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

1. 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

2. theopenstudyowl

I think it's C

3. Michele_Laino

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

4. theopenstudyowl

sure, one sec...

5. theopenstudyowl

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

6. Michele_Laino

ok!

7. theopenstudyowl

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

8. Michele_Laino

ok!

9. 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

10. anonymous

I'm sure it's b

11. theopenstudyowl

prove it

12. Michele_Laino

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

13. Michele_Laino

oops...1- (243/537)

14. Michele_Laino

I think that we have to use the binomial distribution

15. Michele_Laino

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

16. Michele_Laino

whereas the mean is: $\Large m = Np = 243$

17. Michele_Laino

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

18. theopenstudyowl

ok

19. Michele_Laino

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

20. theopenstudyowl

oh ok

21. theopenstudyowl

I think my answer is right! Thanks

22. Michele_Laino

ok! Thanks!

23. theopenstudyowl

is it right?

24. theopenstudyowl

did u get the same result?

25. Michele_Laino

please I continue then

26. theopenstudyowl

ok

27. theopenstudyowl

sorry

28. 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 $\Large \left( {0.412,\quad 0.492} \right)$

29. theopenstudyowl

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

30. Michele_Laino

yes!

31. theopenstudyowl

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

32. Michele_Laino

thanks! :)