Question

a.   How many people must be surveyed to obtain the percent of the population that has health  insurance to within 4% with 90% confidence?

               b.   How many subjects must be sampled to estimate the mean annual expense of a family on cell phones to within $4

                     with 99% confidence. Assume the population standard deviation is $9.

Answer 1

a

Solution :

Given that,

= 0.5

1 - = 1 - 0.5 = 0.5

margin of error = E = 4% = 0.04

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645 ( Using z table )

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

= (1.645 / 0.04)2 * 0.5 * 0.5

=422.816

Sample size = 423

b.

Given that,

standard deviation = =9

Margin of error = E = 4

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58

sample size = n = [Z/2* / E] 2

n = ( 2.58*9 / 4 )2

n =33.96

Sample size = n =34