Question

In a simple random sample of size 56, taken from a population, 22 of the individuals met a specified criteria.

a) What is the margin of error for a 90% confidence interval for p, the population proportion? Round your response to at least 4 decimal places.

b) What is the margin of error for a 95% confidence interval for p? Round your response to at least 4 decimal places. NOTE: These margin of errors are greater than .10 or 10%.

c) How big of a sample is needed to be certain that we have a margin of error less than .10 (or 10%) at 90% confidence?

d) How big of a sample is needed to be certain that we have a margin of error less than .10 (or 10%) at 95% confidence?

Answer 1

Solution :

Given that,

n = 56

x = 22

Point estimate = sample proportion = = x / n = 22/56 = 0.393

1 - = 1 - 0.393 = 0.607

a)

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

Margin of error = E = Z _{/ 2} * √ (( * (1 - )) / n)

= 1.645 * (√((0.393*0.607) / 56)

= 0.1074

b)

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

Margin of error = E = Z _{/ 2} * √ (( * (1 - )) / n)

= 1.96 * (√((0.393*0.607) / 56)

= 0.1279

c)

margin of error = E = 0.10

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 - )

= (1.645/0.10)² * 0.393*0.607

= 64.55

sample size = 65

d)

margin of error = E = 0.10

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 - )

= (1.96/0.10)² * 0.393*0.607

= 91.64

sample size = 92