Q7) Using 6.84 as a planning value for the population standard deviation.

1. What sample size is required to obtain a margin of error of 1.5 and a level of confidence of 0.95? ____________

2. What sample size is required to obtain a margin of error of 2 and a level of confidence of 0.90? ____________

 

Formula for sample size (n):

n = {[(z_α/2)*(σ)]/E}²

where:

z_{α/2 =} confidence interval

σ = standard deviation

E = Margin of Error

1)

Given:

significance level = 1 - confidence interval

= 1 - 0.95

= 0.05

α/2 = significance level/2

= 0.05/2

= 0.025

*find the z-value given p = 0.025

z_α/2 =1.96

σ = 6.84

E = 1.5

n = {[(1.96)*(6.84)]/1.5}²

n = 79.88 or 80

2)

Given:

significance level = 1 - confidence interval

= 1 - 0.90

= 0.10

α/2 = significance level/2

= 0.10/2

= 0.05

*find the z-value given p = 0.05

z_α/2 = 1.645

σ = 6.84

E = 2

n = {[(1.645)*(6.84)]/2}²

n = 31.65 = 32