The Comprehensive Final Examination | Math 1107 | Fall Semester 2009 | CJ Alverson

 

Protocol

 

You will use only the following resources: Your individual calculator; Your individual tool-sheets (two (2) 8.5 by 11 inch sheets); Your writing utensils; Blank Paper (provided by me); This copy of the hourly and

the tables provided by me. Do not share these resources with anyone else.

 

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.

 

When you’re done: Print your name on a blank sheet of paper. Place your toolsheet, test and work under this sheet, and turn it all in to me.

 

Do not share information with any other students during this test.

 

Sign and Acknowledge: 

 

I agree to follow this protocol. Initial:___

 

______________________________________________________________________________________

Name (PRINTED)                                          Signature                                          Date

 

 

Case One | Confidence Interval: Population Mean | Framingham Heart Study

The objective of the Framingham Heart Study was 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

 

Estimate the population mean systolic to diastolic blood pressure ratio R with 95% confidence. That is, compute and discuss a 95% confidence interval for this population mean ratio. Show your work. Fully discuss the results.

 

sample size n=60

sample mean  =  m = 1.696

sd = 0.20511

se = sd/sqrt(60) = 0.026479

from 2.00 0.022750 0.95450, z=2.00

lower95 = m – (z*se) = 1.696 - 2*0.026479 = 1.64304

upper95 = m + (z*se) =  1.696 + 2*0.026479 = 1.74896

Report the interval as [1.64, 1.74].

Our population is the population of Framingham Heart Study subjects and our population mean systolic to diastolic blood pressure ratio.

Our Family of Samples (FoS) consists of every possible random sample of 60 Framingham Heart Study subjects.

From each member sample of the FoS, we compute the sample mean systolic to diastolic blood pressure ratio, the standard deviation sd, se=sd/sqrt(60) and then compute the interval

[m – (2*se), m + (2*se)].

Computing this interval for each member sample of the FoS, we obtain a Family of Intervals (FoI), approximately 95% of which cover the true population mean (for Framingham Heart Study subjects) systolic to diastolic blood pressure ratio.

If our interval is among the approximate 95% super-majority of intervals that cover the population proportion, then the population mean (for Framingham Heart Study subjects) systolic to diastolic blood pressure ratio is between 1.64 and 1.74.

Case Two – 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:

9, 10, 10, 10, 11 | 12, 12, 12, 12, 13 | 15, 16, 17, 18, 18 | 18, 18, 19, 19, 20 | 20, 21, 22, 22, 23

24, 24, 25, 25, 26 | 29, 29, 29, 31, 31| 33, 33, 33, 33, 33 | 34, 39, 40, 40, 40 | 43, 43, 45, 47, 47

50, 56, 56, 56, 60 | 63, 65, 66, 67, 98

Test the following: null (H0): The median number of spots per Fictitious Spotted Toad 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.

Null Hypothesis: Median Fictitious Spotted Toad Spot Count = 35 Spots

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

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

 

9, 10, 10, 10, 11 | 12, 12, 12, 12, 13 | 15, 16, 17, 18, 18 | 18, 18, 19, 19, 20 | 20, 21, 22, 22, 23

24, 24, 25, 25, 26 | 29, 29, 29, 31, 31| 33, 33, 33, 33, 33 | 34, 39, 40, 40, 40 | 43, 43, 45, 47, 47

50, 56, 56, 56, 60 | 63, 65, 66, 67, 98

 

sample error = Number of Sample Toads with Strictly More Than 35 Spots = 19

n = sample size = 60

from 60 19 0.99689, p = 0.99689

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

Each member of the family of samples (FoS) is a single random sample of 60 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 toads with strictly more 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, then more than 99.6% of  member samples from the FoS yield errors as bas as or worse than our error. The sample does not present significant evidence against the null hypothesis.

Case Three – Descriptive Statistics | Glioblastoma Multiforme

 

Glioblastoma multiforme (GBM) is the highest grade glioma tumor and is the most malignant form of astrocytomas. These tumors originate in the brain. GBM tumors grow rapidly, invade nearby tissue and contain cells that are very malignant. GBM are among the most common and devastating primary brain tumors in adults. Suppose that we have a random sample of GBM patients, with survival time (in months) listed below:

 

0, 1, 2, 2, 3| 3, 3, 3, 3, 3| 4, 4, 4, 4, 4| 4, 5, 5, 5, 5| 5, 5, 6, 6, 6| 7, 7, 8, 8, 8| 9, 9, 10, 10, 10| 10, 11, 11, 11 12| 12, 12, 13, 13, 13| 14, 14, 14, 15, 16| 16, 17, 17, 18, 18| 19, 19, 20, 23, 23| 24, 24, 25, 24, 25| 27, 28, 30, 30, 31| 36, 38, 40, 58, 60| 61

Compute and interpret the following statistics: sample size, p00, p25, p50, p75, p100, (p100 – p25),       (p75 – p25), (p75 – p00). 

n=76

minimum = p00 = 0

p25 = 5

p50 = 11

p75 = 19.5

maximum = p100 = 61

range41 = p100 – p25 = 61 – 5 = 56

range31 = p75 – p25 = 19.5 – 5 = 14.5

range30 = p75 – p00 = 19.5 – 0 = 19.5

 

There are 76 patients in our sample who have been diagnosed with and who have died with glioblastoma multiforme (GBM).

 

The GBM patient in our sample with the smallest survival time died less than 1 month after diagnosis.

 

Approximately 25% of the GBM patient in our sample survived 5 months or less after diagnosis.

Approximately 50% of the GBM patient in our sample survived 11 months or less after diagnosis.

Approximately 75% of the GBM patient in our sample survived 19.5 months or less after diagnosis.

 

The GBM patient in our sample with the longest survival time died 61 months after diagnosis.

 

Approximately 75% of the patients in the sample survive between 5 and 61 months after diagnosis. The largest possible difference in survival time between any pair of patients in this upper three-quarter sample is 56 months.

 

Approximately 50% of the patients in the sample survive between 5 and 19.5 months after diagnosis. The largest possible difference in survival time between any pair of patients in this middle half sample is 14.5 months.

 

Approximately 75% of the patients in the sample survive between 0 and 19.5 months after diagnosis. The largest possible difference in survival time between any pair of patients in this lower three-quarter sample is 19.5 months.

 

Case Four – Probability Rules | Pair of Color Slot Machines

Consider a pair of color slot machines, described by the tables below. Assume that the probabilities are correct, and that the machines operate in a mutually independent fashion. Our experiment consists of observing pairs of sequences from the machines.

 

Machine 1

Machine 2

Sequence*

Probability

Sequence*

Probability

BRRYR

.10

BGGRY

.10

GGYBR

.20

YBBYG

.25

GYRYG

.25

GGGYY

.30

BYYBG

.45

RRBBB

.35

Total

1

Total

1

*B-Blue, G-Green, R-Red, Y-Yellow, sequences are numbered from left to right.

Compute Pr{Blue Shows in One or Both of the Machines}, using the Additive and Multiplicative Rules

 

Machine 1

Machine 2

Sequence*

Probability

Sequence*

Probability

BRRYR

.10

BGGRY

.10

GGYBR

.20

YBBYG

.25

GYRYG

.25

GGGYY

.30

BYYBG

.45

RRBBB

.35

Total

1

Total

1

*B-Blue, G-Green, R-Red, Y-Yellow, sequences are numbered from left to right.

Pr{Blue Shows Only on 1st Slot} = Pr{Blue Shows on 1st}*Pr{Blue Does Not Show on 2nd} =

Pr{One of BRRYR, GGYBR or GYRYG Shows}*Pr{GGGYY Shows} =

(Pr{BRRYR}+Pr{GGYBR}+Pr{GYRYG})*Pr{GGGYY Shows} =

(.10+.20+.45)*(.30) = .75*.30 = .225

 

Pr{Blue Shows Only on 2nd Slot} = Pr{Blue Does Not Shows on 1st}*Pr{Blue Shows on 2nd} =

Pr{GYRYG}*Pr{One of BGGRY, YBBYG or RRBBB Shows} =

Pr{GYRYG}*(Pr{BGGRY}+Pr{YBBYG}+Pr{RRBBB}) =

(.25)*(.10+.25+.35) = .25*.70 = .175

 

Pr{Blue Shows on Both Slots} = Pr{Blue Shows on 1st}*Pr{Blue Shows on 2nd} = .75*.70 = .525

 

So Pr{Blue Shows in One or Both of the Machines} =

Pr{Blue Shows Only on 1st Slot} + Pr{Blue Shows Only on 2nd Slot} + Pr{Blue Shows on Both Slots} =

.225 + .175 + .525 = .40 + .525 = .925

 

Compute Pr{Blue Shows in Neither Machine}, using the Complementary and Multiplicative Rules

Other Event = “Blue Shows in Both Machines”

Pr{Blue Shows in Neither Machine} = 1 – Pr{“Blue Shows in Both Machines”} = 1 – .525 = .475

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

60       0     1.00000                                     60       1     1.00000                                     60       2     1.00000                                     60       3     1.00000                                     60       4     1.00000                                     60       5     1.00000                                     60       6     1.00000                                     60       7     1.00000                                     60       8     1.00000                                     60       9     1.00000                                     60      10     1.00000                                     60      11     1.00000                                     60      12     1.00000                                     60      13     0.99999                                     60      14     0.99998                                     60      15     0.99993                                     60      16     0.99980                                     60      17     0.99947                                     60      18     0.99866                                     60      19     0.99689                                     60      20     0.99326                                    

n         error          base p-value

60      21     0.98633                                     60      22     0.97405                                     60      23     0.95377                                     60      24     0.92250                                     60      25     0.87747                                     60      26     0.81685                                     60      27     0.74052                                     60      28     0.65056                                     60      29     0.55129                                     60      30     0.44871                                     60      31     0.34944                                     60      32     0.25948                                     60      33     0.18315                                     60      34     0.12253                                     60      35     0.07750                                     60      36     0.04623                                     60      37     0.02595                                     60      38     0.01367                                     60      39     0.00674                                     60      40     0.00311                                      

n         error          base p-value

60      41     0.00134                                     60      42     0.00053                                     60      43     0.00020                                     60      44     0.00007                                     60      45     0.00002                                     60      46     0.00001                                     60      47     <0.00001                                     60      48     <0.00001                                     60      49     <0.00001
                                 60      50     <0.00001                                   60      51     <0.00001                                   60      52     <0.00001                                   60      53     <0.00001                                   60      54     <0.00001                                   60      55     <0.00001                                   60      56     <0.00001                                   60      57     <0.00001                                   60      58     <0.00001                                        60      59     <0.00001                                            60      60     <0.00001