Question 1084647
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The formula to be used is
n = ( (z*sigma)/E )^2


where,
n = min sample size needed
z = critical value drawn from the standard normal (Z) distribution
sigma = population standard deviation
E = margin of error desired


At 95% confidence, we can find the z critical value to be approximately z = 1.96
Look at the <a href="http://www.sjsu.edu/faculty/gerstman/StatPrimer/t-table.pdf">statistics table</a>. 
Locate "95%" at the bottom where the confidence levels are. The value just above "95%" is "1.960" which is the same as 1.96
A table like the one I have linked is often found in the back of any statistics textbook. 
So this is why z = 1.96, which is approximate.


The variance is given to be 10.8. Take the square root of the variance to get the standard deviation (sigma)
sigma = sqrt(variance)
sigma = sqrt(10.8)
sigma = 3.286335345031
which is approximate


The error desired is E = 0.7 because we want the upper or lower end of the confidence interval to be at most 0.7 units away from the center xbar.


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Summarizing what we have so far:


z = 1.96
sigma = 3.286335345031
E = 0.7


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Plug those three values mentioned above into the formula below


n = ( (z*sigma)/E )^2


n = ( (1.96*3.286335345031)/0.7 )^2


n = ( (6.44121727626076)/0.7 )^2


n = ( 9.2017389660868 )^2


n = 84.6720000000002


n = 85 <font color=blue>Round up to the nearest whole number</font>


Min Sample Size: n = 85


On the last step, we ALWAYS round up regardless of the decimal portion. 
This rounding up is to ensure that we clear the hurdle. 
If we rounded down, then E > 0.7, which is too large. 
The goal is to make {{{E <= 0.7}}}


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Final Answer: <font color=red>85</font>
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