9th November 2009

Summaries

Session 3.5

Working Case List:

 

From http://www.cjalverson.com/_2ndhourlysummer2009verAkey.htm :

Case Five | Hypothesis Test – Population Median | Fictitious Spotted Lizard

 

The Fictitious Spotted Lizard 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 Fictitious Spotted Lizards, in which the number of spots per lizard is noted:

 

8, 11, 13, 14, 15, 15, 16, 16, 17, 19, 19, 20, 20, 20, 20,

21, 21, 22, 22, 22, 22, 26, 26, 26, 27, 24, 24, 24, 25, 27

 

Test the following: null (H0): The median number of spots per Fictitious Spotted Lizard is 22 (h = 22) against the alternative (H1): h > 22. 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

 

Test the following: null (H0): The median number of spots per Fictitious Spotted Lizard is 22 (h = 22) against the alternative (H1): h > 22.

 

Alternative: “Guess is too small”

Error Form: “Count Strictly Above 22”

 

8, 11, 13, 14, 15 | 15, 16, 16, 17, 19 | 19, 20, 20, 20, 20,

21, 21, 22, 22, 22 | 22, 26, 26, 26, 27 | 24, 24, 24, 25, 27

 

sample size = n = 30

sample error = 9

from 30 9 0.99194 , p » 0.99194

 

Interpretation

 

Each member of the family of sample (FoS) is a single random sample of 30 Fictitious Spotted lizards. The FoS consists of all possible samples of this type.

 

From each member sample, compute the sample number of lizards with strictly more than 22 spots. Computing this error for each member sample of the FoS yields a family of errors (FoE).

 

If the true population median spot count for Fictitious Spotted lizards is 22 spots, then approximately 99.1% of the errors equal or exceed our error. The sample does not appear to present significant evidence against the null hypothesis, if the alternative is a higher guess.

 

From http://www.mindspring.com/~cjalverson/_2ndhourlysummer2007versionAkey.htm : 

 

Case Four        

Hypothesis Test – Population Median

Duchenne Muscular Dystrophy: Survival Time

Duchenne muscular dystrophy (DMD) is an inherited disorder characterized by rapidly progressive muscle weakness which starts in the legs and pelvis and later affects the whole body1. Suppose that we follow individuals diagnosed with DMD from diagnosis until death, noting age at  death in months. Consider a random sample of individuals who were diagnosed with, and died with DMD. Age at death (in months) follows below:

37 42 47 50 75 80 85 97 100 127 130 143 150 156 173 177 179 180 181 181 182 184 192 193 193  200 210 235 240 278.

Test the following: null (H0): The median age at death is 180 months (h = 180 months) against the alternative (H1): h < 180 months. 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. 1. http://www.pennhealth.com/ency/article/000705.htm  

Numbers

37 42 47 50 75 80 85 97 100 127 130 143 150 156 173 177 179 | 180 | 181 181 182 184 192 193 193  200 210 235 240 278

n = 30

error = Number of DMD patients with age at death < 180 months = 17

from 30 17 0.29233 , p @ 0.29233

Interpretation

Population: Patients with Duchenne Muscular Dystrophy(DMD)

Population Median: Median Age at Death

Family of Samples(FoS): Each member is a single random sample of 30 deceased DMD patients. The FoS consists of all such samples.

For each member of the FoS, compute the number of sample patients who died at strictly less than 180 months of age. Doing so for every member of the FoS yields the Family of Errors.

If the true population median age at death for DMD patients is 180 months, then the probability of getting a sample as bad or worse than our sample is approximately 29.233%. This sample does not seem to present significant evidence against the Null Hypothesis. We do not reject the Null Hypothesis at either 5% or 1% significance.

From : http://www.mindspring.com/~cjalverson/_2ndhourlysummer2006_key.htm

 

Case Two

Median Test

Traumatic Brain Injury (TBI) and Glasgow Coma Scale (GCS)

 

The Glasgow Coma Scale (GCS) is the most widely used system for scoring the level of consciousness of a patient who has had a traumatic brain injury. GCS is based on the patient's best eye-opening, verbal, and motor responses. Each response is scored and then the sum of the three scores is computed. That is, GCS=Eye+Verbal+Motor.

 

Eye opening: 4. Spontaneous. Indicates arousal, not necessarily awareness; 3. To speech. When spoken to – not necessarily the command to open eyes; 2. To pain. Applied to limbs, not face where grimacing can cause closure and 1. None.

 

Motor response: 6. Obeys commands. Exclude grasp reflex or postural adjustments; 5. Localises. Other limb moves to site of nailbed pressure 4. Withdraws. Normal flexion of elbow or knee to local painful stimulus; 3. Abnormal flexion. Slow withdrawal with pronation of wrist, adduction of shoulder 2. Extensor response. Extension of elbow with pronation and adduction and 1. No movement.

 

Verbal responses: 5. Orientated. Knows who, where, when; year, season, month; 4. Confused conversation. Attends & responds but answers muddled/wrong; 3. Inappropriate words. Intelligible words but mostly expletives or random; 2. Incomprehensible speech. Moans and groans only – no words and 1. None.

 

Glasgow Coma Scale Categories: Mild (13-15); Moderate (9-12) and Severe/Coma (3-8)

 

Traumatic brain injury (TBI) is an insult to the brain from an external mechanical force, possibly leading to permanent or temporary impairments of cognitive, physical, and psychosocial functions with an associated diminished or altered state of consciousness. A patient with mild traumatic brain injury is a person who has had a traumatically induced physiological disruption of brain function, as manifested by a least one of the following: Any period of loss of consciousness; Any loss of memory for events immediately before or after the accident; Any alteration in mental state at the time of the accident (eg, feeling dazed, disoriented, or confused); Any focal neurological deficit(s) that may or may not be transient; but where the severity of the injury does not exceed the following: posttraumatic amnesia (PTA) not greater than 24 hours, after 30 minutes, an initial Glasgow Coma Scale (GCS) of 13-15; and Any loss of consciousness of approximately 30 minutes or less. TBI includes: 1) the head being struck, 2) the head striking an object, and 3) the brain undergoing an acceleration/deceleration movement (ie, whiplash) without direct external trauma to the head.

Consider a random sample of patients with TBI, with GCS at initial treatment and diagnosis listed below:

 

3, 3, 3, 4, 4, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 9, 9, 9, 10, 10, 11, 11, 12, 13, 13, 13, 14, 14

Test the following: null (H0): The median GCS at initial treatment and diagnosis is (h = 9) against the alternative (H1): h ¹ 9. 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.


Key

3, 3, 3, 4, 4, 5, 6, 6, 6, 6, 7, 7, 7, 8, 8, 9, 9, 9, 9, 9, 10, 10, 11, 11, 12, 13, 13, 13, 14, 14

 

n=30

error = maximum of ({# sps < 9}, {# sps > 9}) = maximum of ({15}, {10}) = 15.

 

From 30 15 0.57223, our p-value is approximately 2*0.57223 »1.14446, which we truncate at 100%.

 

This sample produces a nearly perfect split as predicted by the null hypothesis.

 

Our population consists of patients with traumatic brain injury (TBI). Each member of the family of samples is a single random sample of 30 patients with TBI, and the family of samples (FoS) consists of all such samples.

 

Our null hypothesis is that the median Glasgow coma score (GCS) for newly diagnosed cases of TBI is 9, against the alternative hypothesis that the population median is not 9.

 

The null hypothesis suggests that the sample should be evenly split across 9. Each member of the FoS yields a pair of counts: the number of sample patients with GCS < 9, and the number of sample patients with GCS>9. The error is the larger of these counts.

 

Error = Maximum of {# sps < 9}, {# sps > 9})

 

Each member of the FoS yields an error, computed as above, yielding a family of errors(FoE), in which each member is an error computed from a member of the FoS.

 

If the null hypothesis holds true, then approximately 100% of the errors exceed 15. The sample does not present significant evidence against the null hypothesis. In fact, this sample produces a nearly perfect split as predicted by the null hypothesis.

 

 

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

Our error has the form Error = Number of sample lizards who have strictly more than 12 spots. Our computed error is 21, computed from a random sample of 30 Fictitious Spotted Lizards. Using the row

30 21 0.02139, the computed p-value is .02139, or approximately 2.1%.

Discussion

Our population is the population of Fictitious Spotted Lizards.

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).

If the true population median number of spots per lizard for the population of Fictitious Spotted Lizard is 12 spots, then approximately 2.1% of the Family of Samples yield errors as bad as or worse than our single error. The 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 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/_comprehensive%20final%20fall%202006%20Wednesday%20Version%20Key.htm

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 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