Key | Third Hourly | Math 1107| Spring Semester 2011

 

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 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. Do not share information with any other students during this hourly. Neither give nor receive 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 and Your Tool-sheet. Then hand all of this in to me.

 

Before you begin work, sign and acknowledge

 

I agree to follow this protocol. ______ (ÜInitial Here)

 

 

________________________________________________________________________

Name (Print Clearly)                                       Signature                                         Date

 

Case One | Confidence Interval, Mean | Duchenne Muscular Dystrophy

Duchenne muscular dystrophy (DMD) is an inherited disorder characterized by rapidly progressive muscle weakness which starts in the legs and pelvis and later affects the whole body. Suppose that we follow individuals diagnosed with DMD from diagnosis until death, noting age at death in months. Consider a random sample of individuals who were diagnosed with, and died with DMD. Age at death in months follows below:

17 22 37 45 50 65 73 87 93 102  112 123 127 130 137 138 143 150 156 161 169 173 177 179 180 181 181 182 184 185 186 186 188 189 190 190 190 192 193 193 194 194 195 196 196 197 197 199 199 200 200 201 203 204 207 210 213 215 217 219 220 223 227 228 230 231 233 234 235 237 239 240 240 241

Estimate the population mean age at death time for Duchenne muscular dystrophy patients with 97% confidence. That is, compute and discuss a 97% confidence interval for this population mean. Provide concise and complete details and discussion as demonstrated in the case study summaries.

 

Numbers

 

n       m          sd      Lower97    Upper97

74    175.270    55.4837    161.081    189.460

 

From the means/proportions table, row 2.20 0.013903 0.97219, Z ≈ 2.20.

Lower Bound = m – (Z*sd/sqrt(n))  ≈ 175.27 – (2.2*(55.4837/sqrt(74)) ≈ 161.081

Upper Bound = m + (Z*sd/sqrt(n))  ≈ 175.27 + (2.2*(55.4837/sqrt(74)) ≈ 189.460

 

Report the interval as [161.0, 189.4]

 

Interpretation

 

We estimate the population age at death for patients with Duchenne Muscular Dystrophy.

 

Each member of the family of samples is a single random sample of 74 deceased patients with DMD – the family consists of all possible samples of this type.

 

From each member sample, compute the sample mean m and sample standard deviation sd and then an interval as:

 

Lower Bound = m – (2.2*sd/sqrt(n))

Upper Bound = m + (2.2*sd/sqrt(n))

 

Doing this for each member of the family of samples yields a family of intervals, approximately 97% of which capture the true population mean age at death for patients with DMD. If our intervals in this supermajority, then the population mean age at death for patients with Duchenne Muuscular Dystrophy is between 161.1 and 189.4 months.

Case Two | Hypothesis Test, Median | Sea Weaselsä

Sea Weaselsä are instant pets, distributed as a kit - Sea Weasel Eggsä , Sea Weasel Water Conditionerä and Sea Weasel Foodä . A Sea Weaselä Kit is started by placing the Sea Weasel Eggsä , Sea Weasel Water Conditionerä and some water in a container. The eggs will then hatch, producing Sea Weaselsä , and some of them will survive. A random sample of Sea Weaselä Kits is selected, each selected kit is started, and the number of surviving Sea Weaselsä is counted one week after start. Here are the Sea Weaselä counts per kit (kit yields)

27 32 41 47 48 62 65 73 89 91 96 98 104 109 117 124 132 135 140 151 160 165 166 174 176 181 182 183 187 275

 

Test the following: null (H0): The median Sea Weaselä kit yield is 100 (h = 100) against the alternative (H1): h ≠ 100.  Show your work. Completely discuss and interpret your test results, as indicated in class and case study summaries.

 

Numbers

 

27 32 41 47 48 | 62 65 73 89 91 | 96 98 104 109 117 | 124 132 135 140 151 | 160 165 166 174 176 | 181 182 183 187 275

 

n = 30

Sample Count Below 100: 12

Sample Count Above 100: 18

Sample Error for Two-Sided Test = Max(12, 18) = 18

From the row 30 18 0.18080, base p-value = 0.18080

p-value for two-sided test = 2*0.18080 = 0.3616

 

Interpretation

 

Our test concerns the median Sea Weasel kit yield.

 

Each member of the family of samples is a single random sample of 30 kits. 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 kits in the sample whose yields are strictly less than 100 weasels, and the number of kits in the sample whose yields are strictly larger than 100. Computing this error for each member sample yields a family of errors.

 

If the true population median kit yield for the Sea Weasel kits is 100 weasels, then approximately 36.16% of the samples yield errors equal to or more extreme than our sample. The sample does not appear to present significant evidence against the null hypothesis.

 

Case Three | Confidence Interval, Proportion | Sea Weaselsä

 

Using the context and data from case two, consider the proportion of Sea Weaselä kits yielding 100 or more Sea Weaselsä. 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.

 

27 32 41 47 48 | 62 65 73 89 91 | 96 98 104 109 117 | 124 132 135 140 151 | 160 165 166 174 176 | 181 182 183 187 275

n = 30

Sample Count at or Above 100: 18

Sample Event Proportion: p = 18/30 » 0.60

Sample Standard Error for Proportion: sqrt(p*(1 – p)/n) = sqrt((18/30)*(12/30)/30) » 0.089443 

From the row 2.00 0.022750 0.95450 , Z = 2.00

Lower95 = p – (Z*sdp) = 0.60 – (2*0.089443) » 0.421115

Upper95 = p + (Z*sdp) = 0.60 + (2*0.089443) » 0.778885

 

Interpretation

 

We estimate the population proportion of Sea Weasel kits yielding 100 or more Sea Weasels.

 

Each member of the family of samples is a single random sample of 30 kits. The family of samples consists of all possible samples of this type.

 

From each member sample, compute the number e≥100 of kits in the sample whose yields equal or exceed 100 Sea Weasels, then compute an interval:

 

p≥100 = e≥100/30

sdp≥100 = sqrt(p≥100*(1 – p≥100)/30)

lower95 = p≥100 – (2*sdp≥100)

upper95 = p≥100 + (2*sdp≥100)

 

Computing these intervals for each member of the family of samples yields a family of intervals, approximately 95% of which capture the true population proportion of Sea Weasel kits yielding 100 or more Sea Weasels.

If our interval is in the supermajority, then between 42.1% and 77.8% of Sea Weasel kits yield 100 or more Sea Weasels. 

Case Four | Categorical Goodness of Fit | A Color Bowl

We have a color bowl, with black, blue, green, red and yellow marbles. Three hundred and fifty draws with replacement from the bowl yield the following sample counts:

50 black, 33 blue, 60 green, 74 red and 133 yellow.

Our null hypothesis is that the color bowl has the following color distribution: 25% Black 5% Blue, 10% Green, 20% Red and 40% Yellow. Test this Hypothesis. 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. Show all work and detail for full credit.

 

Numbers

n = 50 + 33 + 60 + 74 + 133 = 350

EBlack = 350*PBlack = 350*0.25 = 87.5

ErrorBlack = (nBlack – EBlack )2/ EBlack = (50 – 87.5 )2/ 87.5 » 16.07

 

EBlue = 350*PBlue = 350*0.05 = 17.5

ErrorBlue = (nBlue – EBlue )2/ EBlue = (33 – 17.5 )2/ 17.5 » 13.72

 

EGreen = 350*PGreen = 350*0.10 = 35

ErrorGreen = (nGreen – EGreen )2/ EGreen = (60 – 35 )2/ 35 » 17.86

 

ERed = 350*PRed = 350*0.20 = 70

ErrorRed = (nRed – ERed )2/ ERed = (74 – 70 )2/ 70  » 0.23

 

EYellow = 350*PYellow = 350*0.40 = 140

ErrorYellow = (nYellow – EYellow )2/ EYellow = (133 – 140 )2/ 140  » 0.35

 

ErrorTotal = ErrorBlack + ErrorBlue + ErrorGreen + ErrorRed + ErrorYellow =

16.07 + 13.72 + 17.86 + 0.23 + 0.35 »  48.24 over 5 Categories

From the table row 5 13.2767 0.010, p < .01 

Interpretation

Each member of the family of samples is a single random sample of 350 draws. The family of samples consists of all possible samples of this type.

From each member sample, compute the color counts: nBlack , nBlue , nGreen , nRed and nYellow. Under the null hypothesis, we expect EBlack = 350*PBlack = 350*0.25 = 87.5 black, EBlue = 350*PBlue = 350*0.05 = 17.5 blue, EGreen = 350*PGreen = 350*0.10 = 35 green, ERed = 350*PRed = 350*0.20 = 70 red and EYellow = 350*PYellow = 350*0.40 = 140 yellow.

 

Compute a sample error as

 

ErrorTotal =

{(nBlack – EBlack )2/ EBlack} + {(nBlue – EBlue )2/ EBlue} + {(nGreen – EGreen )2/ EGreen} + {(nRed – ERed )2/ ERed} + {(nYellow – EYellow )2/ EYellow}.

 

Computing this error for each member sample yields a family of errors. If the true color distribution for the population is 25% Black 5% Blue, 10% Green, 20% Red and 40% Yellow, then fewer than 1% of the member samples yield errors equal to or larger than ours. The sample appears to present highly 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 0.14686 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: 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 <0.00001

30 28 <0.00001

30 29 <0.00001

30 30 <0.00001

 

Table: 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