Question

Determine the sample size n needed to construct a 90​% confidence interval to estimate the population proportion for the following sample proportions when the margin of error equals 8​%.

a. p over bar equals 0.10

b. p over bar equals 0.20

c. p over bar equals 0.30

Click the icon to view a table of standard normal cumulative probabilities.

Answer 1

Solution :

Given that,

a) = 0.10

1 - = 1 - 0.10 = 0.90

margin of error = E = 8% = 0.08

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = 1.645

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

= (1.645 / 0.08)² * 0.10 * 0.90

= 38.05

sample size = n = 39

b) = 0.20

1 - = 1 - 0.20 = 0.80

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

= (1.645 / 0.08)² * 0.20 * 0.80

= 67.65

sample size = n = 68

c) = 0.30

1 - = 1 - 0.30 = 0.70

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

= (1.645 / 0.08)² * 0.30 * 0.70

= 88.79

sample size = n = 89