MAT 157 Ch. 8
Transcript of MAT 157 Ch. 8
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Chapter 8
Hypothesis Test
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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
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Steps to a Hypothesis Test
� Can remember the steps by the sentence:
³Happy AuntsMake The Darndest
Cookies´
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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
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Example 1 ± Alpha
� ³At = 0.05´ is given to us in the problem
so we just copy = 0.05
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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
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Example 1 - Model
� The model used for a proportion is the
normal.
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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.
± 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 !!
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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
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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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Example 1 ± Test Statistic and
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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Example 1 ± Test Statistic and
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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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
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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)
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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.