Question 1195763
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Recall that 
E = z*sigma/sqrt(n)
is the margin of error formula for the population mean.


Solve for n to get:
E = z*sigma/sqrt(n)
E*sqrt(n) = z*sigma
sqrt(n) = z*sigma/E
n = (z*sigma/E)^2
This is the minimum sample size formula when dealing with population means.



Use a table like this
<a href = "https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf">https://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf</a>
to determine that z = 2.326 is the critical z value at 98% confidence.
Look at the bottom row in blue. The value 2.326 is just above the 98%


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We have this input info
z = 2.326
sigma = 76.3 = population standard deviation
E = 2.5 = desired margin of error


We wish to have E be 2.5 or smaller


n = min sample size
n = (z*sigma/E)^2
n = (2.326*76.3/2.5)^2
n = 5039.5119498304 ..... this value is approximate
n = <font color=red>5040</font>


We always ROUND UP to the nearest whole number. 
In this case, the number is closer to 5040 than it is to 5039 (since the 0.51 is over 0.5). 
But even if we had something like 5039.00001 we would still round up to 5040.


Why is this? It's because n = 5039 would make the value of E too big. 
We want E to be 2.5 or smaller. As n increases, E decreases and vice versa.


Let's try n = 5039
E = z*sigma/sqrt(n)
E = 2.326*76.3/sqrt(5039)
E = 2.50012699365644
Unfortunately E is too big


Now try n = 5040
E = z*sigma/sqrt(n)
E = 2.326*76.3/sqrt(5040)
E = 2.49987895288062
We are at 2.5 or smaller as desired. 
The threshold has been cleared.


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Final Answer: <font color=red>5040</font>
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