Question

Use 6.89 days as a planning value for the population standard deviation. Assuming 95% confidence, what sample size would be required to obtain a margin of error of 1.5 days (round up to the next whole number)? Assuming 90% confidence, what sample size would be required to obtain a margin of error of 2 days (round up to the next whole number)?

Answer 1

Solution :

Given that,

standard deviation =s =   =6.89

Margin of error = E = 1.5

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.96

sample size = n = [Z_/2* / E] ²

n = ( 1.96* 6.89 / 1.5 )²

n =81

Sample size = n =81

(B)

Solution :

Given that,

Margin of error = E = 2

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

sample size = n = [Z_/2* / E] ²

n = ( 1.645 *6.89 /2 )²

n =32

Sample size = n =32