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

Answer:

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

Step-by-step 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.  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of people who are considered affluent = $\frac{32}{267}$ = 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 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.96) = 0.95

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

P( $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.95

95% confidence interval for p = [ $\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

    = [ $0.12-1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ , $0.12+1.96 \times {\sqrt{\frac{0.12(1-0.12)}{267} } }$ ]

    = [0.081, 0.159]

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