2nd Hourly

Key

2^nd Hourly

Math 1107

Summer Term 2009

Protocol

You will use only the following resources: Your individual calculator; Your individual tool-sheet (one (1) 8.5 by 11 inch sheet); Your writing utensils; Blank Paper (provided by me); This copy of the hourly.

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 six. 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. Do not share information with any other students during this hourly.

When you are finished:

Prepare a Cover Sheet: Print your name on an otherwise blank sheet of paper. Then stack your stuff as follows:

Cover Sheet (Top)

Your Work Sheets

The Test Papers

Your Toolsheet

Then hand all of this in to me.

Sign and Acknowledge: I agree to follow this protocol.

________________________________________________________________________

Name (PRINTED) Signature Date

Work all six (6) cases.

Case One | Descriptive Statistics | Traumatic Brain Injury

Traumatic Brain Injury (TBI) involves the injury of the brain when it involves sudden or intense physical force resulting in the presence of Concussion, Skull Fracture, or Bleeding and Tissue Damage (Contusions, Lacerations, Hemorrhaging) involving the brain. A random sample of TBI cases is acquired, and the age (in years) of the case at injury is determined. The ages at injury are listed below:

4, 5, 5, 6, 6, 7, 8, 9, 10, 12, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 20, 20, 20, 21, 21, 22, 23, 25, 27, 27, 30, 30, 30, 31, 32, 32, 33, 35, 35, 37, 38, 38, 39, 39, 40, 40, 41, 41, 42, 42, 45, 47, 50, 52, 60,63, 65, 70, 70, 71, 71, 72, 72, 73, 73, 74, 74, 75, 75, 75, 76, 76, 77, 79, 80, 81, 89, 105

Compute and interpret the following statistics for age at injury: sample size, p00, p25, p50, p75, p100, (p100-p50), (p75-p25), (p50-p00).

Numbers

n p0 p25 p50 p75 p100 range20 range31 range42

78 4 19 36 70 105 32 51 69

range20 = p50 – p0 = 36 – 4 = 32

range31 = p75 – p25 = 70 – 19 = 51

range42 = p100 – p50 = 105 – 36 = 69

Interpretation

There are 78 Traumatic Brain Injury(TBI) cases in the sample.

The TBI case in the sample with the earliest injury acquired their injury at 4 years of age.

Approximately 25% of the TBI cases in the sample acquired their TBIs at 19 years of age or younger.

Approximately 50% of the TBI cases in the sample acquired their TBIs at 36 years of age or younger.

Approximately 75% of the TBI cases in the sample acquired their TBIs at 70 years of age or younger.

The TBI case in the sample with the latest injury acquired their injury at 105 years of age.

Approximately 50% of the TBI cases in the sample were injured between ages 4 and 36 years. The largest possible difference in age at injury for any pair of cases in the lower half-sample is 32 years.

Approximately 50% of the TBI cases in the sample were injured between ages 19 and 70 years. The largest possible difference in age at injury for any pair of cases in the middle half-sample is 51 years.

Approximately 50% of the TBI cases in the sample were injured between ages 70 and 105 years. The largest possible difference in age at injury for any pair of cases in the upper half-sample is 69 years.

Case Two | Confidence Interval: Population Proportion | 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 yields the following body lengths (excluding legs), in millimeters per spider:

3.2, 4.7, 5.5, 5.7, 5.8, 5.9, 6.1, 6.2, 6.2, 6.2, 6.3, 6.4, 6.5, 6.8, 7.0, 7.2, 7.2, 7.3, 7.4, 7.8, 8.3, 8.5, 8.8, 9.8, 10.1, 10.1, 10.2, 10.4, 10.5, 10.5, 10.5, 10.6, 10.7, 10.9, 11.2, 11.5, 12.1, 12.1, 12.3, 12.4, 12.4, 12.5, 12.6, 12.6, 12.7, 12.8, 12.9, 13.1, 13.2, 13.5, 13.6, 14.2, 14.2, 15.0, 15.5, 16.4, 16.9, 17.1, 19.2, 21.2, 21.9, 22.6, 22.9, 31.3, 32.7, 55.0, 57.1, 60.0

Consider the proportion of green lynx spiders with body lengths of 15mm or longer. Compute and interpret a 95% confidence interval for this population proportion. Show your work. Completely discuss and interpret your test results, as indicated in class and case study summaries.

Numbers

Event = “Spider Body Length 15mm or Longer”

e = sample event count = 15

n = sample size = 68

p = sample event proportion = 15/68 » 0.2206

sdp = sample standard error for p = sqrt(p*(1 – p)/n) = sqrt((15/68)*(53/68)/68) » 0.0503

from 2.00 0.022750 0.95450, z = 2.0

lowerp = p – 2*sdp » (15/68) – 2*0.05283 » 0.12002 » 0.12

upperp = p + 2*sdp » (15/68) + 2*0.05283 » 0.32115 » 0.32

Interpretation

We estimate the population proportion of Green Lynx spiders whose body lengths equal or exceed 15 millimeters with approximate 95% confidence.

Each member of the family of samples (FoS) is a single random sample of 68 Green Lynx spiders. The FoS consists of all possible random samples of this type.

From each member sample, compute the number e of sample spiders whose body lengths equal or exceed 15 millimeters, p = sample event proportion = e/68, sdp = sample standard error for p = sqrt(p*(1 – p)/n) and finally the interval [lowerp = p – 2*sdp,

upperp = p + 2*sdp]. Repeating these calculations for member of the FoS yields a family of intervals, approximately 95% of which cover P₁₅₊, the population proportion of Green Lynx spiders whose body lengths equal or exceed 15 millimeters.

If our interval resides in this supermajority, then between 12.0% and 32.1% of Green Lynx spiders have body lengths equaling or exceeding 15 millimeters.

Case Three | Hypothesis Test – Categorical Goodness-of-Fit | Human Blood Type

In the early 20th century, an Austrian scientist named Karl Landsteiner classified blood according to chemical molecular differences – surface antigen proteins. He was awarded the Nobel Prize for his achievements. Landsteiner observed two distinct chemical molecules present on the surface of the red blood cells. He labeled one molecule "A" and the other molecule "B." If the red blood cell had only "A" molecules on it, that blood was called type A. If the red blood cell had only "B" molecules on it, that blood was called type B. If the red blood cell had a mixture of both molecules, that blood was called type AB. If the red blood cell had neither molecule, that blood was called type O. So the blood types are O, A, B, AB. Suppose that we have a random sample of US human blood donors, whose blood is typed as follows:

O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, O, A, A, A, A, A, A, A, A, A, A, A, A, A, A, A, A, A, A, A, A, A, A, B, B, B, B, B, B, B, B, B, AB, AB, AB, AB, AB

Test the null hypothesis that US blood types are distributed as 46% O, 41% A, 9% B and 4% AB. 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.

Numbers

type o n p e error cumerr

A 22 68 0.41 27.88 1.24011 1.24011

AB 5 68 0.04 2.72 1.91118 3.15129

B 9 68 0.09 6.12 1.35529 4.50659

O 32 68 0.46 31.28 0.01657 4.52316

observed_O = 32

expected_O = n*Pr_O =68*.46 » 31.28

error_O = (observed_O - expected_O )²/ expected_O» (32 – 31.28 )²/ 31.28 » 0.01657

observed_A = 22

expected_A = n*Pr_A=68*.41 » 27.88

error_A = (observed_A – expected_A )²/ expected_A» (22 – 27.88 )²/ 27.88 » 1.24011

observed_B = 9

expected_B = n*Pr_B =68*.09 » 6.12

error_B = (observed_B – expected_B )²/ expected_B» (9 – 6.12 )²/ 6.12 » 1.33529

observed_AB = 5

expected_AB = n*Pr_AB =68*.04 » 2.72

error_AB = (observed_AB – expected_AB )²/ expected_AB» (5 – 2.72 )²/ 2.72 » 1.91118

total sample error = ((observed_O - expected_O )²/ expected_O) + ((observed_A – expected_A )²/ expected_A ) +

( (observed_B – expected_B )²/ expected_B) + ((observed_AB – expected_AB )²/ expected_AB)

» ((32 – 31.28 )²/ 31.28) + ((22 – 27.88 )²/ 27.88) + ((9 – 6.12 )²/ 6.12) + ((5 – 2.72 )²/ 2.72)

» 0.01657 + 1.24011 + 1.33529 + 1.91118

» 4.52316

from rows 4 4.1083 0.250 and 4 4.6416 0.200, p is betweeb .20 and .25

Interpretation

Each member of the family of samples is a single random sample of 68 human sunjects. The FoS consists of all possible samples of this type.

Under the null hypothesis, we expect the following sample counts in random samples of 68 subjects:

expected_O = n*Pr_O =68*.46 » 31.28

expected_A = n*Pr_A=68*.41 » 27.88

expected_B = n*Pr_B =68*.09 » 6.12

expected_AB = n*Pr_AB =68*.04 » 2.72

From each member of the FoS, compute

observed_O

expected_O = n*Pr_O =68*.46 » 31.28

error_O = (observed_O - expected_O )²/ expected_O

observed_A

expected_A = n*Pr_A=68*.41 » 27.88

error_A = (observed_A – expected_A )²/ expected_A

observed_B

expected_B = n*Pr_B =68*.09 » 6.12

error_B = (observed_B – expected_B )²/ expected_B

observed_AB

expected_AB = n*Pr_AB =68*.04 » 2.72

error_AB = (observed_AB – expected_AB )²/ expected_AB

and the total sample error = ((observed_O - expected_O )²/ expected_O) + ((observed_A – expected_A )²/ expected_A) + ( (observed_B – expected_B )²/ expected_B) + ((observed_AB – expected_AB )²/ expected_AB).

Repeating these calculations for each member of the FoS yields a family of errors(FoE). If the true population proportions for US human blood types is .46(O), .41(A), .09(B) and .04(AB), then between 20% and 25% of the sample errors equal or exceed our error. The sample does not appear to present significant evidence against the null hypothesis.

Case Four | Summary Intervals | Green Lynx Spiders

5.9, 6.1, 6.2, 6.2, 6.2, 6.3, 6.4, 6.6, 7.0, 7.2, 7.3, 7.6, 8.4, 8.8, 10.1, 10.2, 10.4, 10.5, 10.5, 10.5, 10.6, 10.7, 10.9, 11.2, 11.5, 12.1, 12.6, 12.6, 12.7, 12.8, 12.9, 13.3, 13.6, 14.2, 14.2, 14.5, 15.5, 16.4, 16.9, 17.1, 19.2, 21.2, 21.9, 22.6, 22.9

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

Numbers

n m sd lower2 upper2 lower3 upper3

45 11.8333 4.69390 2.44553 21.2211 -2.24838 25.9150

lower2=m – 2*sd = 11.8333 – 2*4.69390 » 2.44553

upper2=m + 2*sd = 11.8333 + 2*4.69390 » 21.2211

lower3=m – 3*sd = 11.8333 – 3*4.69390 » -2.24838

upper3=m + 3*sd = 11.8333 + 3*4.69390 » 25.9150

Interpretation

At least 75% of the Green Lynx spiders in the sample have body lengths between 2.4 and 21.1 millimeters.

At least 89% of the Green Lynx spiders in the sample have body lengths between 0 and 25.9 millimeters.

If the Green Lynx spider body lengths cluster symmetrically around a central value, becoming rare as departures from the central value increase, then:

Approxiamtely 95% of the Green Lynx spiders in the sample have body lengths between 2.4 and 21.1 millimeters.

Approximately 100% of the Green Lynx spiders in the sample have body lengths between 0 and 25.9 millimeters.

Case Five | Hypothesis Test – Population Median | Fictitious Spotted Lizard

The Fictitious Spotted Lizard is a native species of Lizard Island, and is noteworthy for the both the quantity and quality of its spots. Consider a random sample of Fictitious Spotted Lizards, in which the number of spots per lizard is noted:

8, 11, 13, 14, 15, 15, 16, 16, 17, 19, 19, 20, 20, 20, 20,

21, 21, 22, 22, 22, 22, 26, 26, 26, 27, 24, 24, 24, 25, 27

Test the following: null (H₀): The median number of spots per Fictitious Spotted Lizard is 22 (h = 22) against the alternative (H₁): h > 22. 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.

Numbers

Test the following: null (H₀): The median number of spots per Fictitious Spotted Lizard is 22 (h = 22) against the alternative (H₁): h > 22.

Alternative: “Guess is too small”

Error Form: “Count Above 22”

8, 11, 13, 14, 15 | 15, 16, 16, 17, 19 | 19, 20, 20, 20, 20,

21, 21, 22, 22, 22 | 22, 26, 26, 26, 27 | 24, 24, 24, 25, 27

sample size = n = 30

sample error = 9

from 30 9 0.99194 , p » 0.99194

Interpretation

Each member of the family of sample (FoS) is a single random sample of 30 Fictitious Spotted lizards. The FoS consists of all possible samples of this type.

From each member sample, compute the sample number of lizards with strictly more than 22 spots. Computing this error for each member sample of the FoS yields a family of errors (FoE). If the true population median spot count for Fictitious Spotted lizards is 22 spots, then approximately 99.1% of the errors equal or exceed our error. The sample does not appear to present significant evidence against the null hypothesis, if the alternative is a higher guess.

Case Six | Confidence Interval: Population Mean | Fictitious Stress Index

The Fictitious Stress Index (FSI) is a measure of a person's stress level. The FSI ranges from 0 to 70. A researcher acquires a random sample of undergraduates at Kennesaw State University, and administers the FSI to these students at the beginning of Fall Semester 2008. The sample FSI scores follow:

10, 10, 12, 13, 14, 15, 16, 16, 17, 18, 19, 20, 21, 22, 22, 23, 24, 24, 25, 26, 26, 27, 29, 29, 30, 32, 32, 33, 34, 36, 37, 38, 39, 40, 40, 40, 40, 41, 41, 42, 42, 42, 43, 43, 44, 44, 44, 44, 44, 45, 45, 45, 46, 66, 68, 70

Estimate the population mean FSI with 97% confidence. That is, compute and discuss a 97% confidence interval for this population mean. Show your work. Fully discuss the results.

Numbers

n m sd se z lower97 upper97

56 33 13.9088 1.85864 2.2 28.9110 37.0890

sample standard error for the mean = se = sd/sqrt(n) » 13.9088/sqrt(56) » 1.85864

from 2.20 0.013903 0.97219, z = 2.2

lower97 = m – 2.2*se » 33 – 2.2*1.85864 » 28.9110

upper97 = m + 2.2*se » 33 + 2.2*1.85864 » 37.0890

Interpretation

We estimate the population mean Fictitious Stress Index (FSI) level for undergraduates enrolled at Kennesaw State University during Fall Semester 2008 with approximate 97% confidence.

Each member of the family of samples (FoS) is a single random sample of 56 undergraduates enrolled at Kennesaw State University during Fall Semester 2008.. The FoS consists of all possible random samples of this type.

From each member sample, compute the mean FSI level, sd = sample standard deviation, the sample standard error for the mean se = sd/sqrt(n) and finally the interval [lower = m – 2.2*se,

upper = m + 2.2*se], where se=sd/sqrt(n). Repeating these calculations for member of the FoS yields a family of intervals, approximately 97% of which cover m_FSI, the population mean FSI level for undergraduates enrolled at Kennesaw State University during Fall Semester 2008

If our interval resides in this supermajority, then the mean FSI level for undergraduates enrolled at KSU during Fall Semester 2008 is between 28.9 and 37.1.

Work all six (6) cases.

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

Table 3. Categories/Goodness of Fit

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

Categories ERROR p-value

6 0.0000 1.000

6 1.6103 0.900

6 2.3425 0.800

6 2.9999 0.700

6 3.6555 0.600

6 4.3515 0.500

6 4.7278 0.450

6 5.1319 0.400

6 5.5731 0.350

6 6.0644 0.300

6 6.6257 0.250

6 7.2893 0.200

6 7.6763 0.175

6 8.1152 0.150

6 8.6248 0.125

6 9.2364 0.100

6 9.5211 0.090

6 9.8366 0.080

6 10.1910 0.070

6 10.5962 0.060

6 11.0705 0.050

6 11.6443 0.040

6 12.3746 0.030

6 13.3882 0.020

6 15.0863 0.010