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

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.  =    ~ N(0,1)

where, = sample proportion of people who are considered affluent = = 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 < < 1.96) = 0.95

P( < < ) = 0.95

P( < p < ) = 0.95

95% confidence interval for p = [ , ]

    = [ , ]

    = [0.081, 0.159]

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

i agree because i see the other persons thinking is clear

answer: 8 years

step-by-step explanation:

the explanation is that in 8 years the child will be 20 and in 8 years the father will be 40 and 20 is half of 40.