Question

In a random sample of 325 students at a university, 260 stated that they were nonsmokers. Based on this sample, compute a 99% confidence interval for the proportion of all students at the university who are nonsmokers. Then find the lower limit and upper limit of the 99% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to two decimal places. (If necessary, consult a list of formulas.) Lower limit: $ ? Upper limit:

Answer 1

Confidence interval is written as

Sample proportion +/- margin of error

The formula for determining margin of error is

`\begin{gathered} \text{margin of error = z }*\text{ }\sqrt[\square]{(pq)/(n)} \\ \end{gathered}`

where

z represents the z score corresponding to the confidence interval

p represents the sample proportion. It also means the probability of success

q represents the probability of failure

q = 1 - p

p = x/n

where

n represents the number of samples

x represents the number of success

From the information given,

n = 325

x = 260

p = 260/325 = 0.8

q = 1 - p = 1 - 0.8 = 0.2

To determine the z score, we would subtract the confidence level from 100 to get alpha, a

Thus

a = 1 - 0.99 = 0.01

a/2 = 0.01/2 = 0.005

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.005 = 0.995

The z score corresponding to the area on the z scores table is 2.576. Thus, the z score for a 99% confidence interval is 2.576

Thus, the 99% confidence interval is

`\begin{gathered} 0.8\text{ }\pm\text{ 2.576 }\frac{\sqrt[\square]{0.8*0.2}}{325} \\ 0.8\text{ }\pm\text{ 2.576 x 0.022} \\ 0.8\text{ }\pm0.057 \end{gathered}`

Thus,

the lower limit is

0.8 - 0.057 = 0.743

the upper limit is

0.8 + 0.057 = 0.857