Question

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 %

Answer #1

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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