Transcript Fundamentals of Research Project Planning: Hypotheses

Estimation of Various Population
Parameters
Point Estimation and Confidence
Intervals
Dr. M. H. Rahbar
Professor of Biostatistics
Department of Epidemiology
Director, Data Coordinating Center
College of Human Medicine
Michigan State University
Important Statistical Terms
• Population: A set which includes all
measurements of interest to the researcher
• Sample: Any subset of the population
• Parameter of interest: Characteristics of
interest in a population are called population
parameters
• (e.g, mean, standard deviation, proportion)
Inferential Statistics
• Estimation includes point and interval
estimation of population parameters
• Mean = 
• Median
• Standard deviation = 
• Proportion = p
• Incidence
Central Limit Theorem
• When large samples, (n>30), are selected
from a population with mean  and
standard deviation , then
• the sampling distribution of the sample
mean is approximately normal with the
same mean ,

and the standard error, SE=
n
Estimation of Population Mean
A 95% confidence interval for  based on a larg
sample, when n>30, is approximately,
x  2(
n
)
Estimation of Population Mean
A (1- )100% confidence interval for 
based on a large sample, when n>30, is
approximately,
x  Z / 2 ( S
n
)
  0.05, Z / 2  1.96
  0.10, Z / 2  1.28
Example # 1
Suppose we want to estimate the average birthweight of children born to non-smoking women in
community X. Assume the standard deviation of
birth-weights is 0.5 Kg and the sample mean birthweight is 3.0 Kg. Find a 95% confidence interval
for .
3.0  2( 0.5
3.0  0.1
(2.9, 3.1)
100
)
Interpretation of CI’s
What do we mean by (2.9 Kg, 3.1 Kg) is a 95%
CI for the mean birth-weight of children born
to non-smoking women in community X?
Does interval (2.9 , 3.1) contain ?
Confidence Interval for P
Suppose we want to estimate the prevalence of
hypertension among adults in community X.
Assume in a random sample of 100 adults from
this community the prevalence of hypertension
is 0.50. Find a 95% confidence interval for P.
0.50  2(
(0.50)(1  0.50)
0.50  0.10
(0.40, 0.60)
100 )
Probability vs Odds of an Event
• An experiment is
repeated for a large
number of times.
Then the probability
of occurrence of an
event E is the
percentage of the time
that that particular
event has occurred in
repeated experiments.
• Suppose P(E)=0.40
• Odds of an event E
=P(E)/{1-P(E)}.
Example,
If P(E)=0.40, then
Odds of E=2/3.
In the following contingency table estimate the
proportion and odds of drinkers among those
who develop Lung Cancer and those without the
disease?
Drinker Yes
No
Lung Cancer
Case
Control
A=33
B=27
C=1667
D= 2273
Total
60
3940
P1=33/1700
P2=27/2300
Odds1=33/1667 Odds2=27/2273
Odds Ratio as a Measure of
Association
• QUESTION: Is
there a difference
in the proportion
of drinkers among
those who develop
Lung Cancer and
those without the
disease?
• Odds Ratio=ad/bc
• OR=1.67
• This means that
the odds of
drinking among
cases is 67% more
than the odds of
drinking among
controls.
Relative Risk as Measure of
Association
• QUESTION:
Are persons
exposed to
“factor X” more
likely to develop
the disease than
those not so
exposed?
Risk among exposed
= a/(a+b)
Risk among non-exposed
= c/(c+d)
Relative Risk
= Risk among exposed
relative to risk among
non-exposed
QUESTION: Are persons exposed to “factor X” more
likely to develop the disease than those not so exposed?
Disease
Yes
No
Factor X Yes A=33
B=27
No C=1667 D=2273
Total
60 (fixed)
3940 (fixed)
Risk among exposed
= a/(a+b)=33/60=0.55
Risk among non-exposed
= c/(c+d)=1667/3940=0.42
Relative Risk
= Risk among exposed relative to risk among nonexposed RR=1.31
95% Confidence Intervals for Odds
Ratios and Relative Risk
• NOTE:
• Confidence
Intervals for Odds
Ratios and Relative
Risk are not
symmetrical
• Odds Ratio=ad/bc
• OR=1.67
• E.g., 95% CI for OR
(1.2, 2.9)
• If the 95% CI for OR
does not include one
then we conclude a
difference in the odds
of drinking between the
two groups
• We use Statistical
packages to calculate
the CI’s for OR or RR
QUESTION: Estimate the difference
between the proportions of drinkers
among Lung Cancer cases and controls!
Drinker Yes
No
Lung Cancer
Yes
No
A=33
B=27
C=1667
D= 2273
P1- P2 = 33/1700 – 27/2300
Total
60
3940
QUESTION: Estimate the difference between
the mean blood levels of cases and controls?
Group 1
Disease
Mean BP
Group 2
No Disease
Mean BP
Factors influencing the
width of the CI’s
Sample size
Variance
Confidence level = (1-)100%
Width of the CI’s
If Sample size =n , then Width 
If Variance , then Width 
If Confidence level  , then Width 
Choose ‘n’ to achieve the
desired width for the CI’s