Question 1194768
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Part (a)


Use this table
<a href = "https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf">https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf</a>
to find the z critical value at 90% confidence is approximately z = 1.645


p = 0.32 is the previous estimate of the proportion
E = 0.03 is the error we want (or smaller)


n = sample size
n = p*(1-p)*(z/E)^2
n = 0.32*(1-0.32)*(1.645/0.03)^2
n = 654.256711111111
<font color=red>n = 655</font>


We always round UP to the nearest whole number when doing min sample size problems. 
It doesn't matter that 654.256711111111 is closer to 654 than it is to 655


We round up to clear the hurdle needed. 
Try n = 654 in the margin of error formula
E = z*sqrt(p*(1-p)/n)
and you'll find that E > 0.03 which isn't what we want.
But if you tried n = 655, then E = 0.03 or E < 0.03 would be the case.


<font color=red>Answer: 655</font>


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Part (b)


This is going to be very similar to part (a). 
However, we'll be using p = 0.50 this time. 
This is the most conservative estimate or guess to make for p if we don't know what it is. 
It's right in the middle of the interval {{{0 <= p <= 1}}}


The other values of z = 1.645 and E = 0.03 remain the same from before.


n = sample size
n = p*(1-p)*(z/E)^2
n = 0.50*(1-0.50)*(1.645/0.03)^2
n = 751.673611111111
<font color=red>n = 752</font>
Once again, always round up.


Since p = 0.5, this means p(1-p) = 0.5*(1-0.5) = 0.5*0.5 = 0.25
The formula above updates to n = 0.25*(z/E)^2 when using p = 0.5


<font color=red>Answer: 752</font>



Further Reading:
<a href = "http://www.ltcconline.net/greenl/courses/201/estimation/ciprop.htm">http://www.ltcconline.net/greenl/courses/201/estimation/ciprop.htm</a>
Scroll down to the section with the subheading of "Finding n to Estimate a Proportion". 
I have no affiliation to the websites in either link mentioned.
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