Question

According to an April 2019 Gallup survey, 112 out of 160 of seniors, defined as those 65 and older, reported being concerned that rising healthcare costs will result in significant and lasting damage to the economy.

a) Construct a 90% confidence interval for the population proportion of seniors concerned about rising costs of healthcare. (2 decimal places)

b) What is the point estimate and margin of error for this confidence interval?

c) Interpret your confidence interval.

Answer 1

Given

x=112

n=160

# 90% confidence interval for the population proportion of seniors concerned about rising costs of healthcare. is

(p^-Zc*sqrt(p^*(1-p^)/n)<P<p^+Zc*sqrt(p^*(1-p^)/n)

p^=X/n=112/160=0.7

Zc=Z0.1/2=1.645

value of z is from standard normal table

(0.7-1.645*sqrt(0.7*(1-0.7)/160)<P<0.7+1.645*sqrt(0.7*(1-0.7)/160))

(0.7-0.06<P<0.7<0.7+0.06)

(0.64<P<0.76)-----is 90 % CI for population proportion

Ansb:

point estimate=P^=0.7

margin of error=Zc*sqrt(p^*(1-p^)/n)

=1.645*sqrt(0.7*(1-0.7)/160)

=0.06

Ansc:

there is 90 % Confidence that true population proportion lies between 0.64 and 0.76