Transcript power

Section 9.4
Day 3
Power
Power of a significance test is the probability
of rejecting the null hypothesis.
Power
Power of a significance test is the probability
of rejecting the null hypothesis.
Best way to get more power to reject a
false null hypothesis is to _______ ____
______.
Power
Power of a significance test is the probability
of rejecting the null hypothesis.
Best way to get more power to reject a
false null hypothesis is to increase
sample sizes.
Power
If we have reason to believe the population
standard deviations are about equal,
make the sample sizes the same.
Power
If we have reason to believe the population
standard deviations are about equal, make
the sample sizes the same.
If we have reason to believe that one
population’s standard deviation is
larger than the other’s, allocate our
resources so you take a larger sample
from the population with the larger
standard deviation.
Page 632, P31
Power
Which level has the greater spread?
Page 632, P31
The previous data indicate that the
measurements from the bottom of the river
have a greater spread, so ….
Page 632, P31
Because the previous data indicate that the
measurements from the bottom of the river
have a greater spread, getting more of the
new measurements at the bottom of the
river would give the test greater power.
Page 636, E60
Page 636, E60
What are the treatments here?
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What are the treatments here?
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What are the experimental units here?
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The experimental units are the panels, not
the individual members.
So, the size of each treatment group is 8
because 8 panels were assigned to each
texture.
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Name:
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This is a two-sided significance test for the
difference between two means.
Question asked “Is there a statistically
significant difference in mean palatability
score between the two texture levels?”
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Check conditions:
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Check conditions:
This is an experiment. Panel members
were randomly assigned to the treatment
groups as they were recruited.
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List 1: scores for coarse texture
List 2: scores for fine texture
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Check conditions: Show your graphs.
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Check conditions: Looking at distribution of
scores for each texture, both distributions
are fairly symmetric with no outliers. It’s
reasonable to assume each sample
came from a normally distributed
population.
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State hypotheses.
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State hypotheses.
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Compute test statistic, find P-value, and
draw sketch.
Page 636, E60
Compute test statistic, find P-value, and
draw sketch.
2-SampTTest
μ 1: ≠ μ 2
Inpt: Data
Pooled: No
List1: L1
Calculate
List2: L2
Freq1: 1
Freq2: 1
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Compute test statistic, find P-value, and
draw sketch.
t  ± 4.929
P-value  0.00022
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If all the panels sampled food with a coarse
texture and all the panels sampled food
with a fine texture, I would reject the null
hypothesis because the P-value of
0.00022 is less than the significance level
of 0.05.
Page 636, E60
If all the panels sampled food with a coarse texture
and all the panels sampled food with a fine
texture, I would reject the null hypothesis
because the P-value of 0.00022 is less than the
significance level of 0.05.
Thus, there is sufficient evidence to support
the claim that there is a statistically
significant difference in the mean
palatability score between the coarse
texture food and the fine texture food.
Page 634, E58
Page 634, E58
a) These are not independent random
samples of men and women but an
observational study conducted by looking
at medical charts in one city.
Page 634, E58
a) These are not independent random
samples of men and women but an
observational study conducted by looking
at medical charts in one city.
The data is fairly symmetrical so the
underlying populations probably are
approximately normal. There is one outlier
for the males.
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a) The populations of men and women are
larger than 10 times their sample sizes
(respectively 10(91) = 910 and 10(84) =
840)
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Although conditions not met, we need to find
the P-value for part b.
Name of test we would use:
Page 634, E58
Although conditions not met, we need to find
the P-value for part b.
Name of test we would use:
One-sided significance test for the
difference of two means
It has been speculated that the mean
amount of calcium in the blood is higher in
women than in men.
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State hypotheses.
Ho:
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State hypotheses.
Ho: μf = μm, where μf and μm are the mean
calcium levels in the blood of women and
men, respectively.
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State hypotheses.
Ho: μf = μm, where μf and μm are the mean
calcium levels in the blood of women and
men, respectively.
Ha : μf > μm
It has been speculated that the mean
amount of calcium in the blood is higher
in women than in men.
E58
2-SampTTest
Inpt: Stats
x1: 2.39690048
Sx1: 0.14049805
n1: 84
x2: 2.3181319
Sx2: 0.12172749
n2: 91
μ1 > μ2
Pooled: No
Calculate
E58
2-SampTTest
Inpt: Stats
x1: 2.39690048 (F)
Sx1: 0.14049805 stdDev
n1: 84
x2: 2.3181319 (M)
Sx2: 0.12172749
n2: 91
μ1 > μ2
Pooled: No
Calculate
t = 3.94938702
p = 0.00005794
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t  3.95; P-value  0.00005
3.95
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Interpret P-value: If these could be
considered independent random samples
of men and women and if there was no
difference between the mean calcium
levels in the blood of men and women, the
probability of a t-statistic of 3.95 or larger
with samples of 84 women and 91 men is
0.00005.
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Interpret P-value: If these could be
considered independent random samples
of men and women and if there was no
difference between the mean calcium
levels in the blood of men and women, the
probability of a t-statistic of 3.95 or larger
with samples of 84 women and 91 men is
0.00005. However, because this was an
observational study, these results are
questionable.
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a) The treatments, raised in long days or
raised in short days, were randomly
assigned to subjects.
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a) Both distributions are moderately skewed,
but neither has any outliers.
Since the distribution of the difference
reduces skewness, and the t-procedure is
robust against non-normality, the conditions
for inference are adequately met to proceed.
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b. Interval?
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b.
(0.50251, 9.46)
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c. The difference between the mean enzyme
concentration of all eight hamsters had they
all been raised in short days and the mean
concentration had they all been raised in
long days.
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d. Because 0 is not in the confidence
interval, and the treatments were
randomly assigned, Kelly has statistically
significant evidence that the difference in
enzyme concentrations between the two
groups of hamsters is due to the difference
in the amount of daylight.
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Page 633, E54
a) Treatments randomly assigned.
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a) Treatments randomly assigned.
Short
Long
Page 633, E54
a) Treatments randomly assigned.
Data for short days is strongly skewed right
and sample size is very small, so not
reasonable to assume data for short days
comes from a normally distributed
population.
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You are 95% confident that the difference
between the mean enzyme concentration of
all eight hamsters had they all been raised in
short days and the mean concentration had
they all been raised in long days is in the
interval (- 8.431, 18.394).
Page 633, E54
You are 95% confident that the difference
between the mean enzyme concentration of
all eight hamsters had they all been raised in
short days and the mean concentration had
they all been raised in long days is in the
interval (- 8.431, 18.394).
Because 0 is in the CI, it’s plausible there is
no difference in the means.
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c) No. Because 0 is in the confidence
interval, Kelly does not have statistically
significant evidence that the two treatments
would result in different mean enzyme
concentrations.
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Page 635, E59
a) No randomization is mentioned in
assigning treatments to experimental units.
(The experimental units here are the 16
dishes.) The distributions are not symmetric,
and the dish with 10 mg of resin is highly
skewed. Experimental group sizes of 8 will
not be enough to compensate for this
degree of skewness. Conditions are not met.
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Page 635, E59
t  ± 2.153; P-value ≈ 0.0588
Page 635, E59
Had the conditions been met, I would not
reject the null hypothesis because the Pvalue of 0.0588 is larger than the
significance level of 0.05.
Page 635, E59
There is insufficient evidence to support
the claim that there is a statistically
significant difference in the mean number of
termites surviving had all dishes received a
dose of 5 mg of resin and the mean number
of termites surviving had all dishes received
a dose of 10 mg of resin.
Page 635, E59
c) The main concerns are the lack of
randomization and the skewness of the distribution
of termites surviving in these rather small
experimental groups—especially in the 10-mg
group. The experimenter should repeat the
experiment, randomly assigning treatments to
the dishes, and using larger number of dishes to
compensate for the skewness.
Page 634, E57
Page 634, E57
Instructor believed that students tend to
make larger errors (in absolute value) ….
Make the necessary changes to the data in
the table so you can test the instructor’s
belief.
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E57
E57
A sample size of 15 is not large enough to
compensate for this large amount of
skewness if we were examining these data
sets individually.
E57
A sample size of 15 is not large enough to
compensate for this large amount of
skewness if we were examining these data
sets individually.
However, because the difference of the
means is a bit more lenient, this amount of
skewness should be okay for inference.
E57
E57
E57
I reject the null hypothesis because the Pvalue of 0.024 is less than the significance
level of α = 0.05.
E57
I reject the null hypothesis because the Pvalue of 0.024 is less than the significance
level of α = 0.05.
There is sufficient evidence to support the
instructor’s claim that students tend to
make larger errors (in absolute value)
when the line segments are vertical than
when the line segments are horizontal.
Questions?