Solutions for
Determining Sample Sizes for the Purpose of Estimation
Operating Assumption: Calculations assume 95% confidence, so a critical value
of 1.96, perhaps rounded up to 2.00.
Mean Estimation for a Simple Random Sample
Example 1: N = 1,850, a 5 point Likert item is used.
Problem 1 Assume that a population variance of .2 has been obtained, and that the
board members want a bound of + .10.
D = B2/4 = (.10)2/4 = .0025
n=
N 2
( N  1 )D  
2
=
1850 (.2 )
= 76.7 or 77
( 1850  1 )D  .2
The board members should send 77 people questionnaires.
Problem 2 Assume that the board members want a bound of + .05, and that you
have no estimate of population variance (use Tchebysheff's theorem for a
5 point Likert item).
D = B2/4 = (.05)2/4 = 0.000625
n=
N 2
( N  1 )D  
2
=
1850 ( 1 )
= 858.2 or 859
( 1850  1 )D 1
The board members should send 859 people questionnaires.
Problem 3 Assume that a population standard deviation of .4 has been obtained,
and that the board members want a bound of + .07.
D = B2/4 = (.07)2/4 = 0.001225
n=
N 2
( N  1 )D  
2
=
1850 (.16 )
= 122.06 or 123
( 1850  1 )D  .16
The board members should send 123 people questionnaires.
Example 2: N = 20,000, a 10-item questionnaire is used (each item scaled 1 to 5) for
the purpose of producing satisfaction scores ranging from 10 to 50.
Problem 4 Assume that a population standard deviation of 8 has been obtained, and
that the financial aid department wants a bound of + 5 points.
D = B2/4 = (5)2/4 = 6.25
n=
N 2
( N  1 )D  
2
=
20000( 64 )
= 10.23 or 11
( 20000  1 )D  64
11 students should receive a questionnaire from the financial aid department
Problem 5 Assume that financial aid department wants a bound of + 2 points
around the mean score, and that you have no estimate of population
variance (use Tchebysheff's theorem).
D = B2/4 = (2)2/4 = 1
n=
N 2
( N  1 )D  
2
=
20000 ( 100 )
= 99.51 or 100
( 20000  1 )D 100
100 students should receive a questionnaire from the financial aid department
Problem 6 Assume that a population variance of 49 has been obtained, and that the
financial aid department wants to assume a standard error of 2.0.
Further assume that financial aid department wants to be roughly 95%
confident in its estimate.
D = B2/4 = (4)2/4 = 4
n=
N 2
( N  1 )D  
2
=
20000( 49 )
= 12.24 or 13
( 20000  1 )D  49
13 students should receive a questionnaire from the financial aid department
Proportion Estimation for a Simple Random Sample
Example 3: How many students should have been interviewed? (Assume N = 2000
and a desired bound of .02). Use the maximum variance possible for a proportion.
D = B2/4 = (.02)2/4 = 0.0001, 2 = p(1-p) = p(1-p) = .5(1-.5) =.25
n=
N 2
( N  1 )D  
2
=
2000(.25 )
= 1111.358 or 1112
( 2000  1 )D  .25
1112 students should have been interviewed.
Mean Estimation for a Stratified Random Sample
Example 4: The sixth grade contains 55 students in track I, 80 students in track II,
and 65 students in track III. A stratified random sample of students,
proportionately allocated, is desired. Assuming a bound of .05 and prior evidence
that the scores have a variance of 10, determine what sample size is needed.






55 3.16 *
N 1 1 

 = 0.275
w 1=
= 
3


55( 3.16 )  80( 3.16 )  65( 3.16 ) 

 Ni  i 
 i 1







80 3.16 *
N 2 2 

 = 0.400
w 2=
= 
3


55( 3.16 )  80( 3.16 )  65( 3.16 ) 

 Ni  i 
 i 1







65 3.16 *
N 3 3 

 = 0.325
w 3=
= 
3


55( 3.16 )  80( 3.16 )  65( 3.16 ) 

 Ni  i 
 i 1

* Please Note: The number 3.16 is the square root of 10. I did not actually
round this number when performing this calculation.
D = B2/4 = .052/4 = 0.000625
55 2 (3.16) 2
80 2 (3.16) 2
65 2 (3.16) 2
3


N i2  i2
55 ( 3.16)
80 ( 3.16)
65 ( 3.16)

Wi
i 1
632
632
632
n=
=
= 197.53 or 198
2
3
2
2



200  0.000625   55 (3.16)  80 (3.16)  65 (3.16) 2
2
2
N D   N i i
i 1
n 1 = n (w 1) = 197.53(0.275) = 54.32075 or 55 students in Track I.
n 2 = n (w 2) = 197.53(0.400) = 79.012 or 80 students in Track II.
n 3 = n (w 3) = 197.53(0.325) = 64.19725 or 65 students in Track III.