Question

According to a health organization in a certain country, 19.6% of the population smoked in 2008 In 2014, a random sample of 650 citizens of this country was selected, 120 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. 

Answer 1

Let p be the sample proportion of people who smoked.

The sample proportion which smoked is 120/650 = 0.1846154= 0.185

Thus, q, the sample proportion which did not smoke is 1-p= 0.8153846= 0.815

The confidence interval can be found using:

p $\pm$ z* sqrt((p*q)/n)

= 0.185 $\pm$ 2.58 * sqrt ( (0.185*0.815) / 650 )

= 0.185 $\pm$ 2.58 * sqrt ( 0.0002319615)

= 0.185 $\pm$ 2.58 * 0.01523028

= 0.185 $\pm$ 0.03929412

Thus, the lower limit = 0.1457059= 0.146

Thus, the upper limit = 0.2242941= 0.224