Key

The Comprehensive Final Examination | Math 1107 | Spring 2010 | 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 | Probability Rules | Color Slot Machine

 Here is our color slot machine – on each trial, it produces a color sequence, using the table below:

 

Color Sequence*

Color Sequence Probability

GRBG

0.25

BBBB

0.05

RYGB

0.25

YYBR

0.20

BYBY

0.25

Total

1.00

 

*B-Blue, G-Green, R-Red, Y-Yellow, sequence is numbered as 1st to 4th, from left to right: (1st 2nd 3rd 4th)

 

Compute the following probabilities. In each of the following, show your intermediate steps and work. If a rule is specified, you must use that rule for your computation. Briefly interpret each probability using a long run or relative frequency argument.

                 

Pr{Green Shows} 

 

Pr{Green Shows} = Pr{One of GRBG or  RYGB Shows} = Pr{ GRBG } + Pr{ RYGB } = 0.25 + 0.25 = 0.50  

 

Pr{Blue and Red Both Show}

 

Pr{ Blue and Red Both Show } = Pr{One of GRBG, RYGB or YYBR Shows} = Pr{ GRBG } + Pr{ RYGB } + Pr{ YYBR }= 0.25 + 0.25 + 0.20 = 0.70

 

Pr{Red or Yellow Shows} - Use the Complementary Rule

 

Other Event = “Neither Red nor Yellow Show”

Pr{“Neither Red nor Yellow Show”} = Pr{BBBB} = 0.05

Pr{Red or Yellow Shows} = 1 - Pr{“Neither Red nor Yellow Show”} = 1 - 0.05 = 0.95

 

Show all work and full detail for full credit.

 

Case Two | Descriptive Statistics | Serum Creatinine

Healthy kidneys remove wastes and excess fluid from the blood. Blood tests show whether the kidneys are failing to remove wastes. Urine tests can show how quickly body wastes are being removed and whether the kidneys are also leaking abnormal amounts of protein. Creatinine is a waste product that comes from meat protein in the diet and also comes from the normal wear and tear on muscles of the body. Creatinine is produced at a continuous rate and is excreted only through the kidneys. When renal dysfunction occurs, the kidneys are impaired in their ability to excrete creatinine and the serum creatinine rises. As kidney disease progresses, the level of creatinine in the blood increases.

 

Suppose that we obtain serum creatinine levels in a random sample of adults. Serum creatinine (as milligrams of creatinine per deciliter of serum) for each sampled subject follows:

12.15, 11.10, 10.05, 9.83, 8.14, 7.30, 5.15, 5.00, 4.93, 4.85, 4.01, 3.50, 3.33, 3.27, 3.24, 2.90, 2.50, 2.30, 2.05, 2.02, 1.99, 1.95, 1.78, 1.61, 1.55, 1.50, 1.47, 1.43, 1.38, 1.36, 1.29, 1.26, 1.25, 1.19, 1.12, 1.09, 1.05, 0.95, 0.92, 0.90

Compute and interpret the following descriptive statistics: p0, p25, p50, p75, p100, p100-p75, p75-p50, p50-p25 and p25-p00.

 

n     p100     p75     p50      p25     p00     R43     R32      R21     R10

40    12.15    4.43    2.005    1.325    0.9    7.72    2.425    0.68    0.425

 

R43 = p100 – p75 = 12.15 – 4.43 = 7.72

R32 = p75 – p50 = 4.43 –2.005 = 2.425

R21 = p50 – p25 = 2.005 – 1.325 = 0.68

R10 = p25 – p00 = 1.325 – 0.9 = 0.425

There are 40 adults in our sample.

The adult in the sample with the lowest serum creatinine has a level of 0.9 mg/dL.

Approximately 25% of the adults in the sample have serum creatinine levels of 1.325 mg/dL or less.

Approximately 50% of the adults in the sample have serum creatinine levels of 2.005 mg/dL or less.

Approximately725% of the adults in the sample have serum creatinine levels of 4.430 mg/dL or less.

Approximately 25% of the adults in the sample have serum creatinine levels of 1.325 mg/dL or less.

The adult in the sample with the highest serum creatinine has a level of12.15 mg/dL.

Approximately 25% of the adults in sample have serum creatinine levels between 0.9 and 1.325 mg/dL. The largest possible difference in serum creatine level between any pair of adults in this lower quarter sample is 0.425.

Approximately 25% of the adults in sample have serum creatinine levels between 1.325 and 2.005 mg/dL. The largest possible difference in serum creatine level between any pair of adults in this lower middle quarter sample is 0.68.

Approximately 25% of the adults in sample have serum creatinine levels between 2.005 and 4.430  mg/dL. The largest possible difference in serum creatine level between any pair of adults in this upper middle quarter sample is 2.425.

Approximately 25% of the adults in sample have serum creatinine levels between 4.430 and 12.150 and 0.9 and 1.325 mg/dL. The largest possible difference in serum creatine level between any pair of adults in this upper quarter sample is 7.720.

Case Three | Confidence Interval Mean | Serum Creatinine

Using the data and context from Case Two, estimate the population mean serum creatinine with 96% confidence. That is, compute and discuss a 96% confidence interval for this population mean. Provide concise and complete details and discussion as demonstrated in the case study summaries.

 

n       m         sd       Z        se      lower96    upper96\

40    3.3665    3.04388    2.1    0.48128    2.35581    4.37719

 

Z = 2.1 from 2.10   0.017864   0.96427

se = sd/sqrt(n) ≈ 3.04388/sqrt(40) ≈ 0.48128

lower96 = m – (Z*se)  ≈  3.3665 – (2.1*0.48128) ≈ 2.35581

upper96 = m + (Z*se)  ≈  3.3665 + (2.1*0.48128) ≈ 4.37719

 

Our population consists of human adults.. Our population mean is the population mean serum creatinine.

 

Each member of the family of samples (FoS) is a single random sample of 40 adults. The FoS consists of all possible samples of this type.

 

From each member of the (FoS), compute: 

 

m = sample mean serum creatinine

sd = sample standard deviation for the sample mean serum creatinine

se = sample standard error = sd/sqrt(40)

 

Z = 2.1 from 2.10   0.017864   0.96427

 

and then compute the interval as:  lower96 = m - (2.10*se), upper96 = m + (2.10*se).

 

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

 

Approximately 96% of the FoI captures the true population mean serum creatinine for human adults.

If our interval resides in this 96% supermajority, then the population mean serum creatinine for human adults is between  2.356 and 4.377 mg/dL.

Case Four | Categorical Goodness of Fit | Gestational Age

Gestational age is the time spent between conception and birth, usually measured in weeks. In general, infants born after 36 or fewer weeks of gestation are defined as premature, and may face significant challenges in health and development. Infants born after 37-40 weeks of gestation are generally viewed as full term, and those born after 41 or more weeks of gestation are generally viewed as post term. Suppose that a random sample of 2005 US resident live born infants yields the following gestational ages (in weeks): 

 

 

 

Test the null hypothesis that the 2005 US resident live births are distributed as 10% Premature, 85% Full Term and 5% Post Term. 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.

 

Premature {36 weeks or less}: 24, 25, 25, 26, 27 | 29, 30, 33, 35, 36 | 36 (11 Observed)

Full Term{37 – 40 weeks}: 37, 37, 37, 37, 37| 37, 37, 37, 37, 37| 37, 37, 37, 37, 38| 38, 38, 38, 38, 38| 38, 38, 38, 38, 38| 38, 39, 39, 39, 39| 39, 40, 40, 40, 40| 40, 40, 40, 40 , 40 | 40  (41 Observed)

Post Term {41 or more weeks }: 41, 41, 41, 42, 42 | 42, 43, 43 (8 Observed)

 

Total = 11 + 41 + 8  = 60

 

Expected Counts from the Null Hypothesis for n=60

 

ePremature = 60*0.10 = 6

eFull Term = 60*0.85 = 51

ePostTerm = 60*0.05 = 3

 

Error Calculations

 

errorPremature = (nPremature – ePremature )2/ ePremature = (11 – 6 )2/ 6 = 4.1675

errorFull Term = (nFull Term – eFull Term )2/ eFull Term = (41 – 51 )2/ 51 = 1.9608

errorPost Term = (nPost Term – ePost Term )2/ ePost Term = (8 – 3 )2/3 = 8.333

 

errorTotal = errorPremature + errorFull Term +errorPost Term = 4.1675 + 1.9608 + 8.333 = 14.461 over 3 categories.

 

 

From 3   9.21030    0.010,  p < 0.01 – the p-value is strictly less than 1%.

 

Interpretation

 

Our population consists of year 2005 US Resident live births. Our categories are based on gestational age: Premature{36 or fewere weeks}, Full Term{37 – 40 weeks} and Post Term {41 or more weeks} Our null hypothesis is that the categories are distributed as: 10% Premature, 85% Full Term and 5% Post Term.

Our Family of Samples (FoS) consists of every possible random sample of 60 year 2005 US resident live births.. Under the null hypothesis, within each member of the FoS, we expect approximately:

ePremature = 60*0.10 = 6

eFull Term = 60*0.85 = 51

ePostTerm = 60*0.05 = 3

 

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

 

errorPremature = (nPremature – ePremature )2/ ePremature

errorFull Term = (nFull Term – eFull Term )2/ eFull Term

errorPost Term = (nPost Term – ePost Term )2/ ePost Term

 

 

Then add the individual errors for the total error as

 

errorTotal = errorPremature + errorFull Term +errorPost Term

 

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

 

If the categories are distributed as: 10% Premature, 85% Full Term and 5% Post Term then strictly less than 1% of the member samples of the Family of Samples yields errors as large as or larger than that of our single sample. Our sample presents highly significant evidence against the null hypothesis.

 

 

 

 

 

 

 

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: Categories

categories error  p-value

3   0.00000   1.000

3   0.21072   0.900

3   0.44629   0.800

3   0.71335   0.700

3   1.02165   0.600

3   1.38629   0.500

3   1.59702   0.450

3   1.83258   0.400

3   2.09964   0.350

3   2.40795   0.300

3   2.77259   0.250

3   3.21888   0.200

3   3.48594   0.175

3   3.79424   0.150

3   4.15888   0.125

3   4.60517   0.100

3   4.81589   0.090

3   5.05150    0.080

3   5.31850    0.070

3   5.62680    0.060

3   5.99150    0.050

3   6.43780    0.040

3   7.01310    0.030

3   7.82400    0.020

3   9.21030    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   4.9566   0.175

4   5.3170   0.150

4   5.7394   0.125

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   6.3423     0.175

5   6.7449     0.150

5   7.2140     0.125

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