Transcript Sample Size and Confidence Intervals for the Population Mean (m)

Sample Size and CI’s for
the Population Mean (m)
Sample Size and CI’s for m


Suppose we wish to estimate a population mean m using
a 95% CI and have a margin of error no larger than E
units. What sample size do we need to use?
Recall the “large” sample CI for m is given by:
X  (z - value) s
95% z = 1.96
n
90% z = 1.645
99% z = 2.576
MARGIN OF ERROR (E)
Note: The z-value should actually be a t-distribution value, but for
sample size planning purposes we will use a standard normal value.
Sample Size and CI’s for m

For a 95% CI if we want margin of error, E we have
E  1.96 s

n
After some wonderful algebraic manipulation
 1.96  s 
n

 E 
2
Oh, oh! We don’t know s !!
1. “Guesstimate”
2. Use sample SD from pilot or prior study.
3. Use fact 95% of observations generally lie
with 2 SD’s of the mean thus
Could also use fact that 99% lie
within 3 SD’s and use 6 instead
of 4 in our crude approximation.
Range
s
4
where Range represents the expected
maximum – minimum we would see
in sample.
Example: Estimating Mean Cholesterol
Level of Females 30 – 40 yrs. of age
Q: What sample size would be necessary to estimate
the mean cholesterol level for the population of
females between the ages of 30 – 40 with a 95%
confidence interval that has a margin of error no
larger than E = 3 mg/dl?
Sample Size and CI’s for m

Suppose from a pilot study we find s = 19.8 mg/dl
We can use this estimate to find the sample size that will
give E = 3 mg/dl.
 1.96  s 
 1.96  19.8 
E 
 
  167.34
3
 E 


2

n  168
2
Standard normal values
90% = 1.645
95% = 1.960
99% = 2.576
Sample Size and CI’s for m
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Suppose we do not have any information about the
standard deviation of the cholesterol levels of
individuals in this population.
We could use the Range/4 or Range/6 as crude
approximations to the standard deviation.
What is the smallest serum cholesterol level we
would expect to see? 100 mg/dl (my guess)
What is the largest? 300 mg/dl (my guess again)
SD approximation = 200/4 = 50 mg/dl
or
SD approximation = 200/6 = 33.33 mg/dl
Sample Size and CI’s for m

Using this crude estimate for the standard deviation we
find the following sample size requirements
 1.96  50 
E 
  1067.11
 3 
2

n  1068
or
 1.96  33.33 
E 
  474.18  n  475
3


2