Question

According to a health organization in a certain? country, 19.6?% of the population smoked in 2008. In? 2014, a random sample of 750 citizens of this country was? selected, 153 of whom smoked. Complete parts a through c below.
a. Construct a 99?% confidence interval to estimate the actual proportion of people who smoked in this country in 2014.
A 99?% confidence interval to estimate the actual proportion has a lower limit of
nothing and an upper limit of
nothing.
?(Round to three decimal places as? needed.)

Answer 1

Solution :

Given that,

n = 750

x = 153

= x / n = 153 /750 = 0.204

1 - = 1 - 0.204 = 0.796

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576

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

= 2.576 * (((0.204 * 0.796) / 750) =0.038

A 99 % confidence interval for population proportion p is ,

- E < P < + E

0.204 - 0.038 < p < 0.204 + 0.038

0.166 < p < 0.242

Lower limit = 0.166

Upper limit = 0.242