Question

1. A population standard deviation is 5.  We want to estimate the population mean within 2, with a 90% level of confidence. How large a sample size is required?

2. The estimate of the population proportion should be within plus or minus 0.05 with a 99% confidence level. The best estimate for the population is 0.42. How large a sample size is required?

Answer 1

Solution :

Given that,

standard deviation = =5

Margin of error = E = 2

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.645

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

n = ( 1.645 * 5/ 2 )2

n =16.91

Sample size = n =17

(B)

Solution :

Given that,

= 0.42

1 - = 1 - 0.42 = 0.58

margin of error = E = 5% = 0.05

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58    ( Using z table )

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

= (2.58 / 0.05)2 * 0.42 * 0.58

=648.5996

Sample size = 649