Question

In this exercise, we examine the effect of the margin of error on determining the sample size needed.
Find the sample size needed to give, with 99% confidence, a margin of error within ±7. Within ±3. Within ±1. Assume that we use σ˜=20 as our estimate of the standard deviation in each case. Round your answers up to the nearest integer.

ME=7 :	n= ?
ME=3 :	n= ?
ME=1 :	n= ?

Answer 1

Solution :

Given that,

standard deviation =s =   =20

Margin of error = E = 7

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* 20 / 7 )2

n =55

Sample size = n =55

b.

standard deviation =s =   =20

Margin of error = E = 3

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* 20 / 3 )2

n =296

Sample size = n =296

c.

standard deviation =s =   =20

Margin of error = E = 1

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* 20 / 1 )2

n =2663

Sample size = n =2663