Key | The Comprehensive Final Examination | Math 1107 | Spring 2011 | 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 toolsheets, test and work under this sheet, and turn it all in to me. Do not share information with any other students during this hourly.

 

Sign and Acknowledge:  I agree to follow this protocol. Initial:___

 

_____________________________________________________________________________________

Name (PRINTED)                                          Signature                                          Date

 

Case One | Color Slot Machine | Random Variables, Conditional Probability

 

Here is our slot machine – on each trial, it produces a 6-color sequence, using the table below:

 

Sequence*

Probability

BBBBBB

.10

BBGGBG

.10

RRYYGG

.15

GBBGGY

.10

RGBYYR

.25

BBYYGR

.10

YBGRYY

.20

Total

1.00

* B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered as 1st to 6th , from left to right: (1st 2nd 3rd 4th 5th6th )

Consider the random variable BlueCount, defined as the number of times that Blue shows in the color sequence. Compute the values and probabilities for BlueCount.

 

Sequence*

Probability

BlueCount

BBBBBB

.10

6

BBGGBG

.10

3

RRYYGG

.15

0

GBBGGY

.10

2

RGBYYR

.25

1

BBYYGR

.10

2

YBGRYY

.20

1

Total

1.00

 

 

Pr{BlueCount = 0} = Pr{ RRYYGG } = 0.15

 

Pr{BlueCount = 1} = Pr{One of RGBYYR  or YBGRYY  Shows} =

Pr{ RGBYYR } + Pr{YBGRYY  Shows} = 0.25 + 0.20 = 0.45

 

Pr{BlueCount = 2} = Pr{One of GBBGGY  or BBYYGR  Shows} =

Pr{ GBBGGY } + Pr{ BBYYGR } = 0.10 + 0.10 = 0.20

 

Pr{BlueCount = 3} = Pr{ BBGGBG } = 0.10

 

Pr{BlueCount = 6} = Pr{ BBBBBB } = 0.10

 

 

Compute Pr{ Blue Shows | Yellow Shows } – This is a conditional probability.

 

Sequence*

Probability

RRYYGG

0.15

GBBGGY

0.10

RGBYYR

0.25

BBYYGR

0.10

YBGRYY

0.20

Total

0.80

 

Pr{ Yellow Shows } =

Pr{One of RRYYGG, GBBGGY, RGBYYR, BBYYGR or YBGRYY Shows} =

Pr{ RRYYGG} + Pr{GBBGGY} + Pr{RGBYYR} + Pr{BBYYGR} + Pr{YBGRYY} =

0.15 + 0.10 + 0.25 + 0.10 + 0.20 = 0.80

 

Sequence*

Probability

GBBGGY

0.10

RGBYYR

0.25

BBYYGR

0.10

YBGRYY

0.20

Total

0.65

 

 

Pr{ Blue and Yellow Show } = Pr{One of GBBGGY, RGBYYR, BBYYGR or YBGRYY Shows} =

Pr{GBBGGY} + Pr{RGBYYR} + Pr{BBYYGR} + Pr{YBGRYY} =

0.10 + 0.25 + 0.10 + 0.20 = 0.65

 

Pr{ Blue Shows | Yellow Shows } = Pr{ Blue and Yellow Show } / Pr{Yellow Shows } = 0.65/0.80

 

Show full work and detail for full credit.

Case Two | Descriptive Statistics | Serum Cotinine in Non-smokers

Nicotine is one of the medically active ingredients in tobacco. It metabolizes as cotinine in the blood. Consider a random sample of US resident adults responding to a health survey who identify as non-smokers. From each sample subject we obtain serum, and determine the serum cotinine level as nanograms cotinine per milliliter serum (ng/mL*). Here is the serum cotinine data:

0.015, 0.015, 0.016, 0.017, 0.019, 0.019, 0.020, 0.021, 0.21, 0.022, 0.022, 0.023, 0.025, 0.027, 0.028, 0.029, 0.029, 0.031, 0.032, 0.034, 0.040, 0.043, 0.044, 0.047, 0.050, 0.54, 0.056, 0.059, 0.062, 0.069, 0.073, 0.075, 0.077, 0.082, 0.084, 0.091, 0.099, 0.101, 0.107, 0.113, 0.250, 0.750, 0.900, 1.005, 1.100

*ng/mL: One billionth of one gram per one thousandth of a liter

Compute and interpret the following statistics: sample size, p00, p25, p50, p75, p100, (p100-p75), (p75-p25) and (p25-p00). Show your work, and discuss completely for full credit.

Numbers

n     p00      p25      p50      p75     p100    Range10    Range31    Range43

45    0.015    0.025    0.047    0.091     1.1      0.01      0.066      1.009

 

R43 = p100 – p75  = 1.100 – 0.091 = 1.009

R31 = p75 – p25 = 0.091 – 0.025 = 0.066

R10 = p25 –p00  = 0.025 – 0.015 = 0.010    

 

Interpretation

 

There are 50 US Resident Adult Self-reported non-smokers in our sample.

The adult in our sample with the lowest serum cotinine level has serum cotinine at 0.015 ng/mL.

Approximately 25% of the adults in our sample have serum cotinine levels at 0.025 ng/mL or less.

Approximately 50% of the adults in our sample have serum cotinine levels at 0.047 ng/mL or less.

Approximately 75% of the adults in our sample have serum cotinine levels at 0.091 ng/mL or less.

The adult in our sample with the highest serum cotinine level has serum cotinine at 1.100 ng/mL.

Approximately 25% of the adults in our sample have serum cotinine levels between 0.091 and 1.1 ng/mL. The largest difference in serum cotinine between any pair of adults in this upper quarter sample is 1.009 ng/mL. 

Approximately 50% of the adults in our sample have serum cotinine levels between 0.025 and 0.091 ng/mL. The largest difference in serum cotinine between any pair of adults in this middle half sample is 0.066 ng/mL.

Approximately 25% of the adults in our sample have serum cotinine levels between 0.015 and 0.025 ng/mL. The largest difference in serum cotinine between any pair of adults in this lower quarter sample is 0.010 ng/mL.

Case Three | Confidence Interval Mean | Serum Cotinine in Non-smokers

Using the context and data from case two, estimate the population mean serum cotinine level with 95% confidence. Show your work. Completely discuss and interpret your test results, as indicated in class and case study summaries.

n       m          sd         se       Z     lower95    upper95

45    0.14602    0.26759    0.039891    2    0.066241    0.22580

Numbers

 

n = 45

m = 0.14602

se = sd/sqrt(n) = 0.26759/sqrt(45) = 0.039891

From 2.00  0.022750  0.95450, Z≈2.00

lower95 = m – (Z*sd/sqrt(n)) ≈ 0.14602 – (2*0.039891) = 0.066241     

upper95 = m + (Z* sd/sqrt(n)) ≈ 0.14602 + (2*0.039891) = 0.22580

 

Interpretation

We estimate the population mean serum cotinine for US resident adult non-smokers.

Each member of the family of samples is a single random sample of 45 adults. The family of samples consists of possible samples of this type.

From each member sample, compute:

m = sample mean serum cotinine, sd  = sample standard deviation, se = sd/sqrt(n) and the interval as [lower95 = m – (Z*sd/sqrt(n)), upper95 = m + (Z* sd/sqrt(n))]. Doing this for each member of the family of samples yields a family of intervals, approximately 95% of which capture the population mean serum cotinine of US resident adult non-smokers. If our interval is in this 95% supermajority, then the population mean serum cotinine for US Resident adult non-smokers is between 0.066241 and 0.22580 ng/mL.

 

Case Four | Median Test | Green Lynx Spiders

The green lynx spider, Peucetia viridans, is a conspicuous, large, bright green spider found on many kinds of shrub-like plants throughout the southern United States and is the largest North American lynx spider. Although it is common throughout Florida and aggressively attacks its insect prey, it very seldom bites humans. Lynx spiders get their name from the way that they sometimes pounce on their prey in a catlike fashion.

 

A random sample of green lynx spiders (female) yields the following body lengths (excluding legs), in millimeters per spider: 

9 9 10 10 10 11 12 12 12 12 13 13 14 14 14 15 15 15 16 16 17 17 18 18 19 20 21 21 22 23

 

Test the following: null (H0): The median body length for female Green Lynx spiders is

18 mm (h = 18) against the alternative (H1): h ≠ 18. Completely discuss and interpret your test results, as indicated in class and case study summaries. Show all work and detail for full credit.

 

Numbers

 

9 9 10 10 10 | 11 12 12 12 12 | 13 13 14 14 14 | 15 15 15 16 16 | 17 17 || 18 18 || 19 20 21 21 22 | 23

 

n = 30

Sample Count Below 18: 22

Sample Count Above 18: 4

Sample Error for Two-Sided Test = Max(22, 4) = 22

From the row 30 22 0.00806, base p-value = 0.00806

p-value for two-sided test = 2*0.00806 = 0.01612

 

Interpretation

 

Our test concerns the median Green Lynx body length (female).

 

Each member of the family of samples is a single random sample of 30 female Green Lynx spiders. The family of samples consists of all possible samples of this type.

 

From each member of the family of samples, compute the sample error as the maximum of the number of spiders in the sample whose body lengths are strictly less than 18 millimeters, and the number of spiders in the sample whose spider legths are strictly larger than 18 millimeters. Computing this error for each member sample yields a family of errors.

 

If the true population Green Lynx spider body length (female) is 18 millimeters, then approximately 1.612% of the samples yield errors equal to or more extreme than our sample. The sample presents significant evidence against the null hypothesis.

 

 

Table: 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   0.146860   0.70628

      1.10   0.135670   0.72867

1.15  0.125070  0.74986

1.20  0.115070  0.76986

1.25  0.105650  0.78870

1.30  0.096800  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  0.0034670 0.99307

2.75  0.0029798 0.99404

2.80  0.0025551 0.99489

2.85  0.0021860 0.99563

2.90  0.0018658 0.99627

2.95  0.0015889 0.99682

3.00  0.0013499 0.99730

 

 

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

Table: Medians

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 25 0.00016

30 26 0.00003

30 27 <0.00001

30 28 <0.00001

30 29 <0.00001

30 30 <0.00001