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
(1 point) A confidence interval for the true mean of the annual medical expenses of a...
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