Question

1.) You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 74%. You would like to be 98% confident that your estimate is within 2% of the true population proportion. How large of a sample size is required?

2.) You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=70.8σ=70.8. You would like to be 95% confident that your estimate is within 1.5 of the true population mean. How large of a sample size is required?

3.) You want to obtain a sample to estimate how much parents spend on their kids birthday parties. Based on previous study, you believe the population standard deviation is approximately σ=79.8σ=79.8 dollars. You would like to be 90% confident that your estimate is within 2.5 dollar(s) of average spending on the birthday parties. How many parents do you have to sample?

Answer 1

Solution :

1)

Given that,

= 74% = 0.74

1 - = 1 - 0.74 = 0.26

margin of error = E = 2% = 0.02

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

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

= (2.326 / 0.02)² * 0.74 * 0.26

= 2602.34 = 2603

sample size = 2603

(2)

Solution :

Given that,

standard deviation = = 70.8

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

= ((1.96 * 70.8) / 1.5)²

= 8558.47 = 8559

Sample size = 8559

3)

Solution :

Given that,

standard deviation = = 79.8

margin of error = E = 2.5

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Sample size = n = ((Z_/2 * ) / E)²

= ((1.645 * 79.8) / 2.5)²

= 2757.13 = 2758

Sample size = 2758