8 8 points Exercise 1436 More about deer mice The 14 deer mi

8. (8 points) Exercise 14.36 More about deer mice. The 14 deer mice described in the previous exercise had an average body length of =91.1 mm. Assume that the standard deviation of body lengths in the population of all deer mice in the rich forest habitat is the same as the = 8 mm for the general deer mouse population.

(a) Following your approach in the previous exercise, now give a 90% confidence interval for the mean body length of all deer mice in the forest habitat.

(b) This confidence interval is shorter than your interval in the previous exercise, even though the intervals come from the same sample. Why does the second interval have a smaller margin of error?

Solution

a)
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=91.1
Standard deviation( sd )=8
Sample Size(n)=14
Confidence Interval = [ 91.1 ± Z a/2 ( 8/ Sqrt ( 14) ) ]
= [ 91.1 - 1.64 * (2.138) , 91.1 + 1.64 * (2.138) ]
= [ 87.594,94.606 ]