Question

Was wondering if someone could go over these homework answers with me before I turn this in.

Answer 1

Which of the following represents the information you need in order to find the sample size necessary for a confidence interval for a single population proportion with a given margin of error and confidence level?

z*, n,

t*, m, p*

t*, n, p*

z*, m, p*

z*, m, s

Answer 2

I think it's C

Answer 3

sorry can you write the meaning of z, m, s, t, p?

Answer 4

sure, one sec...

Answer 5

Z or z refers to a standardized score, also known as a z score.

Answer 6

ok!

Answer 7

its ok, I think I known the answe to this one, could I ask another one in the same thread??

Answer 8

ok!

Answer 9

A company wants to find out what kinds of transportation its employees use to get to work. It conducts a survey of 537 employees, and 243 say they ride the bus. Construct a 95% confidence interval for the proportion of employees who ride the bus to and from work.

(.411, .495)

(.4521, .4539)

(2.168, 2.252)

(.3962, .5098)

(.4316, .4744)
I think it's C

Answer 10

I'm sure it's b

Answer 11

prove it

Answer 12

we have a probability 243/537 for going with bus, and a probability of 1-(243/537) for not going with bus

Answer 13

oops...1- (243/537)

Answer 14

I think that we have to use the binomial distribution

Answer 15

so the standard deviation is:
\[\Large \sigma = \sqrt {Npq} = \sqrt {537\frac{{243}}{{537}}\left( {1 - \frac{{243}}{{537}}} \right)} = ...\]

Answer 16

whereas the mean is:
\[\Large m = Np = 243\]

Answer 17

then we can write:
\[\begin{gathered}
m = Np = 243 \hfill \\
\hfill \\
\sigma = 11.534 \hfill \\
\end{gathered} \]

Answer 18

ok

Answer 19

next, a confidence level of 95% mean a distance about 1.96* sigma

Answer 20

oh ok

Answer 21

I think my answer is right! Thanks

Answer 22

ok! Thanks!

Answer 23

is it right?

Answer 24

did u get the same result?

Answer 25

please I continue then

Answer 26

ok

Answer 27

sorry

Answer 28

referring to our sample of population, we have:
mean value= 243/537=0.452
and
1.96*sigma= 1.96*(11.534/537)=0.04
so our interval, is:
0.452-0.04=0.412 left value
0.452+0.04= 0.492 right value, or

\[\Large \left( {0.412,\quad 0.492} \right)\]

Answer 29

huh, looks like the closest answer is A???

Answer 30

yes!

Answer 31

wow! thanks! And thnxs for walking me through the steps

Answer 32

thanks! :)