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Hypothesis Testing
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Learning Objective
1. Solve Hypothesis Testing ProblemsMean
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Statistical Methods
Statistical
Methods
Descriptive
Statistics
Inferential
Statistics
EstimationHypothesis
Testing
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Hypothesis TestingProcess
Population
I Believe the
Population
Mean Age is 50.
(Hypothesis)
REJECT
IsX = 20 X
= 50? No!
The Sample
Mean Is 20
SampleHypothesis
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Hypothesis TestingSteps
1. State H0
2. State H1
3. Choose
4. Choose n
5. ChooseTest
6. Set Up Critical Values
7. Collect Data8. Compute Test Statistic
9. Make StatisticalDecision
10. Express Decision
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One Population TestsNumerical Data
One
Population
Z Test t TestRuns
Test
xKnown
2 TestSignedRanksTest
Mean Median VarianceRandom-
nessx
Unknown
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t Testfor Mean (
XUnknown)
1. Assumptions Population Is Normally Distributed
If Not Normal, Only Slightly Skewed &Large Sample (n 30)
Taken
2. Parametric Test Procedure
3. t Test Statistic
tX
S
n
x=
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Two-Tailed t TestExample
Does an average box of
cereal contain 368
grams of cereal? Arandom sample of36
boxes had a mean of
372.5 & a standarddeviation of12 grams.
Test at the .05 level. 368 gm.
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t0 2.0301-2.0301
.025
Reject H0
Reject H0
.025
Two-Tailed t TestSolution
H0: X = 368
H1: X 368
=.05df=36 - 1 = 35Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Reject at = .05
There is Evidence Pop.
Average Is Not 368
tX
S
n
x=
=
= +
372 5 368
1236
2 25.
.
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Two-Tailed t TestThinking Challenge
You work for the FTC. A
manufacturer of detergent
claims that the mean weight of
detergent is 3.25 lb. You take
a random sample of64
containers. You calculate the
sample average to be 3.238lb. with a standard deviation of
.117 lb. At the .01 level, is the
manufacturer correct?
3.25 lb.
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Solution Template
H0:
H1:
=df =
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
t0
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Two-Tailed t TestSolution*
H0: X = 3.25
H1: X 3.25
=.01df=64 - 1 = 63Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Do Not Reject at = .01
There is No Evidence
Average Is Not 3.25
t0 2.6561-2.6561
.005
Reject H0
Reject H0
.005
tX
S
n
x=
=
=
3 238 3 25
11764
82. .
..
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One-Tailed t TestExample
Is the average capacity of
batteries at least 140
ampere-hours? A randomsample of20 batteries
had a mean of138.47&
a standard deviation of
2.66. Assume a normaldistribution. Test at the
.05 level.
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One-Tailed t TestSolution
H0: X 140H1: X < 140
= .05df = 20 - 1 = 19
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Reject at = .05
There Is Evidence Pop.
Average Is Less than 140t0-1.7291
.05
Reject
tX
S
n
x=
=
=
138 47 140
2 6620
2 57.
..
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One-Tailed t TestThinking Challenge
Youre a marketing analyst for
Wal-Mart. Wal-Mart had teddy
bears on sale last week. Theweekly sales ($ 00) of bears
sold in 10 stores was:8 11 0
4 7 8 10 5 8 3.
At the .05 level, is thereevidence that the average
bear sales per store is more
than5 ($ 00)?
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Solution Template
H0:
H1:
=df =
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
t0
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One-Tailed t TestSolution*
H0: X 5H1: X > 5
= .05df = 10 - 1 = 9
Critical Value(s):
Test Statistic:
Decision:
Conclusion:
Do Not Reject at = .05
There Is No Evidence
Average Is More than 5
t0 1.8331
.05Reject
tX
S
n
x=
=
= +
6 4 5
337310
131.
..
