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(1 point) A confidence interval for the true mean of the annual medical expenses of a middle-class American family is given as ($ 746, $775). If this interval is based on interviews with 80 families and a standard deviation of $ 130 is assumed. (a) What is the sample mean of annual medical expenses? (b) What is the confidence level of the interval estimate (as a decimal)?

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Answer #1

a)

here sample mean =(746+775)/2=760.5

b)

margin of error ME =(775-746)/2=14.5

hence crtiical value z =ME*sqrt(n)/std deviaiton =14.5*sqrt(80)/130=~1.00

hence confidence interval =0.68

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