SOLUTION: A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a 0.5% margin of error at a 97.5% confidence l
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Question 1202200: A political candidate has asked you to conduct a poll to determine what percentage of people support her. If the candidate only wants a 0.5% margin of error at a 97.5% confidence level, what size of sample is needed? When finding the z-value, round it to four decimal places. Answer by math_tutor2020(3817) (Show Source):
A 97.5% confidence level means the main body has an area of 0.975
Each tail has an area of (1-0.975)/2 = 0.0125
Then use the TI84 to input invNorm(0.0125,0,1) to get the approximate result of -2.2414; we'll apply absolute value to make it positive.
The input =NORMSINV((1-0.975)/2) will give the z value we're after.
So will =NORMSINV(0.0125)
Don't forget about the equal sign up front when executing commands in a spreadsheet.
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The margin of error is 0.5% meaning we have E = 0.005
The political candidate wants to know the value of n to make this confidence interval
phat - E < p < phat + E
phat - 0.005 < p < phat + 0.005
We're not told phat so we assume a conservative estimate of phat = 0.5 which is right in the middle of
The formula for the margin of error of a proportion is
E = z*sqrt(phat*(1-phat)/n)
Solving for n gets this formula
n = phat*(1-phat)*(z/E)^2
which will determine the min sample size needed
The inputs are
phat = 0.5
z = 2.2414
E = 0.005
n = phat*(1-phat)*(z/E)^2
n = 0.5*(1-0.5)*(2.2414/0.005)^2
n = 50238.7396
n = 50239
For problems like this, always round UP to the nearest whole number.
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