construct a confidence interval of the population proportion

construct a confidence interval of the population proportion at the given level of confidence x=120,n=1200,99%. The upper bound of the confidence interval is? Round to the nearest thousandth as needed.

a researchers wishes to estimat the proportion of adults who have high speed internet access.what size sample should be obtain if he wishes the estimate to be within 0.04 with 90% confidence if (a) she uses a previous estimate of 0.34? (b) she does not use any prior estimates?

(a)n=? round to the nearest integer

(b) n=? round to the nearest integer.

Solution

1. Construct a confidence interval of the population proportion at the given level of confidence

x=120, n=1200, 99% confidence (the upper bound and lower bound, round to the nearest thousandth as needed)

Given a=0.01, Z(0.005) = 2.58 (from standard normal table)

p = 120/1200=0.1

So 99% confidence interval is

p +/- Z*sqrt(p*(1-p)/n)

--> 0.1 +/- 2.58*sqrt(0.1*0.9/1200)

--> (0.078, 0.122)

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2. A researcher wishes to estimate the percentage of adults who support abolishing the penny. What size sample should be obtained if he wishes the estimate to be within 4 percentage points with 90% confidence if

(a) he uses a previous estimate of 25%

Given a=0.1, Z(0.05) = 1.65 (from standard normal table)

So n=(Z/E)^2*p*(1-p)

=(1.65/0.04)^2*0.25*0.75

=319.043

Take n=320

(b) he does notuse any prior estimates?

n=(Z/E)^2*p*(1-p)

=(1.65/0.04)^2*0.5*0.5

=425.3906

Take n=426