usage the provided information to discover the minimum sample dimension forced to estimate a population propercent or percent Margin of error: 0.005, confidence level: 99%, p and also q unknown
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Unmuch less the population propercentage is extremely skewed, you can use p=.5 and q=.5

E=.005; confidence level of 99% indicates `z_(alpha/2)=2.58` and

`E=z_(alpha/2)sqrt((pq)/n)`

`.005=2.58sqrt((.5*.5)/n)`

`.0019379845=sqrt((.25)/n)`

`266256=n/.25`

n=66564

Therefore to acquire an error of 1/2% with 99% confidence, you need a sample size of at leastern 66564.

** You might question the...


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Unmuch less the population propercent is exceptionally skewed, you deserve to usage p=.5 and q=.5

E=.005; confidence level of 99% suggests `z_(alpha/2)=2.58` and

`E=z_(alpha/2)sqrt((pq)/n)`

`.005=2.58sqrt((.5*.5)/n)`

`.0019379845=sqrt((.25)/n)`

`266256=n/.25`

n=66564

Therefore to acquire an error of 1/2% via 99% confidence, you require a sample size of at least 66564.

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** You might question the capability to use p=q=.5 As long as the population isn"t very split, the numbers are around the exact same. Using 40% and also 60% yields 266256*.24=63901. Even 70%-30% offers 266256*.21=55914, so the use of 50-50 is justified. **


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