10.3. Determining the Sample Size
Since

 
P X    z  X   P X    z
  1   ,
2
2


n

there is a 1    probability that the value of a sample mean will provide a sampling
error of z

2
n
or less.
Margin of error:
The quantity z

2
is referred to as the margin of error.
n
Note:
 The larger the sample size n is, the smaller the margin of error
is!!
 The larger the population variance  2 is, the larger the margin
of error is!!
Important Result:
The desired margin of error can be obtained by choosing the sample
size.
[Derivation of this result:]
Let E be the desired margin of error. Then,
E  z
z 
2
z2  2

 n  2  n 22
E
E
n
.
Note:
The following procedures can be used to estimate the unknown
population standard deviation 
 Use the sample standard deviation from a previous sample.
 Use judgment or a “best guess”.
1
Example:
15
 x
x  53.87, s 
i 1
i
 x
2
15  1
 6.82, t14, 0.025  2.145
A 95% confidence interval can be obtained by
s
x  t14, 0.025
n
 53.87  2.145 
6.82
15
 53.87  3.78 .
Thus, the margin of error is 3.78.
If we want to obtain a desired margin of error E  2 for a 95% confidence interval,
then
n
z02.05  2
2
2
E
1.96 2  6.82 2

 44.67 ,
22
where  can be estimated by using the sample deviation from the previous sample.
Thus,
n  45 . That is, we need to have a sample of 45 observations in order
to obtain a 95% confidence interval with a desired margin of error E  2 .
Example A (continue):
A random sample of 81 workers at a company showed that they work an average of
100 hours per month with a standard deviation of 27 hours. At 95% confidence, how
many more workers need to be included in the sample to provide a confidence interval
with length 4 (i.e., the margin of error being 2)?
[solution:]
Since E  2 ,
n
z02.05 s 2
E
2
2
1.96 2  27 2

 700.13  n  701 .
22
Thus, we need 701-81=620 more workers.
Online Exercise:
Exercise 10.3.1
2