Question 6 a A manufacturer of golf equipment wishes to esti

Question #6:

a)  A manufacturer of golf equipment wishes to estimate the number of left-handed golfers in the U.S.   How large of a sample is needed in order to be 99% confident that the sample proportion will not differ from the population proportion by more than 3%?

b)  In order to set rates, an insurance company is trying to estimate the number of sick days that full time workers at an auto repair shop take per year. A previous study indicated that the population standard deviation was 2.1 days.   How large a sample must be selected if the company wants to be 92% confident that the population mean differs from the sample mean by no more than 1/2 day?

Solution

(a) Given a=0.01, Z(0.005) = 2.58 (from standard normal table)

We use p=0.5 as estimated

So n=(Z/E)^2*p*(1-p)

=(2.58/0.03)^2*0.5*0.5

=1849

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(b) Given a=1-0.92=0.08, Z(0.08/2)=Z(0.04) =1.75 (from standard normal table)

So n=(Z*s/E)^2

=(1.75*2.1/0.5)^2

=54.0225

Take n=55