Question 1007303
i'm thinking selection A.


here's why:


p = the proportion.
q = 1 minus the proportion.
n = number of sample entries
s = standard deviation of the proportion = sqrt(p*q/n)



you have:


p = .4
q = .6
n = 250
s = sqrt(.4*.6/250) = .0309838668


the 90% two tail confidence interval is equal to the area under the distribution curve between a z-factor of -1.644853626 and a z-factor of +1.644853626.


the z-factor tells you how many standard deviations you are above or below the mean.


the formula for the z-factor is:


z = (x-m) / s


z = the z-factor
x = the raw score which will be a proportion in this case.
m = the mean which will also be a proportion in this case.
s = the standard deviation which will also be a proportion in this case.


you have:


z1 = -1.644853626
z2 = +1.644853626
x1 or x2 = what you want to find.
m = .4
s = .0309838668


the basic formula of z = (x-m)/s becomes:


z1 = (x1-m)/s
z2 = (x2-m)/x


solve for x1 or x2 in each of these formulas andyou get:


x1 = z1*s+m
x2 = z2*s+m


solve for x1 and x2 and you will get:


x1 = .3490360744
x2 = .4509639256


that rounds to .35 to .45


that's selection A.


even if you had rounded off intermediate results, you should still have gotten the same solution.


.0309 rounds to .03


1.645 rounds to 1.65.


x1 = .3505 which rounds to .35
x2 = .4495 which rounds to .45


even though it didn't make a difference in this case, rounding of intermediate results is usually not recommended unless your instructor tells you to do so.


i used a calculator to get the z score.


basically a 90% two tailed confidence interval leaves a 5% tail on each end.
in the z-score table, you look for a .95 area and get the closest z-score to that.
.95 area to the left of that z-score is the same as .05 area to the right of that z-score.


the .95 is the area to the left of the z-score in most full tables.
the z-score would be shown as 1.64 for .9495 area and 1.65 for .9505 area.
that means the z-score is smack in the midle which would make it 1.645.


that takes care of the area to the left of 1.65.
the area to the left of -1.65 would be .05
.95 - .05 = .90 which is the area between those two z-scores.


the z-score table i used is here.


<a href = "http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf" target = "_blank">http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf</a>


visually, the 90% two tailed confidence interval would look like this:


<img src = "http://theo.x10hosting.com/2015/120201.jpg" alt="$$$" </>


the picture shows you that the 90% two tailed confidence interval is between a z-score of -1.645 and 1.645.