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

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theopenstudyowl
 one year ago
Best ResponseYou've already chosen the best response.0Which 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
 one year ago
Best ResponseYou've already chosen the best response.0I think it's C

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3sorry can you write the meaning of z, m, s, t, p?

theopenstudyowl
 one year ago
Best ResponseYou've already chosen the best response.0sure, one sec...

theopenstudyowl
 one year ago
Best ResponseYou've already chosen the best response.0Z or z refers to a standardized score, also known as a z score.

theopenstudyowl
 one year ago
Best ResponseYou've already chosen the best response.0its ok, I think I known the answe to this one, could I ask another one in the same thread??

theopenstudyowl
 one year ago
Best ResponseYou've already chosen the best response.0A 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

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3we have a probability 243/537 for going with bus, and a probability of 1(243/537) for not going with bus

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3oops...1 (243/537)

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3I think that we have to use the binomial distribution

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3so the standard deviation is: \[\Large \sigma = \sqrt {Npq} = \sqrt {537\frac{{243}}{{537}}\left( {1  \frac{{243}}{{537}}} \right)} = ...\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3whereas the mean is: \[\Large m = Np = 243\]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3then we can write: \[\begin{gathered} m = Np = 243 \hfill \\ \hfill \\ \sigma = 11.534 \hfill \\ \end{gathered} \]

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3next, a confidence level of 95% mean a distance about 1.96* sigma

theopenstudyowl
 one year ago
Best ResponseYou've already chosen the best response.0I think my answer is right! Thanks

theopenstudyowl
 one year ago
Best ResponseYou've already chosen the best response.0did u get the same result?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3please I continue then

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.3referring 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: 0.4520.04=0.412 left value 0.452+0.04= 0.492 right value, or \[\Large \left( {0.412,\quad 0.492} \right)\]

theopenstudyowl
 one year ago
Best ResponseYou've already chosen the best response.0huh, looks like the closest answer is A???

theopenstudyowl
 one year ago
Best ResponseYou've already chosen the best response.0wow! thanks! And thnxs for walking me through the steps
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