theopenstudyowl
  • theopenstudyowl
Was wondering if someone could go over these homework answers with me before I turn this in.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
theopenstudyowl
  • theopenstudyowl
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
theopenstudyowl
  • theopenstudyowl
I think it's C
Michele_Laino
  • Michele_Laino
sorry can you write the meaning of z, m, s, t, p?

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More answers

theopenstudyowl
  • theopenstudyowl
sure, one sec...
theopenstudyowl
  • theopenstudyowl
Z or z refers to a standardized score, also known as a z score.
Michele_Laino
  • Michele_Laino
ok!
theopenstudyowl
  • theopenstudyowl
its ok, I think I known the answe to this one, could I ask another one in the same thread??
Michele_Laino
  • Michele_Laino
ok!
theopenstudyowl
  • theopenstudyowl
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
anonymous
  • anonymous
I'm sure it's b
theopenstudyowl
  • theopenstudyowl
prove it
Michele_Laino
  • Michele_Laino
we have a probability 243/537 for going with bus, and a probability of 1-(243/537) for not going with bus
Michele_Laino
  • Michele_Laino
oops...1- (243/537)
Michele_Laino
  • Michele_Laino
I think that we have to use the binomial distribution
Michele_Laino
  • Michele_Laino
so the standard deviation is: \[\Large \sigma = \sqrt {Npq} = \sqrt {537\frac{{243}}{{537}}\left( {1 - \frac{{243}}{{537}}} \right)} = ...\]
Michele_Laino
  • Michele_Laino
whereas the mean is: \[\Large m = Np = 243\]
Michele_Laino
  • Michele_Laino
then we can write: \[\begin{gathered} m = Np = 243 \hfill \\ \hfill \\ \sigma = 11.534 \hfill \\ \end{gathered} \]
theopenstudyowl
  • theopenstudyowl
ok
Michele_Laino
  • Michele_Laino
next, a confidence level of 95% mean a distance about 1.96* sigma
theopenstudyowl
  • theopenstudyowl
oh ok
theopenstudyowl
  • theopenstudyowl
I think my answer is right! Thanks
Michele_Laino
  • Michele_Laino
ok! Thanks!
theopenstudyowl
  • theopenstudyowl
is it right?
theopenstudyowl
  • theopenstudyowl
did u get the same result?
Michele_Laino
  • Michele_Laino
please I continue then
theopenstudyowl
  • theopenstudyowl
ok
theopenstudyowl
  • theopenstudyowl
sorry
Michele_Laino
  • Michele_Laino
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 \[\Large \left( {0.412,\quad 0.492} \right)\]
theopenstudyowl
  • theopenstudyowl
huh, looks like the closest answer is A???
Michele_Laino
  • Michele_Laino
yes!
theopenstudyowl
  • theopenstudyowl
wow! thanks! And thnxs for walking me through the steps
Michele_Laino
  • Michele_Laino
thanks! :)

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