Download 5. A new non-invasive screening test is proposed that is claimed to

5. A new non-invasive screening test is proposed that is claimed to be able to identify patients
with impaired glucose tolerance based on a battery of questions related to health behaviors. The
new test is given to 75 patients. Based on each patient’s responses to the questions they are
classified as positive or negative for impaired glucose tolerance. Each patient also submits a
blood sample and their glucose tolerance status is determined. The results are tabulated below.
Screening Test
Positive
Negative
Impaired Glucose Tolerance
17
8
Not Impaired
13
37
a) What is the sensitivity of the screening test?
b) What is the false positive fraction of the screening test?
Work:
a)
Sensitivity = #(true positives)/(#(true positives)+#(false negatives)) = 17/(17+8) = 17/25 = 0.68
Answer: 0.68
b)
False positive fraction = #(false positives)/(#(false positives)+#(true negatives)) =13/(13+37)
= 13/50 = 0.26
Answer: 0.26
6. BMI in children is approximately normally distributed with a mean of 24.5 and a standard
deviation of 6.2.
a) A BMI between 25 and 30 is considered overweight.
What proportion of children are overweight?
Work:
Let X = BMI in children
X has a normal distribution with mean of 24.5 and sd of 6.2
Z = (X-24.5)/6.2 has a standard normal distribution
P(25<X<30) = P((25-24.5)/6.2 < Z < (30-24.5)/6.2) = P(0.0806 < Z < 0.8871) = 0.2803
Answer: 0.2803
b) A BMI of 30 or more is considered obese.
What proportion of children are obese?
Work:
Let X = BMI in children
X has a normal distribution with mean of 24.5 and sd of 6.2
Z = (X-24.5)/6.2 has a standard normal distribution
P(X > 30) = P( Z > (30-24.5)/6.2) = P(Z > 0.8871) = 0.1875
Answer: 0.1875
c) In a random sample of 10 children, what is the probability that their mean BMI exceeds 25?
Work:
Let X = sample mean of their BMI
X has a normal distribution with mean of 24.5 and sd of 6.2/10
Z = (X-24.5)/(6.2/10)) has a standard normal distribution
P(X > 25) = P( Z > (30-24.5)/ (6.2/10)) = P(Z > 2.8052) = 0.0025
Answer: 0.0025
7. A national survey is conducted to assess the association between hypertension and stroke in
persons over 55 years of age. Development of stroke was monitored over a 5 year follow-up
period. The data are summarized below and the numbers are in millions.
Hypertension
No Hypertension
Develop Stroke
12
4
Did not Develop Stroke
37
26
a) Compute the incidence of stroke in persons over 55 years of age
b) Compute the relative risk of stroke comparing hypertensive to non-hypertensive persons
c) Compute the odds ratio of stroke comparing hypertensive to non-hypertensive persons
a)
incidence of stroke in persons over 55 years of age =
DS / Total = (12+4)/(12+37+4+36) = 16/89
Answer: 16/89
b)
relative risk of stroke comparing hypertensive to non-hypertensive persons
= (DS H/ DS H + Not DS H)/ (DS NH/ DS NH + Not DS NH)
= (12/(12+37))/(4/(4+26)) = (12/49)/(4/30) = (12/49)(30/4) = 90/49
Answer: 90/49
c)
Odds ratio of stroke comparing hypertensive to non-hypertensive persons
= (DS H/ Not DS H)/ (DS NH/ Not DS NH) = (12/37)/(4/26) = (12/37)(26/4) = 78/37
Answer: 78/37
8. Answer True or False to each of the following
a) If there are outliers, then the mean will be greater than the median. False
b) The 90th percentile of the standard normal distribution is 1.645. False
c) The mean is the 50th percentile of any normal distribution. True
d) The mean is a better measure of location when there are no outliers. True