Chapter Eight Tests of Hypothesis Based on a Single Sample

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Chapter EightTests of Hypothesis Based on a
Single Sample
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Hypothesis Testing ElementsNull Hypothesis
H0 gt Prior BeliefAlternative Hypothesis Ha gt
Contradictory BeliefTest Statistic gt Parameter
used to testRejection Region gt Set of values
for rejecting H0
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Hypothesis Testing ErrorsType I Error ? gt
Rejecting H0 when it is true.Type II
Error ? gt Not rejecting H0 when it is
false.
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Example of Type I ErrorsHighway engineers have
found that many factors affect the performance of
reflective highway signs. One is the proper
alignment of the cars headlights. It is thought
that more than 50 of the cars on the road have
misaimed headlights. If this contention can be
supported statistically, then a new tougher
inspection program will be put into operation.
Let p denote the proportion of cars in operation
that have misaimed headlights. Setup a test of
hypothesis to test this statement.
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Type II ErrorsThe reject region for the car test
is R 14, 15, ,20 at ? 0.05. Suppose that
the true proportion of cars with misaimed
headlights is 0.7. What is the probability that
our test is unable to detect this situation?
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Hypothesis Testing ProtocolIdentify parameter of
interestState Null Alternative
HypothesisGive Test StatisticFind Rejection
Region for Level ?Calculate Sample Size for
?(??)Decide if H0 is Rejected or Accepted
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Hypothesis Test about a Population MeanNormal
pdf (known ?)Null Hypothesis H0 u u0Test
Statistic z x u0
?/?nAlternative Hypothesis Reject RegionHa
u gt u0 (Upper Tailed) z ? z?Ha u lt u0 (Lower
Tailed) z ? -z?Ha u ? u0 (Two-Tailed)either
z ? z?/2 or z ? -z?/2
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Hypothesis Testing Mean (Known ?)Automotive
engineers are using more aluminum in the
construction of cars in hopes of improving gas
mileage. For a particular model the number of
miles per gallon obtained currently has a mean of
26.0 mpg with a ? of 5 mpg. It is hoped that a
new design will increase the mean mileage rating.
Assume that ? is not affected by this change. The
sample mean for 49 driving tests with this new
design yielded 28.04 mpg. Use a Hypothesis Test
with ? .05 to make a decision on the validity
of this new design to increase the mean mileage
rating.
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Hypothesis Testing Mean (Known ?)As new
engineering manager, you test the melting point
of 16 samples of hydrogenated vegetable oil from
production, resulting in a sample mean of
94.320F. Your company claims a melting point of
950F to all vendors. Using Hypothesis testing,
test if your production run meets the 950F
specification at level .01. Other evidence
indicates a Normal distribution with ? 1.20 for
the melting point.
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Determining ? (? known)Alternative Type
IIHypothesis Error ?(u?)Ha u gt u0
? z? u0- u? ?/?n Ha u lt
u0 1 - ? -z? u0- u?
?/?nHa u ? u0 ? z?/2 u0- u? - ? -z?/2 u0-
u? ?/?n ?/?n
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Determining n (? known) n ? (z? z?) 2
One Sided u0 - u? n ? (z?/2 z?) 2 Two
Sided u0 - u?
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Example Type II ErrorAt test level .01, what is
the probability of a Type II error when u is
actually 940F?What value of n is necessary to
ensure that ?(94) .10 when ? .01?
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Hypothesis Test (Large Sample) MeanNormal pdf
(Unknown ?)Null Hypothesis H0 u u0Test
Statistic z x u0
s/?nAlternative Hypothesis Reject RegionHa
u gt u0 (Upper Tailed) z ? z?Ha u lt u0 (Lower
Tailed) z ? -z?Ha u ? u0 (Two-Tailed)either
z ? z?/2 or z ? -z?/2 For ? n use
plausible values for ? or use Tables A.17
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Hypothesis Test (Large Sample) MeanOzone is a
component of smog that can injure sensitive
plants even at low levels. In 1979 a federal
ozone standard of 0.12 ppm was set. It is thought
that the ozone level in air currents over New
England exceeds this level. To verify this
contention, air samples are obtained from 64
monitoring stations set up across the region.
When the data are analyzed, a sample mean of
0.135 and a sample SD of 0.03 are obtained. Use a
Hypothesis test at a .01 level of significance to
test this theory.
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Example HT (Large Sample) MeanThe VP of Sales
claims that the salesmen are only averaging 15
sales contacts per week. Looking for ways to
increase this figure, the VP selects 49 salesmen
at random and the number of contacts is recorded
for a week. The sample data reveals a mean of 17
contacts with a sample variance of 9. Does the
evidence contradict the VPs claim at the 5
level of significant?Now the VP wants to detect
a difference equal to 1 call in the mean number
of customer contacts per week. Specifically, he
wants to test u 15 against u 16. With the
same test data, find ? for this test.
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Hypothesis Test (Small Sample) MeanNormal pdf
(Unknown ?)Null Hypothesis H0 u u0Test
Statistic t x u0
s/?nAlternative Hypothesis Reject RegionHa
u gt u0 (Upper Tailed) t ? t?,vHa u lt u0 (Lower
Tailed) t ? -t?,vHa u ? u0
(Two-Tailed)either t ? t?/2,v or t ?
-t?/2,v To find ? Sample Size use Table A.17
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Example HT Mean (Small Sample)A new method for
measuring phosphorus levels in soil is being
tested. A sample of 11 soil specimens with true
phosphorus content of 548 mg/kg is analyzed using
the new method. The resulting sample mean
sample standard deviation for phosphorus levels
are 587 and 10, respectively. Is there evidence
that the mean phosphorus level reported by the
new method differs significantly from the true
value of 548 mg/kg? Use ? .05 assume
measurements of this type are Normal.
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Example HT Mean (Small Sample)The true average
voltage drop from collector to emitter of
insulated gate bipolar transistors is supposed to
be at most 2.5 volts. A sample of 10 transistors
are used to test if the H0 ? 2.5 versus Ha ?
gt 2.5 volts with ? .05. If the standard
deviation of the voltage distribution is ?
0.10, how likely is it that H0 will not be
rejected when in fact ? 2.6?
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Hypothesis Test Population ?2Normal pdf (Unknown
?)Null Hypothesis H0 ?2 ?02Test
Statistic ?2 (n-1)s2 ?02
Alternative Hypothesis Reject RegionHa ?2 gt
?02 (Upper Tailed) ?2 ? ?2?,vHa ?2 lt
?02 (Lower Tailed) ?2 ? ?21-?,v Ha ?2 ?
?02 (Two-Tailed)either ?2 ? ?2?/2,v or
?2 ? ?21-?/2,v
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Example HT Variance Indoor swimming pools are
noted for their poor acoustical properties. The
goal is to design a pool in such a way that the
average time it takes a low frequency sound to
die is at most 1.3 seconds with a standard
deviation of at most 0.6 second. Computer
simulations of a preliminary design are conducted
to see whether these standards are exceeded. The
sample mean was 3.97 seconds and the sample
standard deviation was 1.89 seconds for 30
simulations. Does it appear that the design
specifications are being met at the ? 0.01
level for ??
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Example HT VarianceA new process for producing
small precision parts is being studied. The
process consists of mixing fine metal powder with
a plastic binder, injecting the mixture into a
mold, and then removing the binder with a
solvent.Sample data on parts that should have a
1 diameter and whose standard deviation should
not exceed 0.0025 inch yielded a sample mean of
1.00084 with a sample standard deviation of
0.00282 for 15 measured parts. Test at the ?
.05 level to see if this new process is viable
for the population ?.
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P-ValuesSmallest Level ofSignificance at which
H0 would be rejected when a specified test
procedure is used on a given data set.
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Calculating P-ValuesOne Sided TestsP 1
F(z) Upper-tailedP F(z)
Lower-tailedTwo Sided Test P 2 1 F(z)