6th
July 2009
Summaries
Session
2.4
Hypothesis
Testing: Median Test
Super! Logic! Challenge! Box!FTM II
Objective: Be able to
perform the test of the population median. Be able to fully discuss the testing
procedure and results. This discussion must include a clear discussion
of the population and the null hypothesis, the family of samples, the family of
errors and the interpretation of the p-value.
ThingCorps FTM
manufactures a product, the Super!Logic!Challenge!Box! FTM
A random sample of boxes is
evaluated by a panel of test players. Each player plays the box, and either
beats, or is beaten by the box. The entire panel of players plays each box. The
percentage of test players beaten by each box is listed below:
|
55 57 60 64 65 65
68 69 |
70 71 72 72
73 73 77 |
77 79 79
80 80 87 88 |
88 90 96 97 97
97 100 100 |
Perform the following
tests:
In the following tests,
the following elements are common:
A population of logic
boxes
A Family of Samples, each
member of which is a single random sample of n=30 boxes. The FoS is the collection of all
samples of this type.
Each test leads to a
Family of Errors, based on the individual test.
1. Test a null hypothesis: h
= 50 versus
alternative hypothesis: h ¹
50 ;
The form of the alternative requires us to compute:
#{sps* < 50} = 0 (none of the boxes beat fewer than 50% of the players
and #{sps > 50}=30(every box beats more
than 50% of the players)
*n.b.
sps = Sample Points
The largest of these counts is 30, so refer error =30 to the n=30
part of the median table. Get the row 30 30 0.00000 .
We get an approximate value of 2*.00000, or a p-value of approximately 0.00000.
If the median LogicBox success rate is
50%, then fewer than .0001% of the Family of Samples
will yield an error this severe or worse.
The sample is decidedly incompatible with the null hypothesis.
2. Test a null hypothesis: h
= 65 versus
alternative hypothesis: h > 65 ;
The form of the alternative requires us to compute:
#{sps > 65} = 24 (24 of 30 logic boxes beat at least 65% of players}.
So refer error =24 to the n=30 part of the median table. Get the
row 30 24 0.00072
We get an approximate p-value of 0.00072
If the median LogicBox success rate is
65%, then fewer than .0072% of the Family of Samples
will yield an error this severe or worse.
The sample is decidedly incompatible with the null hypothesis.
3. Test a null hypothesis: h
= 85 versus
alternative hypothesis: h < 85
The form of the alternative requires us to compute:
#{sps < 85} = 20 (20 of 30 logic boxes beat less than 85% of
players}.
So refer error =20 to the n=30 part of the median table. Get the
row 30 20 0.04937
We get an approximate p-value of 0.04937
If the median LogicBox success rate is
85%, then fewer than 4.94% of the Family of Samples will yield an error this
severe or worse.
The sample is kinda' incompatible with
the null hypothesis.
Hypothesis Testing
Median Test
Fictitious Large Fruit
Bats II
Objective: Be able to
perform the test of the population
median. Be able to fully discuss the testing procedure and results. This discussion must include a clear discussion
of the population and the null hypothesis, the family of samples, the family of
errors and the interpretation of the p-value.
Fictitious Large Fruit Bats
Consider the following random sample
of Fictitious Large Fruit Bats. The wingspan, in centimeters, of each bat is
presented below:
5, 6.1, 7.15, 7.32, 8,
8.25, 8.8, 9.2, 10, 11.2,
11.4, 12, 12.8, 13.1, 14.15,
14.25, 15, 15.5, 16, 16.5,
17.2, 18, 18.2, 18.75, 19,
21, 22, 27, 28, 33
Perform the following
tests:
In the following tests,
the following elements are common:
A population of Fictitious
Large Fruit Bats
A Family of Samples, each
member of which is a single random sample of n=30 bats. The FoS is the collection of all samples of this type.
Each test leads to a
Family of Errors, based on the individual test.
Test a null hypothesis: h = 8 versus alternative hypothesis: h
¹ 8
;
The form of the alternative requires us to compute:
#{sps < 8} = 4 (4 of 30 bats have wingspans < 8 cm) and
#{sps > 8}=25(25 of 30 bats have wingspans > 8 cm)
The largest of these counts is 25, so refer error =25 to the n=30
part of the median table. Get the row 30 25 0.00016. We get an approximate value of 2*.00016, or a
p-value of approximately 0.00032.
If the median bat wingspan is 8 cm, then fewer
than .032% of the Family of Samples will yield an error this severe or worse.
The sample is decidedly incompatible with the null hypothesis.
Test a null hypothesis: h
= 6 versus alternative hypothesis: h > 6 ;
The form of the alternative requires us to compute:
#{sps > 6}=29(29 of 30 bats have wingspans > 6 cm)
So refer error =29 to the n=30 part of the median table. Get the
row 30 29 0.00000. We get an
approximate value of .00000.
If the median bat wingspan is 6 cm, then fewer
than .0001% of the Family of Samples will yield an error this severe or worse.
The sample is decidedly incompatible with the null hypothesis.
Test a null hypothesis: h
= 10 versus alternative hypothesis: h < 10
The form of the alternative requires us to compute:
#{sps < 10} = 8 (8 of 30 bats have wingspans < 10 cm)
So refer error =8 to the n=30 part of the median table. Get the row
30 8 0.99739 We get an approximate p-value
of 99.74%.
If the median bat wingspan is 10 cm, then fewer than 99.74% of the
Family of Samples will yield an error this severe or worse.
The sample does not suggest rejection of the null hypothesis.
From: http://www.mindspring.com/~cjalverson/2ndhourlySummer2008Specialkey.htm
Case
Five
Hypothesis Test –
Population Median
Here is another random sample of FHS
adult subjects, listed below:
150/100, 120/82, 154/94, 225/120, 144/88,
142/100, 136/80, 246/110, 135/90, 126/70, 200/130, 204/102, 120/70, 128/90,
240/130, 156/108, 108/70, 176/104, 180/110, 220/110, 145/95, 130/86, 210/94,
160/102, 146/86, 154/96, 150/90, 180/100, 185/115, 145/95
Test the following: null (H0):
The median diastolic blood pressure is 80 mm Hg (h
= 80 mm Hg) against the alternative (H1): h
> 80 mm Hg. Show your work. Completely discuss and interpret your test
results, as indicated in class and case study summaries. Fully discuss the testing procedure and results. This
discussion must include a clear discussion of the population and the null
hypothesis, the family of samples, the family of errors and the interpretation
of the p-value.
Numbers
h > 80 = “Guess Too
Small”
Error = “Number of FHS
Subjects in the Sample whose DBP > 80”
150/100, 120/82, 154/94, 225/120, 144/88, 142/100, 246/110,
135/90, 200/130, 204/102, 128/90, 240/130, 156/108, 176/104, 180/110, 220/110,
145/95, 130/86, 210/94, 160/102, 146/86, 154/96, 150/90, 180/100, 185/115,
145/95
n=30
error
= “Number of FHS Subjects whose DBP > 80” = 26
from
30 26 0.00003 , p » .003.
Interpretation
We estimate the median
diastolic blood pressure (DBP) of Framingham Heart Study (FSH) subjects.
Each member of our family
of samples is a single random sample of 30 Framingham Heart Study (FSH)
subjects, and the family of samples consists of all possible samples of this
type.
From each member of the
family of samples, we compute Error = “Number of FHS Subjects in the Sample
whose DBP > 80”.
If the true population
median diastolic blood pressure for Framingham Heart Study (FSH) subjects is
80, then approximately 0.003%(approximately 3 samples
per thousand) of the member samples yield errors as bad as or worse than our
error. Our sample presents highly significant evidence against the null
hypothesis.
Table 2. Medians
|
n error base p-value 25 1 1.00000 25 2 1.00000 25 3 0.99999 25 4 0.99992 25 5 0.99954 25 6 0.99796 25 7 0.99268 25 8 0.97836 25 9 0.94612 25 10 0.88524 25 11 0.78782 25 12 0.65498 25 13 0.50000 25 14 0.34502 25 15 0.21218 25 16 0.11476 25 17 0.05388 25 18 0.02164 |
n error base p-value 25 19 0.00732 25 20 0.00204 25 21 0.00046 25 22 0.00008 25 23 0.00001 25 23 0.00001 25 24 0.00000 25 25
0.00000 30 1 1.00000 30 2 1.00000 30 3 1.00000 30 4 1.00000 30 5 0.99997 30 6 0.99984 30 7 0.99928 30 8 0.99739 30 9 0.99194 30 10 0.97861 30 11 0.95063 |
n error base p-value 30 12 0.89976 30 13 0.81920 30 14 0.70767 30 15 0.57223 30 16 0.42777 30 17 0.29233 30 18 0.18080 30 19 0.10024 30 20 0.04937 30 21 0.02139 30 22 0.00806 30 23 0.00261 30 24 0.00072 30 24 0.00072 30 25 0.00016 30 26 0.00003 30 27 0.00000 30 28 0.00000 30 29 0.00000 30 30
0.00000 |
From: http://www.mindspring.com/~cjalverson/3rd%20Hourly%20Spring%202007%20Version%20B%20Key.htm
Case One
Hypothesis Test – Population Median
Fictitious Spotted Lizard
The Fictitious Spotted Lizard (FSpL) is
a native species of Lizard Island, and is noteworthy for the both the quantity and
quality of its spots. Consider a random sample of FSpL,
in which the number of spots per lizard is noted:
3, 6, 8, 9, 10, 10, 11, 11, 12, 14, 14, 15, 15,
15, 15, 16, 16, 17, 17, 17, 17, 21, 21, 21, 22, 24, 24, 24, 25, 27
Test the following: null (H0): The
median number of spots per FSpL is 12 (h = 12) against the alternative (H1): h
> 12.
Show your work. Completely discuss and interpret
your test results, as indicated in class and case study summaries. Fully discuss the testing procedure and results. This discussion must include a clear discussion of
the population and the null hypothesis, the family of samples, the family of
errors and the interpretation of the p-value.
Numbers
3, 6, 8, 9, 10, 10, 11, 11| 12 |14, 14, 15, 15, 15, 15,
16, 16, 17, 17, 17, 17, 21, 21, 21, 22, 24, 24, 24, 25, 27
8 below 12
1 at 12
21 above 12
30 total
30 21
0.02139, the computed p-value is .02139,
or approximately 2.1%.
Our Family of Samples (FoS)
consists of every possible random sample of 30 Fictitious Spotted Lizards. From
each member sample of the FoS,
we compute the number of sample lizards who have strictly more than 12 spots.
Computing this error for each member sample of the FoS, we obtain a Family of Errors (FoE).
|
n error base p-value 25 1 1.00000 25 2 1.00000 25 3 0.99999 25 4 0.99992 25 5 0.99954 25 6 0.99796 25 7 0.99268 25 8 0.97836 25 9 0.94612 25 10 0.88524 25 11 0.78782 25 12 0.65498 25 13 0.50000 25 14 0.34502 25 15 0.21218 25 16 0.11476 25 17 0.05388 25 18 0.02164
|
n error base p-value 25 19 0.00732 25 20 0.00204 25 21 0.00046 25 22 0.00008 25 23 0.00001 25 23 0.00001 25 24 0.00000 25 25 0.00000 30 1 1.00000 30 2 1.00000 30 3 1.00000 30 4 1.00000 30 5 0.99997 30 6 0.99984 30 7 0.99928 30 8 0.99739 30 9 0.99194 30 10 0.97861 30 11 0.95063
|
n error base p-value 30 12 0.89976 30 13 0.81920 30 14 0.70767 30 15 0.57223 30 16 0.42777 30 17 0.29233 30 18 0.18080 30 19 0.10024 30 20 0.04937 30 21 0.02139 30 22 0.00806 30 23 0.00261 30 24 0.00072 30 24 0.00072 30 25 0.00016 30 26 0.00003 30 27 0.00000 30 28 0.00000 30 29 0.00000 30 30 0.00000
|
Case Four: Hypothesis Test(Median),
Fictitious Stress Index
The Fictitious Stress Index (FSI) is a measure of a person's
stress level. The FSI ranges from 0 to 70. A researcher acquires a random
sample of undergraduates at Kennesaw State University, and administers the FSI
to these students near the end of Fall Semester 2006. The FSI scores follow:
12, 14, 17, 19, 20; 22, 24, 25, 26, 27; 28,
29, 29, 30, 32; 32, 33, 34, 35, 35; 36, 38, 40, 40, 42; 44, 46, 52, 58, 70
Test the following: null
(H0): The median FSI for the population of undergraduates at KSU is
25 (h = 25 ) against the alternative (H1): h ¹ 25. Show your work.
Completely discuss and interpret your test results, as indicated in class and
case study summaries. Fully discuss the testing procedure and results. This
discussion must include a clear discussion of the population and the null
hypothesis, the family of samples, the family of errors and the interpretation
of the p-value.
Numbers
12, 14, 17, 19, 20; 22, 24, 25, 26, 27; 28,
29, 29, 30, 32; 32, 33, 34, 35, 35; 36, 38, 40, 40, 42; 44, 46, 52, 58, 70
12, 14, 17, 19, 20; 22, 24, 25, 26, 27; 28, 29, 29, 30, 32; 32, 33, 34, 35, 35; 36, 38, 40, 40, 42;
44, 46, 52, 58, 70
n=30
number of sampled KSU undergrads with FSI < 25 = 7
number
of sampled KSU undergrads with FSI > 25 = 22
error
= larger of number of sampled KSU
undergrads with FSI < 25, number of sampled KSU
undergrads with FSI > 25 = 22
from 30 22 0.00806, p =
2*.00806 = .01612 ~ 1.6%
Discussion
Population:
Undergraduates at KSU (Fall 2006)
Population Median: Median
FSI Level for Undergraduates at KSU (Fall 2006)
Family of Samples: Each
member is a single random sample of 30 Undergraduates at KSU (Fall 2006).
For each member of the FoS, compute the number of sample Undergraduates at KSU
(Fall 2006) with FSI levels strictly less than 25 and the number of sample
Undergraduates at KSU (Fall 2006) with FSI levels strictly higher than 25. The
error is the larger of the two numbers. Doing so for every member of the FoS yields the Family of Errors.
If the true population
median FSI level for undergraduates at KSU (Fall 2006) is 25, then the probability
of getting a sample as bad or worse than our sample is
approximately 1.6%. This sample presents significant evidence against the Null
Hypothesis.
Table 2. Medians
|
n error base p-value 25 1 1.00000 25 2 1.00000 25 3 0.99999 25 4 0.99992 25 5 0.99954 25 6 0.99796 25 7 0.99268 25 8 0.97836 25 9 0.94612 25 10 0.88524 25 11 0.78782 25 12 0.65498 25 13 0.50000 25 14 0.34502 25 15 0.21218 25 16 0.11476 25 17 0.05388 25 18 0.02164 |
n error base p-value 25 19 0.00732 25 20 0.00204 25 21 0.00046 25 22 0.00008 25 23 0.00001 25 24 0.00000 25 25
0.00000 30 1 1.00000 30 2 1.00000 30 3 1.00000 30 4 1.00000 30 5 0.99997 30 6 0.99984 30 7 0.99928 30 8 0.99739 30 9 0.99194 30 10 0.97861 30 11 0.95063 |
n error base p-value 30 12 0.89976 30 13 0.81920 30 14 0.70767 30 15 0.57223 30 16 0.42777 30 17 0.29233 30 18 0.18080 30 19 0.10024 30 20 0.04937 30 21 0.02139 30 22 0.00806 30 23 0.00261 30 24 0.00072 30 25 0.00016 30 26 0.00003 30 27 0.00000 30 28 0.00000 30 29 0.00000 30 30
0.00000 |