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. 0.08 to 0.34 0.08 to 0.16 0.24 to 0.34 0.16 to 0.24

0.08 ; 0.16

Step-by-step explanation:

Given :

x = 32 ; n = 267 ; phat = x / n = 32 / 267 = 0.11985

The Zcritical at 95% = 1.96

The confidence interval for proportion :

C. I = Phat ± Z*√(phat(1-phat))/n

C. I = 0.11985 ± 1.96√(0.11985(1-0.11985))/267

C. I = 0.11985 ± 1.96(0.0198765)

C. I = 0.11985 ± 0.0389581

C. I = 0.08 ; 0.158

C. I = 0.08 ; 0.16

i you goofed up when writing this, because hnot is not a word minusf is not either. i cannot answer   because of that.

step-by-step explanation:

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2