Question

Determine the sample size n needed to construct a 95?% confidence interval to estimate the population proportion for the following sample proportions when the margin of error equals 55?%.

a. p? = 0.50

b. p = 0.60

c. p = 0.70

Answer 1

Solution:

Given that,

a ) = 0.50

1 - = 1 - 0.50 = 0.50

margin of error = E = 55% = 0.55

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 = 1.960

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

= (1.960 / 0.55)² * 0.50 * 0.50

= 3.1684

n = sample size = 3

b ) = 0.60

1 - = 1 - 0.60 = 0.40

margin of error = E = 55% = 0.55

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 = 1.960

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

= (1.960 / 0.55)² * 0.60 * 0.40

= 3.042

n = sample size = 3

c ) = 0.70

1 - = 1 - 0.70 = 0.30

margin of error = E = 55% = 0.55

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 = 1.960

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

= (1.960 / 0.55)² * 0.70 * 0.30

= 2.6615

n = sample size = 3