Confidence Intervals for a Single Proportion
Problem 1
At a certain company, the HR department is interested in knowing what percentage of the
employees are content with the medical benefits offered to them. A random sample of 140
employees is obtained, and 74.3% of them say they are content.
(a) Construct and interpret a 90% confidence interval for the proportion of all employees
at this company that would say they are content with the medical benefits.
(b) Construct and interpret a 92.5% confidence interval for the proportion of all
employees at this company that would say they are content with the medical benefits.
(c) Construct and interpret a 97.4% confidence interval for the proportion of all
employees at this company that would say they are content with the medical benefits.
Problem 2
A legislator is interested in knowing what percent of the residents in a certain community
would be in favor of a new law that is being proposed.
(a) If she would like to construct a 90% CI with margin of error no more than 5%, how
many residents must she randomly poll? Assume no previous studies.
(b) If she would like to construct a 90% CI with margin of error no more than 5%, how
many residents must she randomly poll? Assume that a preliminary investigation
yielded that 32% of the community would be in favor of the new law.
(c) If she would like to construct a 95% CI with margin of error no more than 3%, how
many residents must she randomly poll? Assume no previous studies.
(d) If she would like to construct a 95% CI with margin of error no more than 3%, how
many residents must she randomly poll? Assume that a preliminary investigation
yielded that 32% of the community would be in favor of the new law.
Problem 3
We have calculated a confidence interval based upon a sample of n = 100. Now we want to
get a better estimate with a margin of error only one fourth as large. What sample size must
be obtained?
Problem 4
We have calculated a confidence interval based upon a sample of n = 80. Now we want to
get a better estimate with a margin of error only one half as large. What sample size must be
obtained?
Answers
Problem 1
(a) “By Hand”
p̂ ± z*
p̂ (1- p̂)
0.743( 0.257)
= 0.743±1.645
= ( 0.6822, 0.8038)
n
140
“Via Calculator”
1-PropZInt
x = 104
n = 140
C-Level = 0.90
( 0.6821, 0.80362)
We are 90% confident that the proportion of all employees at this company that are content
with the medical benefits is between 68.21% and 80.362%.
(b) “Via Calculator”
1-PropZInt
x = 104
n = 140
C-Level = 0.925
( 0.67709, 0.80862)
We are 92.5% confident that the proportion of all employees at this company that are content
with the medical benefits is between 67.709% and 80.862%.
(c) “Via Calculator”
1-PropZInt
x = 104
n = 140
C-Level = 0.974
( 0.66062, 0.82509)
We are 97.4% confident that the proportion of all employees at this company that are content
with the medical benefits is between 66.062% and 82.509%.
Problem 2
2
æ z* ö
æ 1.645 ö
(a) n = 0.25ç
÷ = 270.6025
÷ = 0.25ç
è 0.05 ø
è ME ø
2
Thus we need 271 residents.
2
æ z* ö
æ 1.645 ö
(b) n = p̂ (1- p̂) ç
÷ = 235.532
÷ = 0.32 ( 0.68) ç
è 0.05 ø
è ME ø
2
2
æ z* ö
æ 1.96 ö
(c) n = 0.25ç
÷ =1067.111
÷ = 0.25ç
è 0.03 ø
è ME ø
Thus we need 236 residents.
2
Thus we need 1068 residents.
2
æ z* ö
æ 1.96 ö
(b) n = p̂ (1- p̂) ç
=
0.32
0.68
÷
( ) çè ÷ø = 928.814
0.03
è ME ø
2
Thus we need 929 residents.
Problem 3
Since sample size is inversely proportional to the square of the ME, multiplying ME by ¼
would correspond to multiplying n by 16. Thus we need a sample of size 1600.
Problem 4
Since sample size is inversely proportional to the square of the ME, multiplying ME by ½
would correspond to multiplying n by 4. Thus we need a sample of size 320.