Transcript Document

T09-00 1 Population Hypothesis Tests
Purpose
Allows the analyst to analyze the results of hypothesis
testing for known large sample means, small or unknown
sample means, and proportion situations based on an a
significance level. The hypothesis tests include the 3
alternative options of "not equal", "greater than", and "less
than" with appropriate conclusion, p-value, test statistic,
and critical value(s).
Inputs
a level
Sample size
Hypothesized value
Hypothesis alternative
Sample statistics (mean, standard deviation, proportion)
Outputs
Test Statistic
Critical Value(s)
p-value
Hypothesis test conclusion
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Critical Values
Reject Ho
Reject Ho
a
a
- CV
CV
-CV = Z/t such that P(<Z/t) = a
CV = Z/t such that P(>Z/t) = a
Reject Ho
Reject Ho
a/2
a/2
- CV
CV
-CV = Z/t such that P(<Z/t) = a/2 . . . . CV = Z/t such that P(>Z/t) = a/2
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p-value (Right Tail Test)
One tail to the right
Ho: 
Ha: >
p-value
Test
statistic
For one tail tests to the right (>) the
p-value is the probability greater
than the test statistic
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p-value (Left Tail Test)
One tail to the left
Ho: 
Ha: <
p-value
- Test
statistic
For one tail tests to the left (<) the
p-value is the probability greater
than the test statistic
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p-value (Two Tail Test)
Two tail
Ho: =
Ha: 
1/2 p-value
- Test
statistic
1/2 p-value
Test
statistic
For two tail tests to the left () the
p-value is the probability greater
than the right test statistic plus the
probability less than the left test
statistic
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Hypothesis Test for Means (Known Variation)
Large samples (n >= 30)
Methodology
Assumptions
Population Variance is known
Test Statistic = Z =
Left Tail: Za
X - o

n
Right Tail: - Za
Two Tail: - Za / 2 and Za / 2
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Hypothesis Test Means (Known Variation) - Example
The Federal Trade Commission (FTC)
periodically conducts studies designed to
test the claims manufacturers make about
their products. For example the label on a
large can of Hilltop Coffee states that the
can contains at least 3 pounds of coffee.
A random sample of 36 cans of coffee were
tested yielding a sample mean of 2.92 lbs
per can. Previous studies have shown the
standard deviation in the lbs/can to be .18.
Is Hilltop Coffee making a valid claim? Test
their claim for a 99% confidence level.
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Hypothesis Test Means (Known Variation) - Example
One tail to the left
Ho:   3
Ha:  < 3
Reject Ho
a
-2.326
Test statistic = -2.667
Facts:
CV = -2.326
test statistic = -2.667
p-value = .004
a = .01
null hypothesis is false
alternative is true
Business Conclusion:
The statistical evidence shows that
Hilltop Coffee is is not living up to
their claim (at least 3 pounds of
coffee in each can).
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Input the a and the sample mean,
standard deviation, sample size and
hypothesized value
Select the alternative hypothesis
The test statistic, critical values, p-value
and decision are automatically calculated
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Hypothesis Test for Means (Unknown Variation)
Small sample (n < 30)
Methodology
Assumptions
Population Variance is unknown
Left Tail: t a ,df
X - o
Test Statistic = t =
s
n
Right Tail: - t a ,df
Two Tail: - t a / 2 ,df and t a / 2 ,df
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Hypothesis Test Means (Unknown) - Example
The manufacturing manager of the Hiney Winery
is responsible for filling the Tiny Hiney wine
bottles with 16 oz of wine. If the wine bottles are
too full the Hiney Winery looses money and if
they are do not contain 16 ounces their
customers get upset.
A random sample of 8 bottles are sampled
yielding the following fill in ounces.
16.02, 16.22, 15.82, 15.92, 16.22, 16.32, 16.12,
and 15.92.
Tiny
Hiney
Using a level of significance of .05 determine
whether the Heiney Winery is filling their bottles
with an average of 16 oz of wine.
16 oz
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Hypothesis Test Means (Unknown) - Example
Two tail
Facts:
Ho:  = 16
Ha:   16
Reject Ho
Reject Ho
a/2
a/2
- 2.3646
2.3646
Test statistic = 1.1186
CV = -2.3646 & +2.3646
test statistic = 1.1186
p-value = .3002
a = .05
null hypothesis is true
alternative is false
Business Conclusion:
The statistical evidence shows that
the Hiney Winery is putting 16
ounces of wine in their Tiny Hiney
bottles.
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Input the a and the sample mean,
standard deviation, sample size and
hypothesized value
Select the alternative hypothesis
The test statistic, critical values, p-value
and decision are automatically calculated
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Hypothesis Test for Proportions (Large)
Sample size is sufficiently large
Methodology
Assumptions


Sufficiently Large (np  5 and n(1- p)  5)
Test Statistic = Z =
Left Tail: Za
p -  o
 o (1   )
n
Right Tail: - Za
Two Tail: - Za / 2 and Za / 2
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Hypothesis Test for Proportions (Large) - Example
Over the past few weeks the Bushwood Golf
Course 20% of the players have been women.
In an effort to attract more women players the
head pro has developed a special “drive time
advertising” promotion to attract women golfers
to Bushwood.
After a week of running the promotion, a
random sample of 400 golfers showed that 300
of them were men and 100 were women.
Use a level of significance of .05 to determine if
promotion has had an impact on the the
number of women playing at Bushwood.
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Hypothesis Test for Proportions (Large) - Example
One tail to the right
Ho:   .20
Ha:  > .20
Facts:
Reject Ho
a
1.6449
Test statistic = 2.50
CV = 1.6449
test statistic = 2.500
p-value = ..0062
a = .05
null hypothesis is false
alternative is true
Business Conclusion:
The statistical evidence shows the
“drive time” advertising promotion
has been effective in attracting
more women to play Bushwood.
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Input the a and the sample
proportion, sample size and
hypothesized value
Select the alternative hypothesis
The test statistic, critical values, p-value
and decision are automatically calculated
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