Transcript Lecture 3

Confidence Intervals
Lecture 3
Confidence Intervals for the
Population Mean (or percentage)
For studies with large samples, “approximately
95% of the time, the population mean will be in
the interval given by the sample mean plus or
minus two standard errors.”
A confidence interval is a range of values
WARNING!
This DOES NOT imply that for a given
experiment the population parameter has a 95%
chance of being in the confidence interval.
Pregnancy Analogy
CORRECT Interpretation: “I am 95%
confident that
Learning to work with the
Gaussian/Normal Distribution
What if I want a 90% confidence interval?
We need to learn how to calculate probabilities
based on the Normal Distribution!
For confidence intervals, we are working with the
sample mean, a random variable which
approximately follows a normal distribution. We
will talk now in general about a random variable,
Y, which follows a normal distribution with mean μ
and standard deviation σ.
Framework for the Calculation
Define z as the value such that (1-α)100% of the
population would fall within z standard deviations of
the population mean (as in the Empirical Rule).
This is equivalent to saying that a single random
variable has a (1-α)100% chance of falling within z
standard deviations of the population mean.
In the context of confidence intervals (and hypothesis
tests), z is called a critical value.
Then (1-α)100% of the time, the following
statements are true:
The Mathematics
FACT:
Y 

follows a “standard normal” distribution.
The Z-score
All normal calculations are based on a z-score:
where Y is a normal random variable (which may
be a single value, an average, etc.)
The z-score measures how many standard
deviations (or standard errors) an observed value
of a normal random variable is away from its
mean.
Z-scores follow a standard normal distribution, a
normal distribution with
Probabilities from
the Standard Normal
Chapter 4 in textbook & Table A5.2 on p.366
Areas under the Normal curve represent
probabilities or proportions of the population.
The total area under the curve equals one.
Column #2 gives the probability of falling within
z standard deviations of the mean, where z is
given in column #1.
May be used to refer to population distributions
or sampling distributions.
Application of a Normal Distribution
to a Population Distribution of IQ’s
Suppose that I am going to measure IQ scores of
UMDNJ –School of Public Health students.
The national average of IQ scores 100 points and
the standard deviation is 16 points. For now
assume that the national average equals the
average for public health students.
According to the Empirical Rule, approximately
68% of scores are between 84 and 116, and
95% of scores are between 68 and 132.
IQ Example continued
Let’s calculate these percentages exactly.
 Plus or minus 1 st. dev.:
 68.27%
= proportion of .6827
 Plus or minus 2 st. dev.:
 95.45%
= proportion of .9545
The probability of a single observed IQ falling
within one standard deviation of the mean is
equivalent to the proportion of all IQ’s that
fall within one standard deviation of the mean.
Probabilities for Single
Observations
Calculate the probability that the IQ score
of one public health student will be within
10 points of the national average.



Distance between mean and
observation=10
Z-score = 10/16 = 0.625 Std. Dev.’s
Probability = 0.4679
Probabilities for Sample Means
(Sampling Distributions)
Suppose I do repeated experiments (studies) in
which I draw 10 students for each study.
Standard error for a mean of 10 measurements
SEM = 16/(square root of 10) = 5.06
Therefore, in approximately
68% of these experiments, the average score will be
between (100-5.06=) 94.94 and 105.06, and
95% of these experiments, the average score will be
between (100-2*5.06=) 89.88 and 110.12.
Equivalently,
For any one experiment, we have a 95% chance that the
sample mean will fall between 89.88 and 110.12.
Sampling Average IQ’s, cont.
Calculate the probability that the average IQ
score of 10 public health students will be within
10 points of the national average given that the
public health distribution of IQ’s is the same as
the national distribution.
Standard error for a mean of 10 measurements
•
•
SEM = 16/(square root of 10) = 5.060
Z-score = 10/5.060 = 1.976
Probability within ± 1.976 st.dev.s
= about 95%
Much larger probability than when looking at at
single measurement!
95% Large Sample Confidence Interval

Suppose that I sample 50 public health students and find that
Sample Mean = 110.7 points
And the Standard Error Estimated from Sample = SEM = 2.5 points
What’s the value of z such that 95% of sample means would fall
within z standard errors of the population mean?
z = 1.96
Therefore a 95% confidence interval is
 110.7 ± 1.96(2.5)

= (105.800, 115.600)
Interpretation: We are 95% confident that the mean IQ for all public
health students falls between 105.8 and 115.6 points.
How likely is it that the mean for the public health students is the
same as that the national mean (100)?
90% Large Sample Confidence Interval
Sample Mean = 110.7 points
Standard Error of the Mean = SEM = 2.5 points
z = 1.645
Therefore a 90% confidence interval is

110.7 ± 1.645(2.5)
= (106.588, 114.813)
Interpretation:
What Confidence Level
should I use?
The more confidence we would like to claim in our
interval estimate
The standard confidence people use is 95%.
Smaller confidence levels (such as 90%) may be
appropriate, especially for pilot studies or other
studies with small sample sizes.
Larger confidence levels may also be appropriate
when costly reforms rest on the conclusions from
those confidence intervals.
Large Sample Confidence
Interval for a Population Mean
Sample mean ± z SEM
Where SEM = square root of s2/n
Typically, 95% confidence intervals are used,
with
z =1.96
Here, 1.96 is the 97.5th percentile of the
standard normal distribution
To be modified for smaller sample sizes shortly

Overweight Status and Eating
Patterns… (AJPH, 2002)
95% CI for mean BMI of girls is
23.3 +/- 1.96*4.9/sqrt(2000) = (23.09, 23.51)
95% CI for mean BMI of boys is
23.0 +/- 1.96*4.8/sqrt(2000) = (22.79, 23.21)
Unfortunately, we have target percentiles for
BMI, not target means.
Large Sample Confidence Intervals
for Population Proportions
p ± z (SE)
Where the estimated variance of the sample
proportion is SE2 = p(1-p)/n
Typically, 95% confidence intervals are
used, with
z =1.96
Actual Percent in the Top Targeted 5th
Percentile – 95% Confidence Interval
For girls,
p = .125, n=2099
SE = sqrt((.125*.875)/2099) = .007219
p ± z (SE)  (0.111, .139)  (11.1%, 13.9%)
For boys,
p = .166, n=2141
SE = sqrt((.166*.834)/2141) = .008041
p ± z (SE)  (0.150, .182)  (15.0%, 18.2%)
I am 95% confident that the percent of boys that is
in the targeted upper 5th percentile is actually
between 15.0% and 18.2% of boys.
“Men Carrying Pollutant
Have More Boys”
For description, see news article.
n =101
p = .57
p(1  p)
.57(.43)
SE =
 .049

n
101
Thus a 95% confidence interval for the
proportion is given by .57 ± 1.96 (.049)
Or equivalently (0.473, 0.667)
Interpretation for Pollutant – Boys E.g.
I am 95 % confident that the true proportion of
boy babies born from parents who both have
detectable PCB levels in their blood is between
0.473 and 0.667.
I.e., if this is one of the 95% of the times that the
true parameter falls in the interval, then the mean
is between 0.473 and 0.667.
If the proportion of of boys in the general
population is 0.51, is there a difference in the
proportion of boys from parents with detectable
PCB levels?
Assumptions for “Large Sample CI’s”
Random sample
•
Every individual in the population has an equal
chance of being selected.
Independent observations
•
Selecting one subject does not alter the chances of
selection of another subject
Large Sample Size – so that the Central Limit Theorem
“kicks in” and the standard error is accurately estimated
•
How large is “large”?
•
Depends!
•
If population distribution is approximately normal,
then “large” is approximately 40 subjects.
•
The farther the population distribution away from
normal, the larger the needed size of the sample.
… E.g., binary variables- at least 30 subjects – 5 in
each category.
Other Assumptions
Other Assumptions
•
•
•
Study Population = Target Population
Variable accurately measures characteristic of
interest
Values are correctly measured and recorded
Quick Re-evaluation of Large
Sample CI’s for Population Means
Why do we use a z-critical value in the CI?
Due to the CLT, we can say that if the sample
size is large, the sample mean will fall within
the specified number (z) of standard errors of
the mean.
More specifically, the 95% CI is derived from
noting z  Y  
 / n falls between –1.96 and +1.96.
• The
standard deviation is assumed to be known.
CI’s for Population Means when
the Sample Sizes are Smaller
ISSUE: The standard deviation is
unknown and, hence, it is estimated.
This estimate is generally OK for large
sized samples, but not for smaller sized
samples.
PROBLEM: There is more variation in
the z-score with an estimated standard
deviation than is allowed for by the Normal
distribution.
Thanks to a smart person!
SOLUTION:
•
•
In 1908, William Sealy Gosset working at the
Guinness Brewery in Dublin, Ireland showed,
mathematically, that the z-score for the sample
average in which the population standard
deviation has been replaced by the sample
standard deviation follows a “t-distribution”
with a specified degrees of freedom (d.f.).
I.e.,
y
t
s/ n
T-Distribution
What does it look like?
•
•
Like the Normal distribution
With larger variance, depending on the d.f.
What are “degrees of freedom.”
•
•
•
It is difficult to define
“Amount of information about the variance.”
Practically, for the sample mean,
d.f. = n-1
Using the T-distribution for CI’s
Assumptions:
•
•
Approximately normal,
Random and independent sample
d.f. = n-1
CI: Y  t *  s 
 n
See Table A5.3 on page 368 for a table of
critical values for 90%, 95% and 99% CI’s.
Cadmium levels (ng/gram) in mothers
The sample sizes are 14 (for smoking) and 18 (for
non-smoking mothers).
Distributions are bimodal.
CI using the t critical value
Means and standard errors are:
•
•
Smoking: 20.414 & 6.814/sqrt(14) ng/gram
Non-smoking: 14.722 & 6.199/sqrt(18) ng/gram
t-critical values for 95% confidence intervals are
2.160 and 2.110 respectively for 13 and 17 degrees
of freedom
Cadmium Confidence Intervals
•
•
•
The confidence intervals are:
Smoking:

20.414 ± 2.160 (1.821)

(16.481, 24.347)
Non-smoking:

14.722 ± 2.110 (1.461)

(11.639, 17.805)
These intervals overlap, suggesting that the mean cadmium
levels for the populations of smoking and non-smoking
mothers are not different.
After the midterm, we will talk about more specific ways to
test this hypothesis. (Comparison of two population means)
HELP: When do I use which CI?
In general for means, use the t-interval.
•
•
•
For means (not proportions), the t-interval is
robust to violations of the Normality
assumption.
If the sample distribution is far from normal (as
determined by, e.g., histograms), non-parametric
or other more “exact” methods may be needed.
These are easier to discuss in the context of
hypothesis testing.
When do I use which CI?, cont…
For proportions with large sample sizes,
use the z-interval.
• “Large” = 30 with between 5 and 25 subjects
in the group of interest
• Is the pollutant-boys sample size OK?
• 101 subjects with at least 43 of each gender.
For proportions with small sample sizes,
there exist “exact” confidence intervals.
See Table A5.1 & pp.16-18
Assumptions for all CI’s
presented so far
Observations within the sample are
independent of one another.
Sample consists of randomly selected
subjects that are representative of the target
population.
•
In particular, each subject in the study
population has an equal chance of being
selected.