mat 157 ch. 8

8/6/2019 MAT 157 Ch. 8

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Chapter 8

Hypothesis Test

8/6/2019 MAT 157 Ch. 8


Steps to a Hypothesis Test

1. Hypotheses

± Null Hypothesis (Ho)

± Alternative Hypothesis (Ha)2. Alpha

3. Distribution (aka model)

4. Test Statistics and P-value5. Decision

6. Conclusion

8/6/2019 MAT 157 Ch. 8


Steps to a Hypothesis Test

� Can remember the steps by the sentence:

³Happy AuntsMake The Darndest

Cookies´

8/6/2019 MAT 157 Ch. 8


Example 1± Hypothesis Testing

� An attorney claims that more than 25% of

all lawyers advertise. A sample of 200

lawyers in a certain city showed that 63 had

used some form of advertising. At = 0.05,

is there enough evidence to support the

attorney¶s claim?

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Hypotheses (Sets up the two

sides of the test)1. Build the Alternative Hypothesis (Ha) first.

± based on the claim you are testing (you get

this from the words in the problem)

� Three choices

± Ha: parameter � hypothesized value

± Ha: parameter < hypothesized value

± Ha: parameter > hypothesized value

2. Build Null Hypothesis (Ho) next.

± opposite of the Ha (i.e. = , � , � )

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Example 1± Constructing

Hypotheses� We need to know what parameter we are

testing and which of the three choices for

alternative hypothesis we are going to use.

± ³An attorney claims that more than 25% of all

lawyers advertise´ tells us that this is a test for

proportions so our parameter is p. ± ³claims that more than 25%´ tells us that

Ha: p > .25 and thereforeHo: p � .25

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Alpha� Alpha = = significance level

± How much proof we are requiring in order to

reject the null hypothesis.

± The complement of the confidence level that

we learned in the last chapter

± Usually given to you in the problem, if not, youcan choose.

� Most popular alphas: 0.05, 0.01, and 0.10

8/6/2019 MAT 157 Ch. 8


Example 1 ± Alpha

� ³At = 0.05´ is given to us in the problem

so we just copy = 0.05

8/6/2019 MAT 157 Ch. 8


Model� The model is the distribution used for the

parameter that you are testing. These are

just the same as we used in the confidence

intervals.

± p and (n � 30) use the normal distribution

± (n < 30) uses the t-distribution ± uses the chi-squared distribution2

W

8/6/2019 MAT 157 Ch. 8


Example 1 - Model

� The model used for a proportion is the

normal.

8/6/2019 MAT 157 Ch. 8


TestS

tatistic� You will have a different test statistic for

each of the four different parameters that we

have learned about.

± p :

± (n � 30) :

n

q p

p p z

oo

o!

Ö

n

x z o

W

Q!

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TestS

tatistic� You will have a different test statistic for

each of the four different parameters that we

have learned about.

± (n < 30) :

± :

n

s

xt o

ndf

Q!

! )1(

2W 2

2

1)n(df 2 )1(

W G

sn !!

8/6/2019 MAT 157 Ch. 8


p-value� This is the evidence (probability) that you

will get off of your chart and then compare

against your criteria (alpha).

� You will need to find the appropriate

probability that goes with your Ha.

± > and < Ha¶s are called one-tailed tests.

± � Ha¶s are called two-tailed tests.

� For z and 2 you have to take the > probability X2

8/6/2019 MAT 157 Ch. 8


Example 1 ± Test Statistic and

p-value� The formula for a test statistic for

proportions is:

� So, from our problem we need a proportion

from a sample (p-hat), the proportion from

our hypothesis (po), and a sample size (n).

n

q p

p p z

oo

o

!

Ö

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p-value� ³A sample of 200 lawyers in a certain city

showed that 63 had used some form of

advertising´ tells us that

± p-hat = 63/200 or 0.315

� From our hypothesis we know

± po = 0.25 (which means that qo = 0.75)

� ³sample of 200´ tells us that

± n = 200

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p-value� So our test statistic and p-value are

value.- pour isnumber this

0.01702.12)(z

isaour insignthe because

2.12)(zneedevalue- por

12.2

200

)75.0)(25.0(

25.0315.0

!"

"

"

!

! z

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Decision ± (always about Ho)� We have two choices for decision

± Reject Ho

± Do Not Reject Ho

� If our evidence (p-value) is less than we

REJECT Ho.

� If our evidence (p-value) is greater than

we DO NOT REJECT Ho.

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Example 1 - Decision

� Our p-value is 0.0170 and our alpha is 0.05

± So, since our p-value is less than our alpha our

decision is: REJECT Ho.

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8/6/2019 MAT 157 Ch. 8


Example 1 -

Summary

1. Ho: p � 0.25

Ha: p > 0.25

2. = 0.053. Model: Normal

4. z = 2.12 and p-value = 0.0170

5. Reject Ho

6. There is enough evidence to suggest that

p>0.25.

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Example 2 ± Hypothesis Testing

A researcher reports that the average salary of

assistant professors is more than $42,000. A

sample of 30 assistant professors has amean of $43,260. At = 0.05, test the claim

that assistant professors earn more than

$42,000 a year. The standard deviation of the population is $5230.

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Example 2 (cont.)

� Hypotheses

± Ho: � $42,000

± Ha: > $42,000 (given claim is ³more than´)

� Alpha

± = 0.05 (given)

� Model

± Normal (n � 30 and it¶s a mean)

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Example 2 (cont.)

� Test statistic and p-value:

0934.09066.011.32)(z:value-

32.1

305230$

000,42$260,43$

!!"

!

!

!

z

n

x z W

Q

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Example 2 (cont.)

� Decision

± 0.0934 > 0.05 (p-value > alpha)

± DO NOT R EJECT Ho

� Conclusion

± We do not have evidence to suggest that

> $42,000.

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Example 3 ± Hypothesis Testing

A physician claims that joggers¶ maximal

volume oxygen uptake is greater than the

average of all adults. A sample of 15 joggers has a mean of 40.6 milliliters per

kilogram (ml/kg) and a standard deviation

of 6 ml/kg. If the average of all adults is36.7 ml/kg, is there enough evidence to

support the physicians claim at = 0.05?

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Example 3 (cont.)

� Hypotheses

± Ho: � 36.7

± Ha: > 36.7

� Alpha

± = 0.05 (given)

� Model

± t(14)

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Example 3 (cont.)

� Test statistic and p-value:

0.025.and0.01 between2.517)P(t

on)distributi-tfor therangea be(willvalue-P

517.2

15

6

7.366.40

(14) !"

!

!

!

t

n s

xt

Q

8/6/2019 MAT 157 Ch. 8


Example 3 (cont.)

� Decision

± (0.01,0.025) < 0.05 (p-value < alpha)

± R EJECT Ho

� Conclusion

± There is evidence to suggest that > 36.7.

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Ex

ample 4 (cont.)� Hypotheses

± Ho: = 16.8

± Ha: � 16.8

� Alpha

± = 0.05 (given)

� Model

± 2(23)

8/6/2019 MAT 157 Ch. 8


Ex

ample 4 (cont.)� Test statistic and p-value:

0.10)(0.05,0.05)(0.025,*212.733))(P(*2

on)distributi-for therangea be(willvalue-P

733.12

)8.16(

)5.12)(124(

)1(

(23)2

2

2

22

2

22

!!{

!

!

!

G

G

G

W

G sn

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Ex

ample 4 (cont.)� Decision

± (0.05,0.10) > 0.05 (p-value > alpha)

± DO NOT R EJECT Ho

� Conclusion

± There is not enough evidence to suggest that

� 16.8.

mat 157 ch. 8

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