Question

We want to estimate the proportion of people who have or are carriers of hepatitis B in a certain country, using 90% confidence and with a desired margin of error of 1.5%. What size sample should be used if:

a. no prior estimate of the desired proportion is known. (6 pts.)

b. a pilot study suggests that the proportion is 6%

Answer 1

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 1.5% = 0.015

At 90% confidence level

= 1 - 0.90 = 0.10

/2 =0.05

Z_/2 = 1.645 ( Using z table )

Sample size = n = (Z_/2 / E)² * * (1 - )

= (1.645 / 0.015)² * 0.5* 0.5

=3006.69

Sample size =3007

(B)

Solution :

Given that,

= 0.06

1 - = 1 - 0.06= 0.94

margin of error = E = 1.5% = 0.015

At 90% confidence level

= 1 - 0.90 = 0.10

/2 =0.05

Z_/2 = 1.645 ( Using z table )

Sample size = n = (Z_/2 / E)² * * (1 - )

= (1.645 / 0.015)² * 0.06* 0.94

=678.3

Sample size = 679