Question

In a survey of 368 people, 81 drink milk straight from the milk carton. For the sample, find the sample proportion, the margin of error, and the interval likely to contain the true population proportion.

a. 22%; b. ±11%; c. 11% to 33%

a. 28%; b. ±5%; c. 23% to 33%

a. 22%; b. ±5%; c. 17% to 27%

a. 28%; b. ±11%; c. 17% to 39%
is this c or a??

Answer 1

@jim_thompson5910

Answer 2

@calculusxy

Answer 3

@agent0smith

Answer 4

?

Answer 5

@agent0smith

Answer 6

Sample proportion should be easy.

Find the formulas for the ME from your notes or something. Confidence interval formulas.

Answer 7

i got this 22%; b. ±11%; c. 11% to 33%

Answer 8

Show work... i'm not about to do the work to check those answers.

Answer 9

phat = 81/368 = 0.2201086957 ≈ 0.22

q = 1 - 0.2201086957 = 0.7798913043 = 0.78

P(0.22 - 1.96*sqrt(0.22*0.78/368) < π < 0.22 + 1.96*sqrt(0.22*0.78/368) ) = 0.95

P(0.22 - 0.04 < π < 0.22 + 0.04) = 0.95

P(0.18 < π < 0.26) = 0.95

sample proportion = 0.18
margin of error = 0.04

Answer 10

so i am guessing on the answer

Answer 11

Looks fine.

Answer 12

so would my answer be correct?

Answer 13

@agent0smith

Answer 14

"Looks fine."

Answer 15

ok i have 4 more could you help me with them?

Answer 16

Possibly.

But if you ask "so would I be correct" after I have already implied that you are, I will not help you again. EVER.

Answer 17

ok