Key

The 3rd Hourly

Math 1107

Spring Semester 2010

Protocol: You will use only the following resources: Your individual calculator; individual tool-sheet (single 8.5 by 11 inch sheet); your writing utensils; blank paper (provided by me) and this copy of the hourly. Do not share these resources with anyone else. In each case, show complete detail and work for full credit. Follow case study solutions and sample hourly keys in presenting your solutions. Work all four cases. Using only one side of the blank sheets provided, present your work. Do not write on both sides of the sheets provided, and present your work only on these sheets. All of your work goes on one side each of the blank sheets provided. Space out your work. Do not share information with any other students during this hourly. Do not use any external resources during this hourly.

Sign and Acknowledge: I agree to follow this protocol. 

 

Name (PRINTED)                                             Signature                                             Date

Case One | Confidence Interval for the Population Proportion | Framingham Heart Study

The objective of the Framingham Heart Study(FHS) is to identify the common factors or characteristics that contribute to Cardiovascular disease (CVD) by following its development over a long period of time (since 1948)  in a large group of participants who had not yet developed overt symptoms of CVD or suffered a heart attack or stroke. Blood pressure is a measurement of the force applied to the walls of the arteries as the heart pumps blood through the body. Blood pressure readings are measured in millimeters of mercury (mm Hg) and usually given as two numbers: the systolic blood pressure (SBP) reading, representing the maximum pressure exerted when the heart contracts and the diastolic blood pressure (DBP) reading, representing the pressure in the arteries when the heart is at rest.

Consider the systolic to diastolic blood pressure ratio R = SBP/DBP. 

A sample of FHS adult subjects yields the following ratios:

1.86        1.71        1.43        1.82        1.62        1.57        1.53        1.42        1.55        1.75        1.95        1.58

2.33        1.69        1.56        1.85        1.51        1.50        1.72        1.50        2.06        1.36        1.78        1.56

1.62        1.75        1.78        1.52        1.83        1.60        1.54        1.80        2.13        1.67        1.86        1.60

1.48        1.47        1.45        1.44        1.50        2.05        1.50        1.79        1.82        1.59        1.74        1.86

1.64        1.54        2.03        1.91        1.92        1.97        1.54        1.67        2.00        1.63        1.82        1.49

 

Define the event “Framingham Heart Study Subject Has Systolic to Diastolic Blood Pressure Ratio of strictly less than 2.00.” Estimate the population proportion for this event with 96% confidence. That is, compute and discuss a 96% confidence interval for this population proportion. Provide concise and complete details and discussion as demonstrated in the case study summaries.

 

“Framingham Heart Study Subject Has Systolic to Diastolic Blood Pressure Ratio of strictly less than 2.00.”

 

1.86        1.71        1.43        1.82        1.62        1.57        1.53        1.42        1.55        1.75        1.95        1.58

2.33        1.69        1.56        1.85        1.51        1.50        1.72        1.50        2.06        1.36        1.78        1.56

1.62        1.75        1.78        1.52        1.83        1.60        1.54        1.80        2.13        1.67        1.86        1.60

1.48        1.47        1.45        1.44        1.50        2.05        1.50        1.79        1.82        1.59        1.74        1.86

1.64        1.54        2.03        1.91        1.92        1.97        1.54        1.67        2.00        1.63        1.82        1.49

 

sample size n = 60

 

event=“Framingham Heart Study Subject Has Systolic to Diastolic Blood Pressure Ratio of strictly less than 2.00.”

 

event count = e = 54

p = e /n = 54/60 = .9

1 – p = 1 – .9 = .1

sdp = sqrt(p*(1 – p)/n) = sqrt(.9*.1/60) ≈ 0.03873

 

from 2.10  0.017864  0.96427, Z ≈ 2.10

 

 lower96 = p – 2.1*sdp ≈ .9 – (2.1*0.03873) ≈ 0.818667

upper96 = p + 2.1*sdp ≈ .9 + (2.1*0.03873) ≈ 0.981333

Our population consists of study subjects in the Framingham Heart Study (FHS). Our population proportion is the population proportion of Framingham Heart Study Subject having systolic to diastolic blood pressure ratios of strictly less than 2.00.

 

Each member of the family of samples (FoS) is a single random sample of 60 FHS study subjects. The FoS consists of all possible samples of this type.

 

From each member of the (FoS), compute: 

 

e = sample count of FHS study subjects with systolic to diastolic blood pressure ratios of strictly less than 2.00.

p = sample proportion of FHS study subjects with systolic to diastolic blood pressure ratios of strictly less than 2.00. = e/60

sdp = square root of (p*(1 – p)/n )  

from 2.10  0.017864  0.96427, Z ≈ 2.10

 

and then compute the interval as: lower96  = p –  (2.10*sdp), upper96 = p +  (2.10*sdp).

 

Computing this interval for each member of the FoS forms a family of intervals (FoI).

 

Approximately 96% of the FoI captures the true population proportion of Framingham Heart Study Subject having systolic to diastolic blood pressure ratios of strictly less than 2.00. If our interval resides in this 96% supermajority, then between 81.8% and 98.1% of Framingham Heart Study Subject have systolic to diastolic blood pressure ratios of strictly less than 2.00.

 

 

Case Two | Confidence Interval for the Population Mean | Framingham Heart Study

 

Using the context and data from Case One, estimate the population mean systolic to diastolic blood pressure ratio with 93% confidence. That is, compute and discuss a 93% confidence interval for this population mean. Show your work. Fully discuss the results. This discussion must include a clear discussion of the population and the population mean, the family of samples, the family of intervals and the interpretation of the interval.

 

n      m         sd         se         Z     lower93    upper93

60    1.696    0.20511    0.026479    1.85    1.64701    1.74499

 

from 1.85   0.032157   0.93569, Z ≈ 1.85

se = sd/sqrt(n) = 0.20511/sqrt(60) ≈ 0.026479;

 

lower93 = m - (z*se) = 1.696 - (1.85*0.026479) ≈ 1.64701

upper93 = m + (z*se) = 1.696 + (1.85*0.026479) ≈ 1.74499

 

Our population consists of study subjects in the Framingham Heart Study (FHS). Our population mean is the population mean systolic to diastolic blood pressure ratio.

 

Each member of the family of samples (FoS) is a single random sample of 60 FHS study subjects. The FoS consists of all possible samples of this type.

 

From each member of the (FoS), compute: 

 

m = sample mean systolic to diastolic blood pressure ratio

sd = sample standard deviation for the sample mean systolic to diastolic blood pressure ratio

se = sample standard error = sd/sqrt(60)

 

from 1.85   0.032157   0.93569, Z ≈ 1.85

 

and then compute the interval as:  lower93 = m - (1.85*se), upper93 = m + (1.85*se).

 

Computing this interval for each member of the FoS forms a family of intervals (FoI).

 

Approximately 93% of the FoI captures the true population mean systolic to diastolic blood pressure ratio for Framingham Heart Study subjects. If our interval resides in this 93% supermajority, then systolic to diastolic blood pressure ratio for Framingham Heart Study subjects.is between  1.647 and 1.745.

 

 

Case Three | Hypothesis Test: Categorical Goodness-of-Fit | Fictitious Striped Lizard

The Fictitious Striped Lizard is a native species of Lizard Mountain, and is noteworthy for the both the quantity and quality of its stripes. Consider a random sample of stripes, in which the number of stripes per lizard is noted:

0, 0, 0, 0, 0, 5, 6, 7, 7, 7, 7, 8, 9, 11, 11, 12, 12, 12, 13, 13, 14, 14, 17, 17, 18, 18, 18, 18, 18, 20, 21, 22, 23, 23, 24, 27, 29, 29, 30, 30

Consider the following stripe count categories

Stripe Count Range

Category

0

Stripe-less

1-10

Sparsely Striped

11-15

Normally Striped

16-25

Heavily Striped

26 or more

Densely Striped

Test the hypothesis that:

Pr{ Fictitious Striped Lizard is Stripe-less} = .05

Pr{ Fictitious Striped Lizard is Sparsely Striped} = 0.25

Pr{ Fictitious Striped Lizard is Normally Striped} = 0.30

Pr{ Fictitious Striped Lizard is Heavily Striped} = 0.25

Pr{ Fictitious Striped Lizard is Densely Striped} = 0.15

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. Show all work and detail for full credit.

stripecat            o     e     error       sum

Densely Striped      5     6    0.16667    0.16667

Heavily Striped     13    10    0.90000    1.06667

Normally Striped     9    12    0.75000    1.81667

Sparsely Striped     8    10    0.40000    2.21667

Stripeless           5     2    4.50000    6.71667

 

Stripeless{0 Stripes}: 0, 0, 0, 0, 0 (5 Observed)

Sparsely Striped{1-10 Stripes}: 5, 6, 7, 7, 7, 7, 8, 9 (8 Observed)

Normally Striped{11-15 Stripes}: 11, 11, 12, 12, 12, 13, 13, 14, 14 (9 Observed)

Heavily Striped{16-25 Stripes}: 17, 17, 18, 18, 18, 18, 18, 20, 21, 22, 23, 23, 24 (13 Observed)

Densely Striped{26 or more stripes}: 27, 29, 29, 30, 30 (5 Observed)

 

Total = 5 + 8 + 9 + 13 + 5 = 22 + 18 = 40

 

Expected Counts from the Null Hypothesis for n=40

 

eStripeless = 40*0.05 = 2

eSparse = 40*0.25 = 10

eNormal = 40*0.30 = 12

eHeavy = 40*0.25 = 10

eDense = 40*0.15 = 6

 

Error Calculations

 

errorStripeless = (nStripeless – eStripeless )2/ eStripeless = (5 – 2 )2/ 2 = 4.5

errorSparse = (nSparse – eSparse )2/ eSparse = (8 – 10 )2/ 10 = 0.4

errorNormal = (nNormal – eNormal )2/ eNormal = (9 – 12 )2/12 = 0.75

errorHeavy = (nHeavy – eHeavy )2/ eHeavy = (13 – 10 )2/ 10 = 0.9

errorDense = (nDense – eDense )2/ eDense = (5 – 6 )2/6 = 0.1667

 

errorTotal = errorStripeless + errorSparse + errorNormal + errorHeavy + errorDense = 4.5 + 0.4 + 0.75 + 0.9 + 0.1667 = 6.71667

 

From 5  5.9886  0.200 and 5  7.7794  0.100,  .10 < p < 0.20 – the p-value is strictly between 10% and 20%.

 

Interpretation

 

Our population consists of Fictitious Striped Lizards. Our categories are based on stripe count: Stripeless{0 Stripes}, Sparsely Striped{1-10 Stripes}, Normally Striped{11-15 Stripes}, Heavily Striped{16-25 Stripes} and Densely Striped{26 or more stripes}. Our null hypothesis is that the categories are distributed as: 5% Stripeless, 25% Sparsely Striped, 30% Normally Striped, 25% Heavily Striped and 15% Densely Striped.

Our Family of Samples (FoS) consists of every possible random sample of 40 Fictitious Striped Lizards. Under the null hypothesis, within each member of the FoS, we expect approximately:

eStripeless = 40*0.05 = 2

eSparse = 40*0.25 = 10

eNormal = 40*0.30 = 12

eHeavy = 40*0.25 = 10

eDense = 40*0.15 = 6

 

From each member sample of the FoS, we compute sample counts and errors for each level of stripe count:

 

errorStripeless = (nStripeless – eStripeless )2/ eStripeless

errorSparse = (nSparse – eSparse )2/ eSparse

errorNormal = (nNormal – eNormal )2/ eNormal

errorHeavy = (nHeavy – eHeavy )2/ eHeavy

errorDense = (nDense – eDense )2/ eDense

 

Then add the individual errors for the total error as

errorTotal = errorStripeless + errorSparse + errorNormal + errorHeavy + errorDense

 

Computing this error for each member sample of the FoS, we obtain a Family of Errors (FoE).

 

If the stripe count categories for Fictitious Striped Lizard are distributed as 5% Stripeless, 25% Sparsely Striped, 30% Normally Striped, 25% Heavily Striped and 15% Densely Striped then between 10% and 20% of the member samples of the Family of Samples yields errors as large as or larger than that of our single sample. Our sample does not appear to present significant evidence against the null hypothesis.

 

 

Case Four | Hypothesis Test: Population Median | Fictitious Spotted Toad

The Fictitious Spotted Toad is a native species of Toad Island, and is noteworthy for the both the quantity and quality of its spots. Consider a random sample of toads, in which the number of spots per toad is noted:

12, 14, 15, 17, 18, 20, 20, 21, 22, 23, 24, 24, 25, 26, 26 , 31, 33, 33, 35, 42, 42, 45, 45, 47, 49, 49, 58, 58, 62, 65

Test the following: null (H0): The median spot count for Fictitious Spotted Toads is 35 (h = 35 ) against the alternative (H1): h < 35.

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.

Null Hypothesis: Median Spot Count = 35 Spotss

Alternative  Hypothesis: Median Spot Count < 35 Spots (Guess is too Large)

Error Function: Number of Sample Toads with Strictly Less Than 35 Spots

 

12, 14, 15, 17, 18 | 20, 20, 21, 22, 23 | 24, 24, 25, 26, 26 | 31, 33, 33, 35, 42 | 42, 45, 45, 47, 49 | 49, 58, 58, 62, 65

 

sample error = Number of Sample Toads with Strictly Less Than 35 Spots = 18

n = sample size = 30

from 30   18   0.18080 , p ≈ .18080 (~18.1%)

 

Our population consists of Fictitious Spotted Toads. Our null hypothesis is that the population median spot count for Fictitious Spotted Toads is 35 spots per toad.

 

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

 

From each member of the (FoS), compute an error as the number of sample toads with strictly less than 35 spots. Computing this error for each member of the FoS forms a family of errors (FoE).

 

If the true population median spot count for Fictitious Spotted Toads is 35 spots per toad, then approximately 18.1% member samples from the FoS yield errors as bas as or worse than our error. The sample does not appear to present significant evidence against the null hypothesis.

 

 

Work all four (4) cases.

 

Table 1. Means and Proportions

 Z(k) PROBRT PROBCENT

0.05 0.48006 0.03988

0.10 0.46017 0.07966

0.15 0.44038 0.11924

0.20 0.42074 0.15852

0.25 0.40129 0.19741

0.30 0.38209 0.23582

0.35 0.36317 0.27366

0.40 0.34458 0.31084

0.45 0.32636 0.34729

0.50 0.30854 0.38292

0.55 0.29116 0.41768

0.60 0.27425 0.45149

0.65 0.25785 0.48431

0.70 0.24196 0.51607

0.75 0.22663 0.54675

0.80 0.21186 0.57629

0.85 0.19766 0.60467

0.90 0.18406 0.63188

0.95 0.17106 0.65789

1.00 0.15866 0.68269

Z(k) PROBRT PROBCENT

1.05 0.14686 0.70628

1.10 0.13567 0.72867

1.15 0.12507 0.74986

1.20 0.11507 0.76986

1.25 0.10565 0.78870

1.30 0.09680 0.80640

1.35 0.088508 0.82298

1.40 0.080757 0.83849

1.45 0.073529 0.85294

1.50 0.066807 0.86639

1.55 0.060571 0.87886

1.60 0.054799 0.89040

1.65 0.049471 0.90106

1.70 0.044565 0.91087

1.75 0.040059 0.91988

1.80 0.035930 0.92814

1.85 0.032157 0.93569

1.90 0.028717 0.94257

1.95 0.025588 0.94882

2.00 0.022750 0.95450

Z(k) PROBRT PROBCENT

2.05 0.020182 0.95964

2.10 0.017864 0.96427

2.15 0.015778 0.96844

2.20 0.013903 0.97219

2.25 0.012224 0.97555

2.30 0.010724 0.97855

2.35 0.009387 0.98123

2.40 0.008198 0.98360

2.45 0.007143 0.98571

2.50 0.006210 0.98758

2.55 0.005386 0.98923

2.60 0.004661 0.99068

2.65 0.004025 0.99195

2.70 .0034670 0.99307

2.75 .0029798 0.99404

2.80 .0025551 0.99489

2.85 .0021860 0.99563

2.90 .0018658 0.99627

2.95 .0015889 0.99682

3.00 .0013499 0.99730

 

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

25 19 0.00732

25 20 0.00204

25 21 0.00046 

 n error base p-value

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

30 12 0.89976

30 13 0.81920

30 14 0.70767

30 15 0.57223

 n error base p-value

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 “< 0.00001”

30 28 “< 0.00001”

30 29 “< 0.00001”

30 30 “< 0.00001”

   Table 3. Categories/Goodness of Fit 

 

Categories ERROR  p-value 3 0.0000 1.000                                  3 0.2107 0.900                                   3 0.4463 0.800                                   3 0.7133 0.700                                   3 1.0217 0.600                                   3 1.3863 0.500                                   3 1.5970 0.450                                   3 1.8326 0.400                                   3 2.0996 0.350                                   3 2.4079 0.300                                   3 2.7726 0.250                                   3 3.2189 0.200                                   3 4.6052 0.100                                   3 4.8159 0.090                                   3 5.0515 0.080                                   3 5.3185 0.070                                   3 5.6268 0.060                                   3 5.9915 0.050                                   3 6.4378 0.040                                   3 7.0131 0.030                                   3 7.8240 0.020                                   3 9.2103 0.010

Categories ERROR p-value

4 0.0000 1.000

4 0.5844 0.900

4 1.0052 0.800

4 1.4237 0.700

4 1.8692 0.600

4 2.3660 0.500

4 2.6430 0.450

4 2.9462 0.400

4 3.2831 0.350

4 3.6649 0.300

4 4.1083 0.250

4 4.6416 0.200

4 6.2514 0.100

4 6.4915 0.090

4 6.7587 0.080

4 7.0603 0.070

4 7.4069 0.060

4 7.8147 0.050

4 8.3112 0.040

4 8.9473 0.030

4 9.8374 0.020

4 11.3449 0.010

 

 Categories ERROR p-value

5 0.0000 1.000

5 1.0636 0.900

5 1.6488 0.800

5 2.1947 0.700

5 2.7528 0.600

5 3.3567 0.500

5 3.6871 0.450

5 4.0446 0.400

5 4.4377 0.350

5 4.8784 0.300

5 5.3853 0.250

5 5.9886 0.200

5 7.7794 0.100

5 8.0434 0.090

5 8.3365 0.080

5 8.6664 0.070

5 9.0444 0.060

5 9.4877 0.050

5 10.0255 0.040

5 10.7119 0.030

5 11.6678 0.020

5 13.2767 0.010