Question

I want to calculate a sample size before I collect CATEGORICAL data. If I want no more than 3% margin of error find the sample size I would need for the following intervals.

a8) 85%

a9) 90%

a10) 95%

Repeat a8 - 10, but this time with a 5% margin of error

a11) 85%

a12) 90%

a13) 95%

Answer 1

Solution :

Given that,

= 0.50

1 - = 1-0.50 = 0.50

1) margin of error = E = 0.03

a8)

At 85% confidence level

= 1-0.85% =1-0.85 =0.15

/2 =0.15/ 2= 0.075

Z/2 = Z_{0.075 = 1.44}

Z_/2 = 1.44

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.44/0.03)² *0.5 *0.50

= 576

sample size = n = 576

a9)

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z_{0.05 = 1.645}

Z/2 = 1.645

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645/0.03)² *0.5 *0.50

= 752

sample size = n = 752

a10)

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z/2 = 1.960

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.960/0.03)² *0.5 *0.50

= 1067

sample size = n = 1067

2)

margin of error = E = 0.05

a11)

At 85% confidence level

= 1-0.85% =1-0.85 =0.15

/2 =0.15/ 2= 0.075

Z/2 = Z_{0.075 = 1.44}

Z_/2 = 1.44

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.44/0.05)² *0.5 *0.50

= 207

sample size = n =207

a12)

At 90% confidence level

= 1-0.90% =1-0.90 =0.10

/2 =0.10/ 2= 0.05

Z/2 = Z_{0.05 = 1.645}

Z/2 = 1.645

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.645/0.05)² *0.5 *0.50

= 271

sample size = n = 271

a13)

At 95% confidence level

= 1-0.95% =1-0.95 =0.05

/2 =0.05/ 2= 0.025

Z/2 = Z_{0.025 = 1.960}

Z/2 = 1.960

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (1.960/0.05)² *0.5 *0.50

= 384

sample size = n = 384