ho since the P-value ≤ , there is sufficient evidence

to support the claim that the mean is less than 98.6°F.


106

98.20

0.62

0.01

n

X

σ

α

1

: 98.60

: 98.60

oH μ

H μ

Stats/Tests/1:Z-Test

Example – Page 405, #10The health of the bear population in Yellowstone National Park

is monitored by periodic measurements take from anesthetized

bears. A sample of 54 bears has a mean weight of 182.9 lb.

Assuming that σ is known to be 121.8 lb, use a 0.10 significance

level to test the claim that the population mean of all such bear

weights is less than 200 lb.

µ = mean weight of bears in Yellowstone National Park

Ho: μ = 200

H1: < 200

Step 1 – Identify the population of interest and the parameter you

want to draw conclusions about.


Use a one sample z-test

Not sure if it’s a SRS – should be representative the population.

Standard deviation of the population is known

Approximately normal distributed since n > 30

Step 2 – Choose the appropriate inference procedure. Verify the



Xxz

n

0.10

1.282

182.9 2001.032

121.8

54

0.1511p value

Step 3 – Carry out the inference procedure.


There is sufficient evidence to fail to reject Ho since (p-value =

0.151 > α = 0.10) and are unable to conclude that weight of

bears in Yellowstone National Park is less than 200 lbs.

Step 4 – Interpret your results in the context of the problem

Underlying Rational of

Hypothesis Testing If the sample results can easily occur when

assumption (null hypothesis) is true, we attribute the

relatively small discrepancy between the assumption

and the sample results to chance.

If the sample results cannot easily occur when that

assumption (null hypothesis) is true, we explain the

relatively large discrepancy between the assumption

and the sample by concluding that the assumption is

not true.


Analysis of the last digits of sample data values sometimes reveals

whether the data have been accurately measured and reported. When

single digits 0 through 9 are randomly selected with replacement,

the mean should be 4.50 and the standard deviation should be 2.87.

Reported data (such as weights or heights) are often rounded so

that the last digits include disproportionately more 0s and 5s. The

last digits in the reported lengths (in feet) of the 73 home runs hit

by Barry Bonds in 2001 are used to test the claim that they come from

a population with mean 4.50 (based on data from USA Today). When

Minitab is used to test the claim, the display is as shown here. Using

a 0.05 significance level, interpret the Minitab results. Does it

appear that the distances were accurately measured?


Test of mu = 4.5 vs mu not = 4.5

The assume sigma = 2.87

Variable N Mean StDev SE Mean

BONDS 73 1.753 2.650 0.336

Variable 95.00% CI Z P

BONDS (1.095, 2.412) – 8.18 0.000

Test statistics = – 8.18, P-value = 0.000. Reject Ho

Lesson 7-5

Testing a Claim about a Mean: σ Not Known

Assumptions for Testing Claims About a

Population Mean: Not Known

Simple random sample

Population standard deviation σ is not known

Population is normally distributed or n > 30

Test Statistic: σ Not Known.

xxt

s

n

Choosing the Appropriate Distribution

σ is known, use a z-distribution

σ is unknown, use a t-distribution

Example – Page 415, Problem #2

Claim: . Sample Data: 75 25, 102, 15.3n x s

Determine whether the hypothesis test involves a normal

distribution, t-distribution or neither.

The sample data appear to come from a population

with a distribution that is very far from normal,

and σ is unknown.

neither

Distribution is not normal and 30n

Example –Page 415, Problem #4

Claim: . Sample Data: 2.80 150, 2.88, 0.24n x s

Determine whether the hypothesis test involves a normal

distribution, t-distribution or neither.

The sample data appear to come from a population

with a distribution that is not normal, and σ is unknown.

t-distribution

σ is unknown and 30n

Example - Page 415, Problem #8

Two-Tailed test with and test statistic

Find the P-value.

9n 1.577t

Use Table A-3 8df

1.577 Falls between: 1.860 and 1.397

P-value is between: 0.10 and 0.20

Using the T-83

2nd/VARS/5:tcdf ( 99, 1.577,8) 0.07672tcdf E

: (0.07672 2) 0.1535two tailed

0.07672

–1.5

77


Find the test statistic, P-value, critical values and

state the final conclusion.

Claim: The mean body temperature of healthy adults is

less than Sample data:

The significance level is 98.6 .F 35,n 98.20 ,x F 0.62.s

0.01.

1

: 98.6

: 98.6

oH

H

, Problem #10

1

: 98.6

: 98.6

oH

H

Find the test statistic and P-Value:3598.200.620.01

nxsα

xxt

s

n

98.20 98.63.812

0.62

35

4( 99, 3.812,34) 2.73 0.000273p value tcdf E E

, Problem #10

1

: 98.6

: 98.6

oH

H

Final Conclusion3598.200.620.01

nXsα

0.000273pv

Reject Ho since P-value ≤ α, there is

sufficient evidence to conclude that

μ < 98.6

, Problem #10

1

: 98.6

: 98.6

oH

H

3598.200.620.01

nxsα

STAT/TESTS/2:T-Test


Heather Carielli is a former student of the author who earned a master’s

degree in statistics at the University of Massachusetts. When she

randomly selected 16 new textbooks in the college bookstore, she found

that they had prices with a mean of $70.41 and a standard deviation of

$19.70. Is there sufficient evidence to warrant rejection of a claim in

the college catalog that the mean price of a textbook at this

college is less than $75?

1

: 75

: 75

oH μ

H μ

75

70.41

19.7

16

0.05

μ

x

s

n

α


1

: 75

: 75

oH

H

75

70.41

19.7

16

0.05

x

s

n

0.932 t 0.1831 p value

Test Statistic P-value


1

: 75

: 75

oH

H

75

70.41

19.7

16

0.05

x

s

n

0.1831 p value

Fail to reject Ho since P-value > α, there is

not sufficient evidence to conclude that

μ < 75


In previous tests, baseballs were dropped 24 ft onto a concrete surface,

and they bounced an average of 92.84 in. In a test of a sample of 40

new balls, they bounced heights had a mean of 92.67 in. and standard

deviation of 1.79 in. Use a 0.05 significance level to determine whether

there is sufficient evidence to support the claim that the new balls have

bounce heights with a mean different from 92.84 in. Does it appear that

the new baseballs are different.

1

: 92.84

: 92.84

oH

H

92.84

92.67

1.79

40

0.05

x

s

n

, Problem #20

1

: 92.84

: 92.84

oH

H

92.84

92.67

1.79

40

0.05

x

s

n

Fail to reject Ho since P-value > α, there is

not sufficient evidence to conclude that

μ ≠ 92.84. No, it does not appear that new

baseballs are significantly different in this

respect.

0.552P value


Patients with chronic fatigue syndrome were tested, than

retested after being treated with fludrocortisone. Listed below

are the changes in fatigue after the treatment. A standard

scale from -7 to 7 was used, with positive values representing

improvements. Use a 0.01 significance level to test the claim

that the mean change is positive. Does the treatment appear

to be effective?

6 5 0 5 6 7 3 3 2 6 5 5 0 6 3 4 3 7 0 4 4


µ = improvements of chronic fatigue syndrome after being treated with fludrocortisone

Ho: μ = 0 H1: > 0

Step 1 – Identify the population of interest and the parameter you

want to draw conclusions about.


Use a one-sample t-test

Not sure if its Random Sample (SRS). Assume itsrepresentative of the population

The population standard deviation σ is not known Sampling distribution is approximately normal since

the normal probability plot show as linear trend withno outliers

Step 2 – Choose the appropriate inference procedure. Verify



xx

ts

n

0.01

2.53

4 08.455

2.168

21

82.45 0.0001p value E

Step 3 – Carry out the inference procedure


There is sufficient evidence to reject Ho since

(p-value = 0.0001 α = 0.01) and conclude that

there was improvements for patients with chronic

fatigue syndrome that was treated with

fludrocortisone.

Step 4 – Interpret your results in the context of the problem

Lesson 7-6

Testing a Claim about a Standard Deviation or

Variance

Main Idea of

Standard Deviation and Variance

The main idea is that the standard deviation

and the variance are measures of consistency.

The less consistent the values of a variable,

the higher the standard deviation of the

variable

Assumptions for Testing Claims

about σ and σ²

The sample is simple random

The population has a normal distribution.

Test Statistic for Testing Claim

about σ and σ²

22

2

( 1)n s

Characteristics of the

Chi-Square Distribution

It is not symmetric

The shape depends upon

the degrees of freedom.

As the number of degrees

of freedom increases, the

distribution becomes more

symmetric.

The values of are

greater than or equal to 0.There is a different distribution for each

number of degrees of freedom.

Chi-Square Distribution for 10

and 20 Degrees of Freedom

2


Find the critical values, test statistic, P-value, and state the final

conclusion.

1 : 12, 0.01, 5, 18H n s

Find the critical value or critical region:

Use table A-4

DF = 4 and 0.01

2

4,0.01 13.277


1 : 12, 0.01, 5, 18H n s

Find the test statistic:

2 22

22

( 1) (5 1)(18)

12

n s

9.00


1 : 12, 0.01, 5, 18H n s

Find the P-value:

Use Table A-4

4df and 9.00

0.05 0.10p value

Use TI-83

2 (9, 99,4)cdf E 0.0611


Critical region = 13.277

Test statistic = 9.00

P-value = 0.0611

4

2

4χ

0.0113

.27

7

9.0

0

Fail to reject Ho since P-value > α, and conclude that σ = 12.


Find the critical values, test statistic, P-value, and state the final

conclusion.

1 : 4, 0.05, 81, 4.7H n s

Find the critical value or critical region:

Use table A-4

DF = 80 and = 0.05

2 2

.025 106.629Rχ χ 2 2

.975 57.153Lχ χ


2 22

22

( 1) (81 1)(4.7)110.45

4

n s

1 : 4, 0.05, 81, 4.7H n s

Find the test statistic:

Find the P-value:

2 (110.45, 99,80) 0.0137cdf E


Critical region = 57.153 and 106.629

Test statistic = 110.450

P-value = 0.0137

80

2

80χ

0.025

10

6.6

29

11

0.4

50

Reject Ho since P-value < , and conclude that σ ≠ 4 (in fact ,

that σ > 4) .

0.025

57

.15

3


1

: 14.1

: 14.1

oH

H

Tests in the author’s past statistics classes have scores with a

standard deviation equal to 14.1. One of his recent classes has

27 test scores with a standard deviation of 9.3. Use 0.01 significance

level to test the claim that this current class has less variation than

past classes. Does lower standard deviation suggest that current

class is doing better?

Identify the population of interest and parameter

you want to Draw conclusions about. State the

hypothesis in words and symbols

σ = standard deviation of class test scores


Choose the appropriate inference procedure. Verify the

Conditions for using the selected procedure.

Use a σ test

Conditions

Simple Random Sample

Population is normally distributed


27

14.1

9.3

0.01

n

s

2 ( 99,11.31,26) 0.0056p value cdf E

2 22

22

( 1) (27 1)(9.3)11.311

14.1

n s

If the conditions are met, carry out the inference procedure


Reject Ho since P-value ≥ α, there and conclude that the standard

deviation of test scores is less than 14.1.

No, a lower standard deviation does not suggest that the current

class is doing better, but they are more similar to each other and

show less variation. Standard deviation speaks only about the

spread of the scores not their location.

Interpret your results in the context of the problem

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