Question

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 70 cookies. The mean is 22.85 and the standard deviation is 2.16 construct a 80% confidence interval estimate of the standard deviation of the numbers in all such cookies.

Answer 1

Sol:

n=70

s=2.16

c=0.80

alpha=1-0.80=0.20

alpha/2=0.20/2=0.10

chisq alpha/2

==CHISQ.INV(0.1,69)

1-alpha/2=1-0.80/2=0.90

=54.4381

chisq 1-alpha/2

=CHISQ.INV(0.9,69)

=84.41787

80% confidence interval for standard deviation is

sqrt(n-1* s/chisq 1-alpha/2)<sigma<,sqrt( n-1* s/chisq alpha/2)

sqrt((70-1)*2.16^2/84.41787)<sigma<sqrt((70-1)*2.16^2/54.4381)

1.952815<sigma< 2.431794

80% lower limit sigma=1.952815

80% upper limit sigma=2.431794