Question 930088: Can you please help me with this one? THANK YOU!
The estimate of the population proportion is to be within plus or minus .02, with a 90 percent level of confidence. The best estimate of the population proportion is .16. How large a sample is required? (Round up your answer to the next whole number.)
Sample size
Found 2 solutions by ewatrrr, jim_thompson5910:
Answer by ewatrrr(24785)   (Show Source):
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website! We use the formula n = p(1-p)(z/E)^2 where
n = min sample size
p = sample proportion
z = critical number from the standard normal distribution
E = margin of error
In this case,
p = 0.16 (sample proportion)
z = 1.64485362695147 (based on the 90% confidence interval)
E = 0.02 (Margin of error)
Note: I used a calculator to compute z. You can use a table to find z.
Let's plug those values into the formula and solve for n
n = p(1-p)(z/E)^2
n = 0.16(1-0.16)(1.64485362695147/0.02)^2
n = 909.062600576059 ... use a calculator here
n = 910 ... Always round UP (to the nearest whole number)
Why do we round up? If we rounded down, then we would come up short and the margin of error would be too big. If we used n = 909, then the margin of error would exceed E = 0.02. If we use n = 910, then E would be less than 0.02 as desired.
So the min sample size required is 910
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