Transcript Biostatistics General Practice

Biostatistics
Practice2
Example: A study shows that 70% of all patients coming to a
certain medical clinic have to wait at least fifteen minutes to
see their doctor. What is the probability that among ten
patients coming to this clinic
(a) 3 patients will wait
(b) more than 8 patients will wait at least fifteen minutes to
see their doctor?
P(A) = p = 0,70
10
3
10 3
 0,009
(a) P(A  3)    0,70 (1  0,70)
3 
10
P (A  8)     0,70i (1  0,70)10 i
i 9  i 
10
10
9
1
  0,70 (1  0,70)   0,7010 (1  0,70) 0
9 
10
10
(b)
 0,121 0,028  0,149
Example: The number of emergency calls which an
ambulance service gets per day is a random variable
having th Poisson distribution with l5,5. What is the
probability that on any given day it will receive
(a) only 4 emergency calls,
(b) More than 4 emergency calls.
P( X  A) 
(a)
(b)
A l
le
A!
5,54 e 5,5
P( X  4) 
 0.156
4!

4
l Ae  l
lAe  l
P( X  4)  
A 5
A!
 1 
A 0
A!
Example: A group of patients have a mean body mass index
(BMI) of 22 and a standard deviation 4. BMIs are
approximately normally distributed. A subject is classified as
obese if his BMI has greater than 30 BMI. In this group of
patients what is the propotion of obese patients? What is the
proportion of patients whose BMI less 18?
=22
=4
P( x  30)  ?
x
z

0
z=2
30  22
P( x  30)  P( z 
)  P( z  2)
4
 0.0228
18  22
P( x  18)  P( z 
)  P( z  1)
4
 0,5  0,3413
 0,1587
Example: It is known that 25% of the persons in the
study were obese. In this study, 200 individuals were
examined and mean BMI score was found as 23 and std.
deviation was found as 5. What is the cut-off point for
obesity? What is the number of the obese individuals?
P(Z  z)  0,25  z  0,68
x
z

x  23
0,68 
5
x  23  0,68* 5  26,4
Number of obese individuals = 200*0,25 = 50
Example: Scoring system was changed in ÖSS in 2003.
The lowest score of the students who attended Hacettepe
University School of Medicine was 195 in 2002, and 310
in 2003. It is also known that, in 2002 mean=160 & s=10,
and in 2003 mean=270 & s=15. Has the attendance score
increased in 2003?
x
195  160
z 2002 

 3,5

10
x   310  270
z2003 

 2,67

15
Sampling Distribution
P
e
r
c
e
n
t
n=10
n=25
n=100
P
e
r
c
e
n
t
n=10
n=25
n=100
Example: If the mean and standard deviation
of serum cholesterol values for male patients
who was diagnosed as having myocardial
infarction are 240 and 40 mg/dl, what is the
probability that a random sample of 100 men
will have a mean greater than 250 mg/dl?
z
z
x
P( x  250)  ?
/ n
250  240
40 / 100
 2,5
P( z  2,5)  0,0062
In the previous example, instead of 100 men, if
we randomly select 16 men, what is the
probability of obtaining a sample mean greater
than 250 mg/dl?
z
250  240
40 / 16
1
P( x  250)  P( z  1)  0,1587
Example: A researcher wants to estimate the
average serum cholesterol level of healty men.
Mean and standart deviation of the randomly
selected 100 men are 160 and 30 mg/dl.
95% CI for the population mean:
xi  t( n1; )
s
s
   xi  t( n1; )
n
n
160  1.98(3)    160  1.98(3)
154.06    165.94
When the researcher selected 36 men at random,
95% CI for the population mean:
xi  t( n1; )
160  1.98(
s
s
   xi  t( n1; )
n
n
30 ) 
36
  160  1.98(
30 )
36
150.1    169.9
90% CI for the population mean:
160  1.66(
30 ) 
36
  160  1.66(
152.7    168.3
30 )
36
Example: A researcher wants to estimate the
prevalence of osteoporosis in women over fifty
years of age. Among 100 randomly selected
women, 15 women have a diagnosis of
osteoporosis. What is the 95% confidence
interval for the proportion of osteoporosis in
women?
15
p
 0.15
100
Sp 
Point estimate of the proportion of
osteoporosis
p(1  p)
0.15x0.85

 0.036
n
100
The standard error of the estimate is 0.036.
95% CI for the population proportion
p  t (n1; )Sp  P  p  t (n1; )Sp
0.15  1.98  0.036  P  0.15  1.98  0.036
0.08  P  0.22
If a researcher wants to calculate interval for the
population proportion with 90% confidence:
0.15  1.66  0.036  P  0.15  1.66  0.036
0.09  P  0.21