Question 1015179: Determining Sample Size. Use the given data to find the minimum sample size required to estimate a population proportion or percentage. Find the sample size needed to estimate the percentage of robberies in Texas that result in arrests. Use a 0.04 margin of error, use a confidence level of 90%, and assume that P and Q(P and Q with the little hat on top) are unknown.
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! since we do not know the values of P and Q, we assume that P=0.50 and then Q=0.5
********************************************************************************
Margin of Error(ME) = z * sqrt( (P * Q) / n ), where n the sample size and z is the z-score, now
:
Our level of confidence is 90% or 0.90 which is a
:
1 - a = 1 - 0.90 = 0.10
:
divide 0.10 / 2 = 0.05 and
:
z of (0.05) is 1.645 from z-tables
(look for 0.05 probability and choose corresponding z-value, note that for this problem we use -1.645 as 1.645 since we are interested in how many standard deviations we are from the mean)
:
therefore,
0.04 = 1.645 * sqrt( (0.5 * 0.5) / n )
:
(0.5 * 0.5) / n = (0.04/1.645)^2
:
0.25/n = 0.000591273
n = 0.25 / 0.000591273 = 422.81640625 approx 423
:
Our sample size is 423
:
|
|
|
