Question

A market research company conducted a survey to find the level of affluence in a city. They defined the category "affluence" for people earning $100,000 or more annually. Out of 267 persons who replied to their survey, 32 are considered affluent. What is the 95% confidence interval for this population proportion? Answer choices are rounded to the hundredths place

Answer 1

0.08 to 0.16

Explanation:

Answer 2

A 95% confidence interval for this population proportion is [0.081, 0.159].

Explanation:

We are given that a market research company conducted a survey to find the level of affluence in a city.

Out of 267 persons who replied to their survey, 32 are considered affluent.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

P.Q. =
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` ~ N(0,1)

where,
`\hat p` = sample proportion of people who are considered affluent =
`(32)/(267)` = 0.12

n = sample of persons = 267

p = population proportion

Here for constructing a 95% confidence interval we have used One-sample z-test for proportions.

So, 95% confidence interval for the population proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 <
`\frac{\hat p-p}{\sqrt{(\hat p(1-\hat p))/(n) } }` < 1.96) = 0.95

P(
`-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` <
`{\hat p-p}` <
`1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.95

P(
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` < p <
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ) = 0.95

95% confidence interval for p = [
`\hat p-1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ,
`\hat p+1.96 * {\sqrt{(\hat p(1-\hat p))/(n) } }` ]

= [
`0.12-1.96 * {\sqrt{(0.12(1-0.12))/(267) } }` ,
`0.12+1.96 * {\sqrt{(0.12(1-0.12))/(267) } }` ]

= [0.081, 0.159]

Therefore, a 95% confidence interval for this population proportion is [0.081, 0.159].