Question

Find the sample size needed to give, with 99% confidence a margin of error within +/- 5

σ = 80 n = ____

σ = 30 n =____

σ =10 n = _____

Answer 1

Solution :

Given that,

standard deviation = =80

Margin of error = E = +/-5

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )  

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

n = ( 2.58*80 /5 )2

n =1704.03

Sample size = n =1704

(B)

Solution :

Given that,

standard deviation = =30

Margin of error = E = +/-5

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )  

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

n = ( 2.58*30 /5 )2

n =239.6

Sample size = n =240

(C)

Solution :

Given that,

standard deviation = =10

Margin of error = E = +/-5

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z/2 = 2.58 ( Using z table ( see the 0.005 value in standard normal (z) table corresponding z value is 2.58 )  

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

n = ( 2.58*10 /5 )2

n =26.6

Sample size = n =27