Key
The 3rd Hourly
Math 1107
Spring 2009
Protocol
You will use only the
following resources: Your individual calculator; individual tool-sheet (one (1)
8.5 by 11 inch sheet), writing utensils, blank paper
(provided by me) and 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.
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
Case One | Confidence
Interval: Population Mean | The
The objective of the
Framingham Heart Study was to identify the common factors or characteristics
that contribute to Cardiovascular disease (CVD) by following its development
over a long period of time (since 1948)
in a large group of participants who had not yet developed overt
symptoms of CVD or suffered a heart attack or stroke. Blood pressure is a
measurement of the force applied to the walls of the arteries as the heart
pumps blood through the body. Blood pressure readings are measured in
millimeters of mercury (mm Hg) and usually given as 2 numbers. The top
number is the systolic blood pressure reading. It represents the maximum
pressure exerted when the heart contracts. The bottom
number is the diastolic blood pressure reading. It represents the pressure in
the arteries when the heart is at rest. A
sample of FHS adult subjects yields the following readings: (top/bottom)
130/70, 175/75, 136/84, 124/84, 144/88,
154/90, 164/97, 210/120, 110/75, 166/108, 100/70, 172/110, 160/90, 122/84,
162/80, 155/85, 120/65, 128/84, 130/90, 210/110, 110/68, 160/106, 132/72,
120/80, 200/104, 165/105, 132/88, 134/84, 152/74, 138/70, 122/80, 155/90,
166/108, 120/80, 200/130, 121/85, 150/100, 135/75, 140/78, 142/85, 146/94,
185/90, 166/78, 193/106, 160/80, 140/80, 150/110, 140/84, 130/82, 130/80,
238/122, 128/72, 220/118, 165/95, 208/114, 126/80, 140/90, 166/104, 130/70,
131/88
Estimate the population mean systolic
blood pressure with 95% confidence. That is, compute and discuss a
95% confidence interval for this population mean. Show
your work. Fully discuss the results. This discussion must include a clear discussion of
the population and the population mean, the family of samples, the family of
intervals and the interpretation of the interval.
Numbers
n m sd
se Z lower95
upper95
60 150.967
29.6573 3.82875 2
143.309 158.624
n = 60
m » 150.967
sd » 29.6573
se = sd/sqrt(n) = sd/sqrt(60) » 29.6573/sqrt(60) »
3.82875
Z = 2 from the mean/proportion row: 2.00 0.022750 0.95450
lower95 = m –
Z*se » 150.967 – 2*(29.6573/sqrt(60)) »
150.967 – 2*3.82875
»
143.309
upper95 = m +
Z*se » 150.967 + 2*(29.6573/sqrt(60)) »
150.967 + 2*3.82875
»
158.624
Report the interval as [143.3,158.6].
Interpretation
Our population is the population of
Framingham Heart Study subjects and our population mean is the
mean systolic blood pressure (SBP, mm Hg).
Our Family of Samples (FoS) consists of
every possible random sample of 60 Framingham Heart Study subjects.
From each member
sample of the FoS, we compute the sample mean (m) and standard deviation (sd)
for FHS SBP, and then compute the interval
[m – 2.00*( sd/sqrt(n)), m + 2.00*(
sd/sqrt(n))].
Computing this
interval for each member sample of the FoS, we obtain a Family of Intervals (FoI),
approximately 95% of which cover the true population mean systolic blood
pressure for Framingham Heart Study subjects.
If our interval, [143.3,158.6] is among the approximate 95%
super-majority of intervals that cover the population mean, then the true
population mean systolic blood pressure for Framingham Heart Study subjects is between 143.3 and 158.6 mm Hg.
Case Two | Confidence
Interval: Population Proportion | Prenatal Care
A random
sample of Year 2000 Georgia resident live births are checked for prenatal care
status, in the following categories:
|
Prenatal
Care Status |
Number in
Sample |
|
Prenatal
Care Began 1st Trimester (Months 1-3 of Pregnancy) |
410 |
|
Prenatal
Care Began 2nd Trimester (Months 4-6 of Pregnancy) |
52 |
|
Prenatal
Care Began 3rd Trimester (Months 7-9 of Pregnancy) |
12 |
|
No Prenatal
Care |
6 |
|
Prenatal
Care Status Unknown |
20 |
|
Total |
500 |
Compute and
interpret a 99% confidence interval for the population proportion of year 2000
Numbers
n = 500
event = “Prenatal Care Begins in 1st or 2nd
Trimester”
nevent = 410 + 52 = 462
pevent =nevent/n = 462/500 = .924
sdp = sqrt(pevent *(1 – pevent )/ n) »
sqrt(.924*(1 – .924 )/ 500) » .01185
Z = 2.60 from row: 2.60 0.004661 0.99068
lower99 = pevent – Z*sdp » .924 – 2.6*.01185 » .893187
upper99 = pevent + Z*sdp » .924 + 2.6*.01185 » .954812
Report the
interval as [.893, .954].
Interpretation
Our population is the population of
Our Family of Samples (FoS) consists of
every possible random sample of 500 year 2000
From each member
sample of the FoS, we compute the sample proportion p of live births where prenatal care began in the 1st or
2nd trimester, sdp = sqrt(p*(1–p)/n) and then compute the
interval
[p – (2.60*sdp), p + (2.60*sdp)].
Computing this
interval for each member sample of the FoS, we obtain a Family of Intervals (FoI),
approximately 99% of which cover the true population proportion of year 2000
If our interval, [.893, .954] is among the approximate
99% super-majority of intervals that cover the population proportion, then
between 89.3% and 95.4% of year 2000 Georgia resident live births had prenatal
care beginning in the 1st or 2nd trimester.
Case Three |
Hypothesis Test: Population Median | The
Using the
Numbers
The data
from Case One…The bottom number is the diastolic blood pressure reading.
It represents the pressure in the arteries when the heart is at rest. A sample of
FHS adult subjects yields the following readings: (top/bottom)
130/70, 175/75, 136/84,
124/84, 144/88, 154/90, 164/97, 210/120, 110/75, 166/108, 100/70, 172/110,
160/90, 122/84, 162/80, 155/85, 120/65, 128/84, 130/90, 210/110, 110/68,
160/106, 132/72, 120/80, 200/104, 165/105, 132/88, 134/84, 152/74, 138/70,
122/80, 155/90, 166/108, 120/80, 200/130, 121/85, 150/100, 135/75, 140/78,
142/85, 146/94, 185/90, 166/78, 193/106, 160/80, 140/80, 150/110, 140/84,
130/82, 130/80, 238/122, 128/72, 220/118, 165/95, 208/114, 126/80, 140/90,
166/104, 130/70, 131/88
n = 60
Error Form:
“Guess is too small” – Count strictly above the guess:
136/84, 124/84,
144/88, 154/90, 164/97, 210/120, 166/108,
172/110, 160/90, 122/84, 155/85, 128/84, 130/90, 210/110, 160/106, 200/104,
165/105, 132/88, 134/84, 155/90, 166/108, 200/130, 121/85, 150/100, 142/85,
146/94, 185/90, 193/106, 150/110, 140/84, 130/82, 238/122, 220/118, 165/95,
208/114, 140/90, 166/104, 131/88
error =
#FHS subjects in the sample whose diastolic blood pressure strictly exceeds 80
= 38
From the
row: 60 38 0.01367, p-value » 1.367%
Interpretation
Our population
is the population of Framingham Heart Study subjects.
Our Family of
Samples (FoS) consists of every possible random sample of 60 Framingham
Heart Study subjects.
From each member
sample of the FoS, we compute Error = Number of sample subjects whose
diastolic blood pressure strictly exceeds 80 mm Hg. Computing this error
for each member sample of the FoS, we obtain a Family of Errors (FoE).
If the true
population diastolic blood pressure for Framingham Heart Study subjects is 80
mm Hg, then approximately 1.367% of the Family of
Samples yield errors as bad as or worse than our single error. The sample
presents significant evidence against the null hypothesis.
Case Four | Hypothesis
Test: Categorical Goodness of Fit | Prenatal Care
A random sample
of Year 2000 Georgia resident live births are checked for prenatal care status,
in the following categories – consider the portion of the sample reporting
prenatal care:
|
Prenatal
Care Status |
Number in
Sample |
|
Prenatal
Care Began 1st Trimester (Months 1-3 of Pregnancy) |
410 |
|
Prenatal
Care Began 2nd Trimester (Months 4-6 of Pregnancy) |
52 |
|
Prenatal
Care Began 3rd Trimester (Months 7-9 of Pregnancy) |
12 |
Test the hypothesis that:
Pr
Pr
Pr
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
|
Prenatal Care Status |
n |
P |
E=nP |
Error |
|
PNC T1 |
410 |
0.7 |
331.8 |
18.4305003 |
|
PNC T2 |
52 |
0.2 |
94.8 |
19.32320675 |
|
PNC T3 |
12 |
0.1 |
47.4 |
26.43797468 |
|
Total Sample with PNC |
474 |
1 |
474 |
64.19168174 |
n=474
Prenatal Starts 1st Trimester
Observed=410
Expected = n*PT1= 474*.70 » 331.8
Error = (Observed – Expected)2/Expected » (410 –
331.8)2/331.8 » 18.4305003
Prenatal Starts 2nd Trimester
Observed=52
Expected = n*PT2= 474*.20 » 94.8
Error = (Observed – Expected)2/Expected » (52 –
94.8)2/94.8 » 19.32320675
Prenatal Starts 3rd Trimester
Observed=12
Expected = n*PT3= 474*.10 » 47.4
Error = (Observed – Expected)2/Expected » (12 –
47.4)2/47.4 » 26.43797468
Total Error » 18.4305003 +
19.32320675 + 26.43797468 » 64.19168174 over three categories.
From row: 3 9.2103 0.010, p-value < .01, since our error
exceeds 9.2103.
Interpretation
Our population is the population of Year 2000
Our Family of Samples (FoS) consists of
every possible random sample of 474 Year 2000
ET1 = N*PT1 = 474*.70 » 331.8
ET2 = N*PT2 = 474*.20 » 94.8
ET3 = N*PT3
= 474*.10 » 47.4
From each member sample of the FoS, we compute sample counts and
errors for each level of survival:
ET1 = N*PT1 = 474*.70 » 331.8
ErrorT1 = (OT1 ─ ET1)2/
ET1
ET2 = N*PT2 = 474*.20 » 94.8
ErrorT2 = (OT2
─ ET2 )2/
ET2
ET3 = N*PT3
= 474*.10 » 47.4
ErrorT3 = (OT3 ─ ET3)2/
ET3
Then add the individual errors for the total error as Total Error =
ErrorT1 + ErrorT2 + ErrorT3
Computing this error for each member sample of the FoS, we obtain a
Family of Errors (FoE).
If the prenatal
care categories are distributed as: 70%
T1, 20% T2 and 10% T3, then fewer 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.
Work all four (4)
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 60
0 1.00000 60 1
1.00000 60 2
1.00000 60 3
1.00000 60 4
1.00000 60 5
1.00000 60 6
1.00000 60 7
1.00000 60 8
1.00000 60 9
1.00000 60 10
1.00000 60 11
1.00000 60 12
1.00000 60 13
0.99999 60 14
0.99998 60 15
0.99993 60 16
0.99980 60 17
0.99947 60 18
0.99866 60 19
0.99689 60 20
0.99326 |
n error base p-value 60
21 0.98633 60 22
0.97405 60 23
0.95377 60 24
0.92250 60 25
0.87747 60 26
0.81685 60 27
0.74052 60 28
0.65056 60 29
0.55129 60 30
0.44871 60 31
0.34944 60 32
0.25948 60 33
0.18315 60 34
0.12253 60 35
0.07750 60 36
0.04623 60 37
0.02595 60 38
0.01367 60 39
0.00674 60 40
0.00311 |
n error base p-value 60
41 0.00134 60 42
0.00053 60 43
0.00020 60 44
0.00007 60 45
0.00002 60 46
0.00001 60 47
<0.00001 60 48
<0.00001 60 49
<0.00001 |
Table 3. Categories/Goodness of Fit
|
Categories
ERROR p-value 3 0.0000 1.000 3 0.2107
0.900
3 0.4463 0.800 3 0.7133
0.700
3 1.0217 0.600 3 1.3863
0.500
3 1.5970 0.450 3 1.8326
0.400
3 2.0996 0.350 3 2.4079
0.300
3 2.7726 0.250 3 3.2189 0.200 3 4.6052
0.100
3 4.8159 0.090 3 5.0515 0.080 3 5.3185
0.070
3 5.6268 0.060 3 5.9915
0.050
3 6.4378 0.040
3 7.0131 0.030 3 7.8240
0.020
3 9.2103 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 |