A shipment of 1000 small lab mice arrives at the animal care

A shipment of 1000 small lab mice arrives at the animal care facility with the weight following a skewed distribution (mu = ? and sigma =2.5). A sample of 40 mice is selected and weighed. The sample average weight is 10.2 g, and the sample standard deviation is 2 g.

In order to find a 90% confidence interval for the mean weight of all the mice in the shipment.

                   The margin error is?

                   The 90% confidence interval is ?

Solution

Since the sample size is larger than 30, we can use the normal distribution approximated based on central limit theorm.

Given a=1-0.9=0.1, Z(0.05) = 1.645 (from standard normal table)

So The margin error is

Z*s/vn = 1.645*2/sqrt(40) =0.5201947

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So the lower bound is

xbar - Z*s/vn =10.2 - 0.5201947 =9.679805

So the upper bound is

xbar + Z*s/vn =10.2 + 0.5201947 =10.72019