In a survey of 349 teenagers in the United States, 14% said
thatt they worked during the their summer vacation.
What is the margin of error for the survey?
±±
Give an interval that is likely to contain the exact percent of all
U.S. teenagers who worked during their summer vacation?
Between %
and %
here
P=proportion worker say that they are interested in doing work in summer vacation out of total number of survey n=349
Here
p=0.14 n=349
What is the margin of error for the survey?
And
#here we considere the 95%CI
ME=margin of error
ME=±
%
=1-0.95=0.05
=Z0.025=1.96
value of z is obtained from standard normal table
ME=0.019
95% interval that is likely to contain the exact percent of all U.S. teenagers who worked during their summer vacation is
(P-ME,P+ME)
(0.14-0.019,0.14+0.019)
(0.121,0.159)
Between 12.1 % and 15.9 %
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