Homework Problem #4
The Second Task Force Report on Blood Pressure Control in Children reports bloodpressure norms for children by age and sex group. The mean ± standard deviation for
17-year-old boys for diastolic blood pressure is 63.7 ± 11.4 mmHg, based on a large
sample.
5.65. One approach for defining elevated blood pressure is to use 90 mmHg – the
standard for elevated adult diastolic blood pressure –as the cutoff. What percentage of
17-year-old boys would be found to have elevated blood pressure, using this approach?
: = 63.7 mmHg
F = 11.4 mmHg
cutoff = 90 mmHg
We wish to find Pr(X $ 90 mmHg). Note in graph above it says Pr(X > 90) but > and $
give the same answer with continuous distributions.
To use STATA we will convert 90 mm Hg to the N(0,1) in terms of number of SDs from
the mean or
z = (90 - 63.7)/11.4 = 2.307
STATA will allow us to use the normal function with 2.307, which is in terms of the N(0,1),
to calculate the desired area.
normal(2.307) = the area to the left of 2.307 (i.e. the complement of the area we want).
So we will subtract normal(2.307) from 1 to get the area to the right of 2.307.
di 1 - normal((90 - 63.7)/11.4)
.01052692
Thus 1% of the 17 year old boys would be expected to be hypertensive using the above
definition.
5.66
Suppose 25 of the 2000 17 year old boys in the 11th grade have hypertension.
Given the data above, would this be considered unusual.
We are still assuming the DBP of the 17 year old boys is distributed N(63.7, 11.42). So
we would expect approximately 1% of the boys to have DBP in the hypertensive range.
Or the probability that a 17 old boy would be abnormal is 0.01.
We can now use the binomial distribution (the boys are either normal or abnormal) to
find the probability that 25 or more boys out of 2000 would be abnormal.
. di 1 - binomial(2000, 24,0.01) = 1 - Pr(X <= 24) = .15565101
The probability is 16% that 25 or more out of 2000 would be abnormal. This is not
considered rare.
Hypertension
A doctor diagnoses a patient as hypertensive and prescribes an antihypertensive
medication. To assess the clinical status of the patient, she takes n replicate bloodpressure measurements before the patient starts the drug (baseline measurement) and
n replicate blood-pressure measurements 4 weeks after starting the drug (4 week
follow-up). She will use the average of the n replicates at baseline minus the average of
the n replicates at follow-up to assess the status of the patient (i.e. you use the same
number of replicates at baseline and follow-up).
She knows, based on previous clinical experience with the drug, that the mean change
in diastolic blood pressure over the 4 week period over a large number of patients tends
to be normally distributed with mean 5 mm Hg and with variance 33/n, where n is the
number of replicates at baseline.
5.75
What is the probability that the decline in DBP will be at least 5 mm Hg if there is
exactly one measurement at each of baseline and follow-up.
This says we are talking about the difference of the baseline and follow-up
measurements (i.e. the decline) following a normal distribution with
mean = 5 mm Hg
and
variance = 33/1 = 33.
So the probability the DBP will decline by at least 5 is 50% or 0.5.
5.76
The mean changes from the 5 of problem 5.75 to 2. However, the variance (or SD)
remains the same. So the normal curve that goes with this problem is the same shape
as the curve for 5.75. The center for the graph for this problem is just shifted down to 2.
The probability we are looking for is the hatched area in the graph above (i.e. the area
under the curve N(2,33) and to the right of 5.
To transform 5 to the N(0,1) curve we subtract 2 and divide by the square root of 33.
Then we can use the STATA function normal.
But
is the area to the left of
and under the N(0,1) curve.
Notice that "norm" above should be normal.
So we calculate 1 - normal to get the correct area.
. di 1 - normal((5 - 2)/sqrt(33))
.30075407
So the probability associated with the area hatched above is 0.30.
Notice as the variance of the two curves gets smaller the curves will become taller and
more narrow. The area to the right of 5 under the N(5,33) curve will remain 0.5 since
the area to the right of the mean is 0.5 regardless of the shape of the curve. However,
the area to the right of 5 and under the N(2,33) curve will get smaller as the variance
gets smaller.
5.77
So we are looking for the n that will make the variance (33/n) such that the area to the
right of 5 and under the N(2,33) curve is 0.1 since 5*0.1 = 0.5.
We can look at the tables in the back of Rosner page 752 under column B for 0.1. The
z on the N(0,1) curve that cuts off approximately 0.1 is 1.28. So we solve the equation
below for n:
(I just squared both sides of the equation)
9n = 33(1.6384)
n=6
You can use the inverse normal function of STATA because you know the area under
N(0,1) is = 0.1 and what you need to know is what is the cutoff on the N(0,1) that will
give you that area. normal is the cumulative distribution so to use the invnorm we will use
1 - 0.10 or 0.90.
. di invnormal(0.90)
1.2815516
So 1.28 is again found to be the cutoff. Then you use the formula above to calculate n.