Key

Comprehensive Final Examination: Wednesday Version

Math 1107

Spring Semester 2008

 

Protocol

 

You will use only the following resources: Your individual calculator; Your individual tool-sheets (two (2) 8.5 by 11 inch sheet); 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 hourly.

 

Sign and Acknowledge:  I agree to follow this protocol.

 

 

Name (PRINTED)                                          Signature                                          Date

Case One | Conditional Probability | Color Slot Machine

Here is our slot machine – on each trial, it produces a 10-color sequence, using the table below:

 

Sequence*

Probability

RRBBRRYRRR

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

1.00

*B-Blue, G-Green, R-Red, Y-Yellow, Sequence is numbered from left to right: (1st 2nd 3rd 4th 5th6th7th 8th 9th 10th )

Compute the following conditional probabilities:

 

Pr{ Blue Shows Twice | Yellow Shows}

Pr{ Green Shows Strictly More Than Once | “BR” Shows }

Pr{ Red Shows | Yellow Shows}

 

Pr{ Blue Shows Twice | Yellow Shows}= Pr{ Blue Shows Twice and Yellow Shows}/ Pr{Yellow Shows}

 

 

Sequence*

Probability

RRBBRRYRRR

.10

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

.40

 

Pr{ Blue Shows Twice and Yellow Shows} =

Pr{One of RRBBRRYRRR , RRYYGRRBBY or YYGBYYBGRR Shows} =

Pr{RRBBRRYRRR} + Pr{RRYYGRRBBY} + Pr{YYGBYYBGRR Shows} = .10+.10+.20 = .40

 

 

 

Sequence*

Probability

RRBBRRYRRR

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

.90

 

Pr{Yellow Shows} = Pr{One of RRBBRRYRRR, BBYYGGYGBR, GRRGGYBRGG, GYGYRYGYY, RRYYGRRBBY or YYGBYYBGRR shows} = Pr{RRBBRRYRRR } + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} +  Pr{GYGYRYGYY} + Pr{RRYYGRRBBY} + Pr{ YYGBYYBGRR} = .10 + .15 + .10 + .25 + .10 + .20 = .90

Pr{ Blue Shows Twice | Yellow Shows}= Pr{ Blue Shows Twice and Yellow Shows}/ Pr{Yellow Shows} = .20/.90 @ .2222

 

Pr{Green Shows Strictly More Than Once | “BR” Shows } = Pr{ Green Shows Strictly More Than Once and “BR” Shows }/ Pr{ “BR” Shows }

 

 

 

Sequence*

Probability

RRBBRRYRRR

.10

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

Total

.45

 

Pr{ “BR” Shows } = Pr{One of RRBBRRYRRR, RRGGRGBRRB, BBYYGGYGBR or GRRGGYBRGG Shows} = Pr{RRBBRRYRRR} + Pr{RRGGRGBRRB} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} = .10 + .10 + .15 + .10 = .45

 

Sequence*

Probability

RRGGRGBRRB

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

Total

.35

 

Pr{ Green Shows Strictly More Than Once and “BR” Shows } = Pr{One of RRGGRGBRRB, BBYYGGYGBR or GRRGGYBRGG Shows} = Pr{RRGGRGBRRB} + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} =.10 + .15 + .10 = .35

Pr{Green Shows Strictly More Than Once | “BR” Shows } = Pr{ Green Shows Strictly More Than Once and “BR” Shows }/ Pr{ “BR” Shows } = .35/.45 = 7/9 @ .7778

 

Pr{ Red Shows | Yellow Shows}

 

 

Sequence*

Probability

RRBBRRYRRR

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

.90

 

Pr{Yellow Shows} = Pr{One of RRBBRRYRRR, BBYYGGYGBR, GRRGGYBRGG, GYGYRYGYY, RRYYGRRBBY or YYGBYYBGRR shows} = Pr{RRBBRRYRRR } + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} +  Pr{GYGYRYGYY} + Pr{RRYYGRRBBY} + Pr{ YYGBYYBGRR} = .10 + .15 + .10 + .25 + .10 + .20 = .90

 

Sequence*

Probability

RRBBRRYRRR

.10

BBYYGGYGBR

.15

GRRGGYBRGG

.10

BGYGYRYGYY

.25

RRYYGRRBBY

.10

YYGBYYBGRR

.20

Total

.90

 

Pr{ Red and Yellow Show} = Pr{One of RRBBRRYRRR, BBYYGGYGBR, GRRGGYBRGG, GYGYRYGYY, RRYYGRRBBY or YYGBYYBGRR shows} = Pr{RRBBRRYRRR } + Pr{BBYYGGYGBR} + Pr{GRRGGYBRGG} +  Pr{GYGYRYGYY} + Pr{RRYYGRRBBY} + Pr{ YYGBYYBGRR} = .10 + .15 + .10 + .25 + .10 + .20 = .90

Pr{ Red Shows | Yellow Shows} = Pr{ Red and Yellow Show} / Pr{Yellow Shows} = .90/.90 = 1

 

Case Two | Summary Intervals | 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):

 

25,26, 27, 29, 30, 32, 33, 34, 34, 35, 35, 36, 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, 40, 40, 40, 40, 40, 40, 40, 40, 40, 40, 41, 41, 41, 42, 42, 42, 43, 43

 

Let m denote the sample mean, and sd the sample standard deviation. Compute and interpret the intervals m±2sd and m±3sd, using Tchebysheff’s Inequalities and the Empirical Rule. Be specific and complete. Show your work, and discuss completely for full credit.

 

n       m          sd      Lower2sd    Upper2sd    Lower3sd    Upper3sd

56    37.3393    3.92788     29.4835     45.1950     25.5556     49.1229

 

There are 56 year 2005 US resident live born infants in the sample.

 

n @ 56

m @ 37.3393

sd @ 3.92788

 

m-2*sd @ 37.3393-2*3.92788 @ 29.48    

m+2*sd @ 37.3393+2*3.92788 @ 45.19   

    

m-3*sd @ 37.3393-3*3.92788 @ 25.56     

m+3*sd @ 37.3393+3*3.92788 @ 49.12    

 

We have a random sample of 56 year 2005 US resident live born infants in the sample.      

 

At least 75% of the year 2005 US resident live born infants in our sample have gestational ages between 29.48 and 45.19 weeks.

At least 89% of the year 2005 US resident live born infants in our sample have gestational ages between 25.56 and 49.12 weeks.

If the year 2005 US resident live born infant gestational ages cluster symmetrically around a central value, with extreme values on either side of the central value becoming increasingly rare, then:

Approximately 95% of the year 2005 US resident live born infants in our sample have gestational ages between 29.48 and 45.19 weeks.

Approximately 100% of the year 2005 US resident live born infants in our sample have gestational ages between 25.56 and 49.12 weeks.

 

Case Three | Confidence Interval for Proportion | Gestational Age

Consider the proportion of Year 2005 US Resident Live Births that are “Full Term,” that is births with  [37,40] weeks of gestation at birth. Using the data from Case Two, compute and interpret a 98% confidence interval for this population proportion.

Numbers

 

From 2.35 0.009387 0.98123, z=2.35

n = 56

e = 36

p = 36/56 @ 0.64286

sdp = sqrt(p*(1-p)/n) = sqrt((36/56)*(20/56)/56) @ 0.064030

 

lowCI = p − z*sdp = 0.64286 − 2.35*0.064030 @ 0.49239

highCI = p + z*sdp = 0.64286 + 2.35*0.064030 @ 0.79333

 

Report the interval as [.492, .793 ].

 

Interpretation

 

Our population is the population of year 2005 US resident live born infants and our population mean is the mean gestational age (weeks). Our event is that the live born infant was born with between 37 and 40 weeks of gestation.

Our Family of Samples (FoS) consists of every possible random sample of 56 year 2005 US resident live born infants. From each individual sampled live born infant, gestational age in weeks is obtained.

From each member sample of the FoS, we compute the sample proportion p of infants in the sample with between 37 and 40 weeks of gestation at birth and sdp, where sdp=sqrt(p*(1-p)/56), and then compute the interval

[p – 2.35*sdp, p + 2.35*sdp].

Computing this interval for each member sample of the FoS, we obtain a Family of Intervals (FoI), approximately 98% of which cover the true population proportion of year 2005 US resident live born infants with between 37 and 40 weeks of gestation.

If our interval, [.492, .793] is among the approximate 98% super-majority of intervals that cover the population mean, then between 49.2% and 79.3% of year 2005 US resident live born infants have gestation ages between 37 and 40 weeks.

 

Case Four | Hypothesis Test for Median | Green Lynx Spiders

 

A random sample of male green lynx spiders yields the following lengths, in millimeters per spider:

5.20, 4.70, 5.70, 5.65, 5.75, 4.70, 4.80, 6.20, 5.50, 5.95 5.75, 5.95, 5.40, 5.65, 5.90

7.50, 5.20, 6.20, 5.85, 7.00 6.45, 6.35, 5.85, 5.75, 6.10, 6.55, 6.95, 6.80, 6.35, 5.80

 

Test the following: null (H0): The median length (in mm) for male green lynx spiders is 6 (h = 6) against the alternative (H1): h  < 6.  Show your work. Completely discuss and interpret your test results, as indicated in class and case study summaries.

 

Numbers

 

Error Form = “Guess is too large”

 

5.20, 4.70, 5.70, 5.65, 5.75 | 4.70, 4.80, 6.20, 5.50, 5.95 | 5.75, 5.95, 5.40, 5.65, 5.90

7.50, 5.20, 6.20, 5.85, 7.00 | 6.45, 6.35, 5.85, 5.75, 6.10 | 6.55, 6.95, 6.80, 6.35, 5.80

 

Error = 19

n=30

 

Our error has the form Error = Number of male green lynx spiders with length strictly less than 6 mm}. Our computed error is 7, computed from a random sample of 30 male green lynx spiders. Using the row 30 19 0.10024, the computed p-value is 0.10024 or approximately 10.0%.

 

Interpretation

Our population is the population of male green lynx spiders.

Our Family of Samples (FoS) consists of every possible random sample of 30 male green lynx spiders.

From each member sample of the FoS, we compute Error = Number of male green lynx spiders with length less than 6 mm. Computing this error for each member sample of the FoS, we obtain a Family of Errors (FoE).

If the true population median length in millimeters for male green lynx spiders is 6, then approximately 10.0% of the Family of Samples yield errors as bad as or worse than our single error. The sample does not appear to present significant evidence against the null hypothesis (when the alternative is to choose a smaller guess).

Case Three | Confidence Interval, Proportion | C-reactive Protein

 

 

 

 

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

 

 

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 0.00000

30 28 0.00000

30 29 0.00000

30 30 0.00000