SOLUTION: Really Big Medical Center (RBMC) is testing patients who have spent time in grassy areas this summer and fall for Lyme disease. Suppose that the probability of someone with exposur

Question 1200621: Really Big Medical Center (RBMC) is testing patients who have spent time in grassy areas this summer and fall for Lyme disease. Suppose that the probability of someone with exposure to tall grass having Lyme is .3 (or 30 percent). You sample of 15 patients. You expect that 95% of the time the number of patients with Lyme disease will be between
and
. (HINT: Rememer the Empirical Rule.) Where applicable, keep 4 decimal places on intermediate calculations. Round your answers to 2 decimal places and put the numbers in carefully with no leading or trailing spaces. You may use leading zeros if needed.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
population ratio is assumed to be .3
population standard error is equal to sqrt(.3 * .7 / 15) = .1183215957.
critical z-score at 95% confidence interval is plus or minus z = 1.959963986.
use the z-score formula to find the raw score.
z = (x-m)/s is the z-score formula.
z is the z-score
x is the sample mean proportion.
m is the assumed population mean proportion.
s is the standard error.
on the low side of the 95% confidence interval, the formula becomes:
-1.959963986 = (x - .3) / .1183215957.
solve for x to get x = -1.959963986 * .1183215957 + .3 = .0680939337.
on the high side of the 95% confidence interval, the formula becomes:
1.959963986 = (x - .3) / .1183215957.
solve for x to get x = 1.959963986 * .1183215957 + .3 = .5319060663.
your 95% confidence interval is from .07 to .53.
here's what it looks like on the z-score graph at https://davidmlane.com/hyperstat/z_table.html

the 95% confidence interval is a two tailed confidence interval with an alpha of .05 / 2 = .025 on each end.

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