Question

Determine the sample size n needed to construct a 99​% confidence interval to estimate the population mean for the following margins of error when σ=77.

​a) 5

​b) 8

​c) 10

​a)n=

​(Round up to the nearest​ integer.)

​b)n=

​(Round up to the nearest​ integer.)

​c)n=

​(Round up to the nearest​ integer.)

Answer 1

Solution :

Given that,

standard deviation = $\sigma$ = 77.

margin of error = E = 5

a) At 99% confidence level the z is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

Sample size = n = ((Z_$\alpha$/2 * $\sigma$ ) / E)²

= ((2.576* 77.) / 5)²

= 1573.7 = 1574

Sample size = 1574

b)  margin of error = E = 8

At 99% confidence level the z is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

Sample size = n = ((Z_$\alpha$/2 * $\sigma$ ) / E)²

= ((2.576*77. ) / 8)²

= 614.7 = 615

Sample size = 615

c) margin of error = E = 10

At 99% confidence level the z is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$ / 2 = 0.01 / 2 = 0.005

Z_$\alpha$/2 = Z_0.005 = 2.576

Sample size = n = ((Z_$\alpha$/2 * $\sigma$ ) / E)²

= ((2.576 *77. ) / 10)²

= 393.4 = 393

Sample size = 393