Question 1158347: in conducting a hypothetical survey of 650 Kansas city residents, are a researcher found that 45% are fans of the Kansas city Jayhawks
A. Calculate the margin of error with 95% confidence level survey.
B. If the researcher wanted to get a margin of error of 2.5 how many additional Kansas city residents would need to survey? Assume the number of residents who are fans of the Kansas Jayhawks stays at 45%
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 95% CI half-interval for a one sample proportion test is z(0.975)*sqrt(p*(1-p)/n)
=1.96*sqrt(0.45*0.55/650); sqrt term=0.0195
=0.038 is half-interval, and this is the margin of error ANSWER
CI is (0.412, 0.488), adding and subtracting the error to and from the mean
for an error of 0.025 (presumed that is what is being asked for
1.96*sqrt(.45*.55/n)=0.025
1.96*sqrt(0.45*0.55)/0.025=sqrt(n)=39.00
n=1521.27 round to 1522
by rounding earlier, the sample size will be 1521.
There is slight error in the CI in that 45% of 650 is not an integer, so mathematically the CI can be done but on a calculator an integer must be used, so the exact percentage won't be 45%
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