Please show your work Please show your work 2 Suppose that i

Please show your work

Please show your work 2. Suppose that in Problem 1 only 25 of 36 randomly selected family caregivers responded, so we have the following data: Assuming that these data come from a population that has a normal distribution, (a) Determine a 99% confidence interval for the mean ages of all US caregivers. (b) Determine a one-sided lower and one-sided upper 99% confidence interval for the mean ages of all US caregivers.

Solution

(a) sample mean=46.56

sample standard deviation = 3.697747

Given a=1-0.99=0.01, Z(0.005) = 2.58 (from standard normal table)

So the lower bound is

xbar - Z*s/vn =46.56 -2.58*3.697747/sqrt(25) =44.65196

So the upper bound is

xbar +Z*s/vn =46.56 +2.58*3.697747/sqrt(25) =48.46804

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(b) Given a=0.01, Z(0.01)=2.33 (from standard normal table)

So the lower bound is

xbar - Z*s/vn =46.56 -2.33*3.697747/sqrt(25) =44.83685

So the upper bound is

xbar + Z*s/vn =46.56 +2.33*3.697747/sqrt(25) =48.28315