Question

Help Please!
1,200 employees of a company were surveyed to find out whether they were satisfied with the company’s insurance policy. The survey showed that 80% of the respondents were not satisfied with the policy. The standard error of the proportion is ____% The number of people who are satisfied with the policy is between ____% and ___%
I'll post below what I have done so far which isn't much.

Answer 1

p=0.8
n=1200
The standard error of proportion is +/- 1.15

I used online calculator but still unsure if I even used that correctly.
Struggling with the z-table so maybe if someone could show me what to look for.

The number of people who are satisfied with the policy is between 18.85% and 21.15%
I got this answer from the difference of the standard error.

Answer 2

@amistre64 are u busy? I could use some help please. :)

Answer 3

im always busy ... even when im not

Answer 4

sqrt(pq/n) is the formula for standard error of a proportion, which is not what you have. can you prove me wrong?

Answer 5

lol no I trust you way better than me.

Answer 6

there is no 'confidence' value to determine an interval with.

Answer 7

so I have to find the confidence value it wasn;t given

Answer 8

what would I go with then

Answer 9

it should be stated, or some margin of error expressed. you can define an interval that i know of without basing it on something

Answer 10

They gave me choices could I figure it out through process of elimination?

Answer 11

maybe

Answer 12

ok one sec

Answer 13

so which function would you be more impressed with? the boring line, or the cool stepping one?

http://assets.openstudy.com/updates/attachments/5557f409e4b0db8ecf8fb455-amistre64-1431828754599-untitled.jpg

Answer 14

1,200 employees of a company were surveyed to find out whether they were satisfied with the company’s insurance policy. The survey showed that 80% of the respondents were not satisfied with the policy. The standard error of the proportion is ( +/-1.15 or +/-2.3 )% The number of people who are satisfied with the policy is between (78.85 or 18.85)_% and (81.15 or 21.15) %

Answer 15

that really doesnt look like anything that would correspond the the information to me,

can you take a screen shot by chance?

Answer 16

hold on

Answer 17

hmm, yeah, thats not making sense to me. srry

Answer 18

ok thanks

Answer 19

btw I am more impressed with the stepping stone. :)

Answer 20

\[p\pm E\]
\[p\pm z\sqrt{\frac{pq}{n}}\]
lol, me too ... hits all the required points that the line hits

Answer 21

i just dont see how they are figuring their spread.

Answer 22

I have been working for days to get ready for the exam and I have lots of faulty questions in my packet. Happens all the time. If I find out I will let you know.

Answer 23

those are percents

sqrt(.8*.2/1200) = .0115.... or 1.15%

Answer 24

I still need to figure out and make sure i know the z table/score

Answer 25

or rather

1200(.8) = +- sqrt(1200(.8)(.2))

Answer 26

yeah but those were the only choices I had.

Answer 27

Thanks for trying. Have a good night or day wherever in the world you might happen to be. :)

Answer 28

80 -+ 1.15 = 78.85 and 81.15

80 -+ 2.30 = 77.70 and 82.30

using ( +/-1.15 or +/-2.3 )

(78.85 or 18.85) and (81.15 or 21.15)
?????? ??????

those other 2 dont even come up in the setup

Answer 29

my educated guess is 1.15 and 78 and 81

Answer 30

the only problem with this is, we can only be 68% confident of this interval, which isnt very good. we would rather be 95% or 99% confident.

Answer 31

Thank you