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 – Hypothesis Test: Population Median | 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, 4, 4, 4| 4, 5, 5, 5, 5| 5, 5, 6, 6, 6 | 7, 7, 8, 8, 8| 9, 10, 10, 10, 11| 12, 13, 13, 13, 14

14, 14, 15, 16, 16 | 17, 17, 18, 18, 19 | 19, 20, 23, 23, 24 | 27, 28, 30, 30, 31 | 36, 38, 40, 58, 60

Test the following: null (H0): The median survival time (in months after diagnosis) for patients who have died with glioblastoma multiforme is 25 (h = 25) against the alternative (H1): h < 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.

Null Hypothesis: Median Survival Time = 25 Months

Alternative  Hypothesis: Median Survival Time < 25 Months (Guess is too Large)

Error Function: Number of Sample Patients Surviving Strictly Less Than 25 Months

 

0, 1, 2, 2, 3 | 3, 3, 4, 4, 4| 4, 5, 5, 5, 5| 5, 5, 6, 6, 6 | 7, 7, 8, 8, 8| 9, 10, 10, 10, 11| 12, 13, 13, 13, 14

14, 14, 15, 16, 16 | 17, 17, 18, 18, 19 | 19, 20, 23, 23, 24 | 27, 28, 30, 30, 31 | 36, 38, 40, 58, 60

 

sample error = Number of Sample Patients Surviving Strictly Less Than 25  Months = 50

n = sample size = 60

from 60 50 <0.00001, p < 0.00001 < .01 (p is strictly less than 1/100,000)

Our population consists of patients who have been diagnosed with and who have died with glioblastoma multiforme (GBM). Our null hypothesis is that the population median survival time for this population is 25 months.

Each member of the family of samples (FoS) is a single random sample of 60 patients who have been diagnosed with and who have died with glioblastoma multiforme (GBM). The FoS consists of all possible samples of this type.

From each member of the (FoS), compute an error as the number of sample GBM patients surviving strictly less than 25 months. Computing this error for each member of the FoS forms a family of errors (FoE).

If the true population median survival time for GBM patients is 25 months, then fewer than .001% of  member samples from the FoS yield errors as bas as or worse than our error. The sample presents highly significant evidence against the null hypothesis.

Case Two | Confidence Interval: Population Proportion | 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        2.57        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        2.44        2.50        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

 

Consider the proportion of Framingham Heart Study subjects with systolic to diastolic blood pressure ratios of  2 or greater (R ³ 2). Compute and interpret a 96% confidence interval for this population proportion. Show your work. Completely discuss and interpret your test results, as indicated in class and case study summaries.

 

2.57, 2.33, 2.06, 2.13, 2.44 | 2.50, 2.05, 2.03, 2.00

 

 

sample size n=60

event = “diastolic blood pressure ratios of  2 or greater (R ³ 2)”

sample event count = e = 9

p = e/n = 9/60 = .15

1 – p  = 1 – .15  = .85

sdp = sqrt( (p*(1 – p) ) = sqrt( (.15*.85) ) = 0.046098

from 2.10 0.017864 0.96427, z=2.10

lower96 = p – (z*sdp) = .15 – (2.1*0.046098) = 0.053195

upper96 = p + (z*sdp) = .15 + (2.1*0.046098) = 0.24681

Report the interval as [0.053,0.246].

Interpretation

Our population is the population of Framingham Heart Study subjects and our population proportion is the proportion of Framingham Heart Study subjects with systolic to diastolic blood pressure ratios of  2 or greater (R ³ 2).

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 proportion p of  Framingham Heart Study subjects with systolic to diastolic blood pressure ratios of  2 or greater (R ³ 2), sdp = sqrt( (p*(1 – p)/60 ) and then compute the interval

[p – (2.1*sdp), p + (2.1*sdp)].

Computing this interval for each member sample of the FoS, we obtain a Family of Intervals (FoI), approximately 96% of which cover the true population proportion of Framingham Heart Study subjects with systolic to diastolic blood pressure ratios of  2 or greater (R ³ 2).

If our interval is among the approximate 96% super-majority of intervals that cover the population proportion, then between 5.3% and 24.6% of Framingham Heart Study subjects with systolic to diastolic blood pressure ratios of  2 or greater (R ³ 2).

Case Three – 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{Both Blue and Red Show in the Pair}, 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.

Identify the qualifying pairs: Both Blue and Red Show in the Pair: (BRRYR , BGGRY ),( BRRYR, YBBYG ),( BRRYR , RRBBB ), (GGYBR , BGGRY ),( GGYBR , YBBYG ),( GGYBR , RRBBB ), (GYRYG , BGGRY ),( GYRYG , YBBYG ),( GYRYG , RRBBB ), (BYYBG , BGGRY ),( BYYBG , RRBBB ).

 

Pr{One of (BRRYR , BGGRY ),( BRRYR, YBBYG ),( BRRYR , RRBBB ), (GGYBR , BGGRY ),( GGYBR , YBBYG ),( GGYBR , RRBBB ), (GYRYG , BGGRY ),( GYRYG , YBBYG ),( GYRYG , RRBBB ), (BYYBG , BGGRY ),( BYYBG , RRBBB ) Shows} =  

 

Pr{(BRRYR , BGGRY ) } + Pr{ ( BRRYR, YBBYG ) } + Pr{ ( BRRYR , RRBBB ) } +

Pr{(GGYBR , BGGRY ) } + Pr{ ( GGYBR , YBBYG ) } + Pr{ ( GGYBR , RRBBB ) } +

Pr{(GYRYG , BGGRY ) } + Pr{ ( GYRYG , YBBYG ) } + Pr{ ( GYRYG , RRBBB ) } +

Pr{ (BYYBG , BGGRY ) } + Pr{ ( BYYBG , RRBBB )} =

 

(.10*.10) + (.10*.25) + (.10*.35) + (.20*.10) + (.20*.25) + (.20*.35) + (.25*.10) + (.25*.25) + (.25*.35) +

(.45*.10) + (.45*.35) =  (.1*.70) + (.2*1.4) + (.45*.45) = .07 + .028 + .2025 = .3005 

 

Compute Pr{Red Shows in the Pair}, using the Complementary 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.

 

Other Event = “Red Does Not Show in the Pair” = “Red Shows in Neither Slot”

Qualifying Pairs: (BYYBG , YBBYG), ( BYYBG, GGGYY)

 

Pr{ Other Event } = Pr{ “Red Does Not Show in the Pair” } = Pr{ “Red Shows in Neither Slot” } =

Pr{One of (BYYBG , YBBYG) or ( BYYBG, GGGYY)} = Pr{(BYYBG , YBBYG) } + Pr{( BYYBG, GGGYY)} = (.45*.25) + (.45*.30) = .45*.55 = .2475

 

So then

 

Pr{Red Shows in the Pair} = 1 – Pr{ Other Event } = 1 – .2475 = .7525

 

Case Four – Descriptive Statistics | 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

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

n=60

 

min = p00 = 9

p25 = 18

p50 = 27.5

p75 = 41.5

max = p100 = 98

 

range42 = (p100 – p50) = 98 – 27.5 = 70.5

range31 = (p75 – p25) = 41.5 – 18 = 23.5

range20 = (p50 – p00) =  27.5 – 9 = 18.5

 

There are 60 Fictitious Spotted Toads in the sample.

The toad in the sample with the fewest spots has 9 spots.

Approximately 25% of the toads in the sample have 18 or fewer spots.

Approximately 50% of the toads in the sample have 27.5 or fewer spots.

Approximately 75% of the toads in the sample have 41.5 or fewer spots.

The toad in the sample with the most spots has 98 spots.

 

Approximately 50% of the toads in the sample have between 9 and 27.5 spots. The largest possible difference in spot count between any pair of toads in the lower half sample is 18.5 spots.

 

Approximately 50% of the toads in the sample have between 18 and 41.5 spots. The largest possible difference in spot count between any pair of toads in the middle half sample is 23.5 spots.

 

Approximately 50% of the toads in the sample have between 27.5 and 98 spots. The largest possible difference in spot count between any pair of toads in the upper half sample is 70.5 spots.

 

 

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