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Chapter Outline

7.1 Introduction to Hypothesis Testing7.2 Hypothesis Testing for the Mean (Large Samples)7.3 Hypothesis Testing for the Mean (Small Samples)7.4 Hypothesis Testing for Proportions7.5 Hypothesis Testing for Variance and Standard

Deviation

2 Larson/Farber 4th ed.

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Section 7.1Introduction to Hypothesis Testing


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Section 7.1 Objectives

State a null hypothesis and an alternative hypothesisIdentify type I and type I errors and interpret the levelof significanceDetermine whether to use a one-tailed or two-tailedstatistical test and find a p-valueMake and interpret a decision based on the results of

a statistical testWrite a claim for a hypothesis test


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Hypothesis Tests

S tatistical hypothesisA statement, or claim, about a population parameter.

Need a pair of hypothesesone that represents the claimthe other, its complement

When one of these hypotheses is false, the other must be true.


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Stating a Hypothesis

Nu ll hypothesisA statistical hypothesisthat contains a statement

of equality such as e , =,or u .Denoted H 0 read Hsubzero or H naught.

A lternative hypothesisA statement of inequality such as >, { ,

or

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Stating a Hypothesis

To write the null and alternative hypotheses, translatethe claim made about the population parameter froma verbal statement to a mathematical statement.

Then write its complement. H 0: k H a: > k

H 0: k H a: < k

H 0: = k H a: k

R egardless of which pair of hypotheses you use, youalways assume = k and examine the samplingdistribution on the basis of this assumption.


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Ex ample: Stating the Null and AlternativeHypotheses

Write the claim as a mathematical sentence. State the nulland alternative hypotheses and identify which representsthe claim.

1. A university publicizes that the proportion of itsstudents who graduate in 4 years is 82%.

Equality condition

Complement of H 0

H 0:

H a:

(Claim) p = 0.82

p 0.82

S olu tion:


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2.5 gallons per minute



2. A water faucet manufacturer announces that the meanflow rate of a certain type of faucet is less than 2.5gallons per minute.

Inequalitycondition

Complement of H a H 0:

H a: (Claim)< 2.5 gallons per minute

Sol

ution:


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20 ounces



3. A cereal company advertises that the mean weight of the contents of its 20-ounce size cereal boxes is morethan 20 ounces.

Inequalitycondition

Complement of H a H 0:

H a: (Claim)> 20 ounces

S olu tion:


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Types of E rrors

No matter which hypothesis represents the claim,always begin the hypothesis test ass u ming that theeq u ality condition in the n u ll hypothesis is tr u e.A t the end of the test, one of two decisions will bemade:

reject the null hypothesis

fail to reject the null hypothesisB ecause your decision is based on a sample, there isthe possibility of making the wrong decision.


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Types of E rrors

A type I error occurs if the null hypothesis is rejected

when it is true.A type II error occurs if the null hypothesis is notrejected when it is false.

Act u al Tr u th of H 0Decision H 0 is true H 0 is false

Do not reject H 0 Correct Decision Type II ErrorR eject H 0 Type I Error Correct Decision


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Ex ample: Identifying Type I and Type IIE rrors

The USD A limit for salmonella contamination for chicken is 20%. A meat inspector reports that thechicken produced by a company exceeds the USD A limit. You perform a hypothesis test to determinewhether the meat inspectors claim is true. When will atype I or type II error occur? Which is more serious?(S ource: United S tates De partment of Agriculture)


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Let p represent the proportion of chicken that iscontaminated.

Solution: Identifying Type I and Type IIE rrors

H 0:

H a:

p 0.2 p > 0.2

Hypotheses:

(Claim)

0.16 0.18 0.20 0.22 0.24 p

H 0: p 0.20 H 0: p > 0.20Chicken meetsUSD A limits.

Chicken exceedsUSD A limits.


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A type I error is rejecting H 0 when it is true.The actual proportion of contaminated chicken is lessthan or equal to 0.2, but you decide to reject H 0.

A type II error is failing to reject H 0 when it is false.The actual proportion of contaminated chicken isgreater than 0.2, but you do not reject H 0.

H 0: H a:

p 0.2 p > 0.2

Hypotheses:(Claim)


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H 0: H a:

p 0.2 p > 0.2

Hypotheses:(Claim)

With a type I error, you might create a health scareand hurt the sales of chicken producers who wereactually meeting the USD A limits.With a type II error, you could be allowing chickenthat exceeded the USD A contamination limit to besold to consumers.A type II error could result in sickness or even death.


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L evel of Significance

L evel of significanceYour maximum allowable probability of making atype I error.

Denoted by E , the lowercase Greek letter alpha.B y setting the level of significance at a small value,you are saying that you want the probability of rejecting a true null hypothesis to be small.Commonly used levels of significance:

E = 0.10 E = 0.05 E = 0.01P (type II error) = (beta)


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Statistical Tests

A fter stating the null and alternative hypotheses andspecifying the level of significance, a random sampleis taken from the population and sample statistics arecalculated.The statistic that is compared with the parameter inthe null hypothesis is called the test statistic .

2

x

2 (Section 7.5) s2 z (Section 7.4) p

t (Section 7.3 n < 30) z (Section 7.2 n u 30)

S tandardized test

statistic

Test statisticPop u lation

parameter

p


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P -values

P -val u e (or probability val u e)The probability, if the null hypothesis is true, of obtaining a sample statistic with a value as extreme or more extreme than the one determined from thesample data.Depends on the nature of the test.


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Nature of the Test

Three types of hypothesis testsleft-tailed testright-tailed testtwo-tailed test

The type of test depends on the region of thesampling distribution that favors a rejection of H 0.

This region is indicated by the alternative hypothesis.


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L eft - tailed Test

The alternative hypothesis H a contains the less-thaninequality symbol (

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The alternative hypothesis H a contains the greater-than inequality symbol (>).

z0 1 2 3-3 -2 -1

R ight - tailed Test

H 0: k

H a: > k

Teststatistic

P is the areato the rightof the teststatistic.


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Two - tailed Test

The alternative hypothesis H a contains the not equalinequality symbol (). Each tail has an area of P .

z0 1 2 3-3 -2 -1

Test

statistic

Test

statistic

H 0: = k

H a: { k P is twice thearea to the left of the negative teststatistic.

P is twice thearea to the rightof the positivetest statistic.


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Ex ample: Identifying The Nature of a Test

F or each claim, state H 0 and H a. Then determinewhether the hypothesis test is a left-tailed, right-tailed,or two-tailed test. Sketch a normal sampling distributionand shade the area for the P -value.

1. A university publicizes that the proportion of itsstudents who graduate in 4 years is 82%.

H 0: H a:

p = 0.82 p 0.82

Two-tailed test z0-z z

P -valuearea P -valuearea

S olu tion:


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2. A water faucet manufacturer announces that themean flow rate of a certain type of faucet is less than2.5 gallons per minute.

H 0: H a:

Left-tailed test z0-z

P -valuearea

2.5 gpm< 2.5 gpm

S olu tion:


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3. A cereal company advertises that the mean weight of the contents of its 20-ounce size cereal boxes ismore than 20 ounces.

H 0: H a:

R ight-tailed test z0 z

P -valuearea 20 oz> 20 oz

S olu tion:


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M aking a Decision

Decision R u le Based on P -val u eCompare the P -value with E .

If P e E , then reject H 0.

If P > E , then fail to reject H 0.C laim

Decision Claim is H 0 Claim is H a

R eject H 0

F ail to reject H 0

There is enough evidence toreject the claim

There is not enough evidenceto reject the claim

There is enough evidence tosupport the claim

There is not enough evidenceto support the claim


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Ex ample: Interpreting a Decision

You perform a hypothesis test for the following claim.How should you interpret your decision if you reject H 0? If you fail to reject H 0?

1. H 0 (Claim) : A university publicizes that the proportion of its students who graduate in 4 years is82%.


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Solution: Interpreting a Decision

The claim is represented by H 0.If you reject H 0 you should conclude there issufficient evidence to indicate that the universitysclaim is false.

If you fail to reject H 0, you should conclude there is

insufficient evidence to indicate that the universitysclaim (of a four-year graduation rate of 82%) isfalse.


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Ex ample: Interpreting a Decision

You perform a hypothesis test for the following claim.How should you interpret your decision if you reject H 0? If you fail to reject H 0?

2. H a (Claim) : C onsumer Re ports states that the meanstopping distance (on a dry surface) for a HondaCivic is less than 136 feet.

Sol

ution:

The claim is represented by H a.H 0 is the mean stopping distanceis greater than or equal to 136 feet.


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Solution: Interpreting a Decision

If you reject H 0 you should conclude there is enoughevidence to support C onsumer Re ports claim that thestopping distance for a Honda Civic is less than 136

feet.

If you fail to reject H 0, you should conclude there isnot enough evidence to support C onsumer Re ports claim that the stopping distance for a Honda Civic isless than 136 feet.


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z0

Steps for Hypothesis Testing

1. State the claim mathematically and verbally. Identifythe null and alternative hypotheses.

H 0: ? H a: ?2. Specify the level of significance.

= ?3. Determine the standardized

sampling distribution anddraw its graph.

4. Calculate the test statisticand its standardized value.A dd it to your sketch. z

0Test statistic

This sampling distribution

is based on the assumptionthat H 0 is true.


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Steps for Hypothesis Testing

5. F ind the P -value.6. Use the following decision rule.

7. Write a statement to interpret the decision in thecontext of the original claim.

Is the P -value lessthan or equal to thelevel of significance?

F ail to reject H 0.

Yes

R eject H 0.

N o


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Section 7.1 Summary

Stated a null hypothesis and an alternative hypothesisIdentified type I and type I errors and interpreted thelevel of significanceDetermined whether to use a one-tailed or two-tailedstatistical test and found a p-valueMade and interpreted a decision based on the results

of a statistical testWrote a claim for a hypothesis test


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Section 7.2Hypothesis Testing for the M ean

(L arge Samples)


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F ind P -values and use them to test a mean Use P -values for a z-testF ind critical values and rejection regions in a normaldistributionUse rejection regions for a z-test


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U sing P -values to M ake a Decision

Decision R u le Based on P -val u eTo use a P -value to make a conclusion in a hypothesistest, compare the P -value with E .1. If P e E , then reject H 0.2. If P > E , then fail to reject H 0.


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Ex ample: Interpreting a P -value

The P -value for a hypothesis test is P = 0.0237. What isyour decision if the level of significance is1. 0.05?

2. 0.01?

Sol

ution:B ecause 0.0237 < 0.05, you should reject the n u ll

hypothesis .

S olu tion:B ecause 0.0237 > 0.01, you should fail to reject then u ll hypothesis .


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F inding the P -value

A fter determining the hypothesis tests standardized teststatistic and the test statistics corresponding area, do oneof the following to find the P -value.

a. F or a left-tailed test, P = (A rea in left tail). b. F or a right-tailed test, P = (A rea in right tail).c. F or a two-tailed test, P = 2( A rea in tail of test statistic).


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Ex ample: F inding the P -value

F ind the P -value for a left-tailed hypothesis test with atest statistic of z = -2.23. Decide whether to reject H 0 if the level of significance is = 0.01.

z0-2.23

P = 0.0129

S olu tion:F or a left-tailed test, P = (A rea in left tail)

B ecause 0.012 9 > 0.01, you should fail to reject H 0


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z0 2.14

Ex ample: F inding the P -value

F ind the P -value for a two-tailed hypothesis test with atest statistic of z = 2.14. Decide whether to reject H 0 if the level of significance is = 0.05.

S olu tion:F or a two-tailed test, P = 2( A rea in tail of test statistic)

B ecause 0.0324 < 0.05, you should reject H 0

0.9 838

1 0.9 838= 0.0162 P = 2(0.0162)

= 0.0324


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Z -Test for a M ean

Can be used when the population is normal and W isknown, or for any population when the sample size nis at least 30.

The test statistic is the sample meanThe standardized test statistic is z

When n u 30, the sample standard deviation s can besubstituted for W.

x z

n

Q

W

! standard error xn

W W ! !

x


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U sing P -values for a z -Test for M ean

1. State the claim mathematicallyand verbally. Identify the nulland alternative hypotheses.

2. Specify the level of significance.

3. Determine the standardized teststatistic.

4. F ind the area that correspondsto z.

State H 0 and H a.

Identify E .

Use Table 4 inA ppendix B .

x zn

Q

W !


In Words In Symbols

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U sing P -values for a z -Test for M ean

R eject H 0 if P -valueis less than or equal

to E . O therwise,fail to reject H 0.

5. F ind the P -value.a. F or a left-tailed test, P = (A rea in left tail).

b. F or a right-tailed test, P = (A rea in right tail).c. F or a two-tailed test, P = 2(A rea in tail of test

statistic).6. Make a decision to reject or

fail to reject the null hypothesis.

7. Interpret the decision in thecontext of the original claim.


In Words In Symbols

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Ex ample: Hypothesis Testing U sing P -values

In an advertisement, a pizza shop claims that its meandelivery time is less than 30 minutes. A randomselection of 36 delivery times has a sample mean of

28.5 minutes and a standard deviation of 3.5 minutes. Isthere enough evidence to support the claim at E = 0.01?Use a P -value.


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Solution: Hypothesis Testing U sing P -values

H 0: H a:

E = Test S tatistic:

30 min

< 30 min0.01

28.5 30

3.5 36

2.57

x z

n

Q!

!

!

Decision:

A t the 1% level of significance,you have sufficient evidence toconclude the mean delivery timeis less than 30 minutes.

z0-2.57

0.0051

P -val u e

0.0051 < 0.01Reject H 0


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Ex ample: Hypothesis Testing U sing P -values

You think that the average franchise investmentinformation shown in the graph is incorrect, so yourandomly select 30 franchises and determine the

necessary investment for each. The sample meaninvestment is $135,000 with astandard deviation of $30,000. Isthere enough evidence to support

your claim at E = 0.05? Use a P -value.


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Solution: Hypothesis Testing U sing P -values

H 0:

H a:

E = Test S tatistic:

= $143,260

$143,260

0.05

135, 000 143, 26030,000 30

1.51

x z

n

Q

W !

!

!

Decision:

A t the 5% level of significance,there is not sufficient evidence toconclude the mean franchiseinvestment is different from

$143,260.

P -val u e P = 2(0.0655)

= 0.1310

0.1310 > 0.05F ail to reject H 0

z0-1.51

0.0655


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R ejection R egions and Critical Values

Rejection region (or critical region )The range of values for which the null hypothesis isnot probable.If a test statistic falls in this region, the nullhypothesis is rejected.A critical value z0 separates the rejection region from

the nonrejection region.


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R ejection R egions and Critical Values

F inding C ritical Val u es in a N ormal D istrib u tion1. Specify the level of significance E .2. Decide whether the test is left-, right-, or two-tailed.3. F ind the critical value(s) z0. If the hypothesis test is

a. left-tailed , find the z-score that corresponds to an areaof E ,

b. right-tailed , find the z-score that corresponds to an areaof 1 E ,

c. two-tailed , find the z-score that corresponds to E and1 E .4. Sketch the standard normal distribution. Draw a vertical

line at each critical value and shade the rejection region(s).


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Ex ample: F inding Critical Values

F ind the critical value and rejection region for a two-tailed test with E = 0.05.

z0 z0 z0

= 0.025 = 0.025

1 = 0.9 5

The rejection regions are to the left of - z0 = -1. 9 6 andto the right of z0 = 1.9 6.

z0 = 1.9 6-z0 = -1.9 6

S olu tion:


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Decision R ule Based on R ejectionR egion

To use a rejection region to conduct a hypothesis test,calculate the standardized test statistic, z. If thestandardized test statistic1. is in the rejection region, then reject H 0.

2. is not in the rejection region, then fail to reject H 0.

z0 z0

F ail to reject H 0.

R eject H 0.

L eft -Tailed Test z < z

0

z0 z0

R eject H o.

F ail to reject H o.

z > z0Right -Tailed Test

z0 z0

Two -Tailed Test z0 z < - z0 z > z0

R eject H 0

F ail to reject H 0

R eject H 0


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U sing R ejection R egions for a z -Test for aM ean

6. F ind the standardized teststatistic.

7. Make a decision to reject or failto reject the null hypothesis.


or if 30

use

x z nn

s

QW

W

! u

} .If z is in the rejectionregion, reject H 0.O therwise, fail to

reject H 0.


In Words In Symbols

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Solution: Testing with R ejection R egions

H 0:

H a:

E =

Rejection Region:

$45,000

< $45,000

0.05

43,500 45,000

5200 30

1.58

x z

n

Q

W

! !

!

Decision:A t the 5% level of significance,there is not sufficient evidenceto support the employees claimthat the mean salary is less than$45,000.

Test S tatistic

z0-1.645

0.05

-1.58

-1.64 5

F ail to reject H 0


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Ex ample: Testing with R ejection R egions

The U.S. Department of A griculture reports that themean cost of raising a child from birth to age 2 in a ruralarea is $10,460. You believe this value is incorrect, soyou select a random sample of 9 00 children (age 2) andfind that the mean cost is $10,345 with a standarddeviation of $1540. A t = 0.05, is there enoughevidence to conclude that the mean cost is different

from $10,460? ( Ada pted from U. S . De partment of AgricultureC enter for Nutrition P olicy and P romotion)


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Solution: Testing with R ejection R egions

H 0:

H a:

E =

Rejection Region:

= $10,460

$10,460

0.05

10 345 10 460

1540 900

2 24

x

n

Q

W

! !

!

Decision:A t the 5% level of significance,you have enough evidence toconclude the mean cost of raising a child from birth to age2 in a rural area is significantlydifferent from $10,460.

Test S tatistic

z0-1.9 6

0.025

1.9 6

0.025

-1.9 6 1.9 6

-2.24

Reject H 0


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Section 7.2 Summary

F ound P -values and used them to test a mean Used P -values for a z-testF ound critical values and rejection regions in anormal distributionUsed rejection regions for a z-test


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Section 7.3

Hypothesis Testing for the M ean(Small Samples)


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F inding Critical Values in a t -Distribution

1. Identify the level of significance E .2. Identify the degrees of freedom d.f. = n 1.3. F ind the critical value(s) using Table 5 in A ppendix B in

the row with n 1 degrees of freedom. If the hypothesistest isa. left-tailed , use O ne Tail, E column with a negative

sign,b. right-tailed , use O ne Tail, E column with a

positive sign,c. two-tailed , use Two Tails, E column with a

negative and a positive sign.


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Ex ample: F inding Critical Values for t

F ind the critical value t 0 for a left-tailed test givenE = 0.05 and n = 21.

S olu tion:The degrees of freedom ared.f. = n 1 = 21 1 = 20 .Look at = 0.05 in the O ne Tail, E column.B ecause the test is left-tailed, the critical value isnegative.

t 0-1.725

0.05


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Ex ample: F inding Critical Values for t

F ind the critical values t 0 and - t 0 for a two-tailed testgiven E = 0.05 and n = 26.

S olu tion:The degrees of freedom ared.f. = n 1 = 26 1 = 25 .Look at = 0.05 in theTwo Tail, E column.B ecause the test is two-tailed, one critical value isnegative and one is positive.

t 0-2.060

0.025

2.060

0.025


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t -Test for a M ean ( n < 30, W U nknown)

t -Test for a MeanA statistical test for a population mean.The t -test can be used when the population is normal

or nearly normal, W is unknown, and n < 30.The test statistic is the sample meanThe standardized test statistic is t .

The degrees of freedom are d.f. = n 1.

xt s n

Q!

x


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U sing the t -Test for a M ean (Small Sample)



3. Identify the degrees of freedomand sketch the samplingdistribution.

4. Determine any critical value(s).

State H 0 and H a.

Identify E .


d.f. = n 1.


In Words In Symbols

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U sing the t -Test for a M ean (Small Sample)

5. Determine any rejectionregion(s).


7. Make a decision to reject or fail to reject the nullhypothesis.


xt s n Q!

If t is in the rejectionregion, reject H 0.O therwise, fail toreject H 0.


In Words In Symbols

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Ex ample: Testing with a Small Sample

A used car dealer says that the mean price of a 2005Honda Pilot LX is at least $23, 9 00. You suspect thisclaim is incorrect and find that a random sample of 14

similar vehicles has a mean price of $23,000 and astandard deviation of $1113. Is there enough evidence toreject the dealers claim at = 0.05? A ssume the

population is normally distributed. ( Ada pted from Kelley

Blue Boo k )


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Ex ample: Testing with a Small Sample

A n industrial company claims that the mean pH level of the water in a nearby river is 6.8. You randomly select19 water samples and measure the pH of each. The

sample mean and standard deviation are 6.7 and 0.24,respectively. Is there enough evidence to reject thecompanys claim at = 0.05? A ssume the population isnormally distributed.


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Ex ample: U sing P -values with t -Tests

The A merican A utomobile A ssociation claims that themean daily meal cost for a family of four traveling onvacation in F lorida is $118. A random sample of 11 such

families has a mean daily meal cost of $128 with astandard deviation of $20. Is there enough evidence toreject the claim at = 0.10? A ssume the population isnormally distributed. ( Ada pted from American Automobile

Association)


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Solution: U sing P -values with t -Tests

H 0: H a:

Decision:

= $118 $118

TI-83/84set up: Calculate: Draw:

F ail to reject H 0. A t the 0.10 level of significance, thereis not enough evidence to reject the claim that themean daily meal cost for a family of four traveling onvacation in F lorida is $118.


0.1664 > 0.10

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Section 7.3 Summary

F ound critical values in a t -distributionUsed the t -test to test a mean Used technology to find P -values and used them witha t -test to test a mean


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Section 7.4

Hypothesis Testing for Proportions


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Use the z-test to test a population proportion p


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z -Test for a Population Proportion

z -Test for a Pop u lation ProportionA statistical test for a population proportion.Can be used when a binomial distribution is givensuch that n p u 5 and nq u 5.The test statistic is the sample proportion .The standardized test statistic is z.

p p

p p p z pq n

QW

! !


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U sing a z -Test for a Proportion p



3. Sketch the sampling distribution.


State H 0 and H a.

Identify E .


Verify that n p 5 and nq 5


In Words In Symbols

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U sing a z -Test for a Proportion p

5. Determine any rejectionregion(s).


7. Make a decision to reject or fail to reject the null

hypothesis.


If z is in the rejectionregion, reject H 0.O therwise, fail toreject H 0.

p p z pq n!


In Words In Symbols

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Ex ample: Hypothesis Test for Proportions

Z ogby International claims that 45% of people in theUnited States support making cigarettes illegal withinthe next 5 to 10 years. You decide to test this claim andask a random sample of 200 people in the United Stateswhether they support making cigarettes illegal within thenext 5 to 10 years. O f the 200 people, 4 9 % support thislaw. A t = 0.05 is there enough evidence to reject the

claim?S olu tion:Verify that n p 5 and nq 5.n p = 200(0.45) = 9 0 and nq = 200(0.55) = 110


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Ex ample: Hypothesis Test for Proportions

The Pew R esearch Center claims that more than 55% of U.S. adults regularly watch their local television news.You decide to test this claim and ask a random sample of

425 adults in the United States whether they regularlywatch their local television news. O f the 425 adults, 255respond yes. A t = 0.05 is there enough evidence tosupport the claim?

S olu tion:Verify that n p 5 and nq 5.n p = 425(0.55) 234 and nq = 425 (0.45) 1 9 1


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Solution: Hypothesis Test for Proportions

H 0:

H a:

E =

Rejection Region:

p 0.55

p > 0.55

0.05

255 425 0 .55

(0 .55 )(0 .45 ) 425

2 .07

p p z

pq n

! !

!

Decision:A t the 5% level of significance,there is enough evidence to

support the claim that morethan 55% of U.S. adultsregularly watch their localtelevision news.

Test S tatistic

z0 1.645

0.05

2.071.64 5

Reject H 0


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Section 7.4 Summary

Used the z-test to test a population proportion p


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Section 7.5

Hypothesis Testing for Variance andStandard Deviation


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F inding Critical Values for the 2-Test

1. Specify the level of significance E .2. Determine the degrees of freedom d.f. = n 1.3. The critical values for the 2-distribution are found in Table 6

of A ppendix B . To find the critical value(s) for aa. right-tailed test , use the value that corresponds to d.f. and

E .b. left-tailed test , use the value that corresponds to d.f. and

1 E .c. two-tailed test , use the values that corresponds to d.f. and

E and d.f. and 1 E .


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F inding Critical Values for the 2-Test

2

12 E

2 L G

2 R G

12 E

1

2

20 G

E

1

2

E

20 G

1

R ight-tailed Left-tailed

Two-tailed


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Ex ample: F inding Critical Values for 2

F ind the critical 2-value for a left-tailed test whenn = 11 and E = 0.01.

S olu tion:Degrees of freedom: n 1 = 11 1 = 10 d.f.The area to the right of the critical value is1 E = 1 0.01 = 0. 99 .

F rom Table 6, the critical value is .

20 2.558 G !

2

0.01

20 G

20 2.558 G !

9 0 Larson/Farber 4th ed.

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Ex ample: F inding Critical Values for 2

F ind the critical 2-value for a two-tailed test whenn = 13 and E = 0.01.

S olu tion:

Degrees of freedom: n 1 = 13 1 = 12 d.f.The areas to the right of the critical values are

F rom Table 6, the critical values are and

02

051

.0E !

11 92 0.9 5E ! 2

L G 2

R G 2

1 0.0052 E !1 0.0052 E !

2 3.074 L G !2 28.299 R G !

2 3.074 L G !2 28.299 R G !


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The Chi -Square Test

2-Test for a Variance or S tandard DeviationA statistical test for a population variance or standarddeviation.

Can be used when the population is normal.The test statistic is s2.The standardized test statistic

follows a chi-square distribution with degrees of freedom d.f. = n 1.

22

2( 1)n s

G W !


U i th 2 T t f V i

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U sing the 2-Test for a Variance or Standard Deviation



3. Determine the degrees of freedom and sketch the samplingdistribution.


State H 0 and H a.

Identify E .


d.f. = n 1


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Ex l H th i T t f th

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Ex ample: Hypothesis Test for thePopulation Variance

A dairy processing company claims that the variance of the amount of fat in the whole milk processed by thecompany is no more than 0.25. You suspect this is

wrong and find that a random sample of 41 milk containers has a variance of 0.27. A t = 0.05, is thereenough evidence to reject the companys claim? A ssumethe population is normally distributed.


Sol tion: H pothesis Test for the

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Solution: Hypothesis Test for thePopulation Variance

H 0: H a:

= df = Rejection Region:

Test S tatistic:

Decision:

2 0.252 > 0.25

0.05

41 1 = 40

22

2

( 1) (41 1)(0.27)0.25

43.2

n s G

W

! !

!

F ail to Reject H 0A t the 5% level of significance,there is not enough evidence toreject the companys claim that thevariance of the amount of fat in thewhole milk is no more than 0.25.

20.05E !

55.75855.75843.2


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Ex ample: Hypothesis Test for the

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Ex ample: Hypothesis Test for thePopulation Variance

A sporting goods manufacturer claims that the varianceof the strength in a certain fishing line is 15. 9 . A randomsample of 15 fishing line spools has a variance of 21.8.A

t = 0.05, is there enough evidence to reject themanufacturers claim? A ssume the population isnormally distributed.


Solution: Hypothesis Test for the

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5.62 9 2

1 0.0252 E !

26.11 9

Solution: Hypothesis Test for thePopulation Variance

H 0: H a:

=

df = Rejection Region:

Test S tatistic:

Decision:

2 = 15.92 15.9

0.05

15 1 = 14

22

2

( 1) (15 1)(21.8)15.9

19 .19 4

n s G

W

! !

}

F ail to Reject H 0A t the 5% level of significance,there is not enough evidence toreject the claim that the variance inthe strength of the fishing line is15.9 .

5.62 919 .19 4

26.11 9


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Section 7.5 Summary

F ound critical values for a 2-testUsed the 2-test to test a variance or a standarddeviation

les4e ppt ch07

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