use the offered data to uncover the minimum sample size forced to calculation a population proportion or percent Margin the error: 0.005, confidence level: 99%, p and q unknown
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uneven the population proportion is really skewed, you have the right to use p=.5 and q=.5

E=.005; to trust level of 99% means `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

Thus to obtain an error that 1/2% through 99% confidence, you require a sample size of at the very least 66564.

** You may question the...


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Unless the populace proportion is an extremely skewed, you deserve to use p=.5 and q=.5

E=.005; to trust 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

Thus to gain an error the 1/2% v 99% confidence, you require a sample dimension of at least 66564.

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** You may question the ability to use p=q=.5 As lengthy as the population isn"t really divided, the number are around the same. Utilizing 40% and 60% yields 266256*.24=63901. Also 70%-30% gives 266256*.21=55914, so the use of 50-50 is justified. **


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