Question

The tensile strength in MPa (megapascals) measured for a random sample of 20 pieces of die-cast aluminium is displayed in the following stem-and-leaf plot: 17 007 leaf unit = 1 18 3 19 47 20 125 21 22 23 23 04448 24 014 25 26 2 The sample mean and standard deviation are: y = 215.05 s=27.19225 Assuming that the tensile strength measurements are Normally distributed, what is the width (upper limit minus lower limit) of the 90% confidence interval for the population standard deviation? Select one: O a. 15.676 O b. 12.000 O c. 20.145 O d. 11.207 O e. 21.028

Answer 1

Sol:

n=20

df=n-1=20-1=19

alpha=0.10

alpha/2=0.10/2=0.05

critical value

==CHISQ.INV.RT(0.05,19)

=30.14352721

90% lower limit for standard deviation

sqrt((n-1)*s^2/chi sqalpha/2)

=sqrt((20-1)*27.19225^2)30.14352721)

=21.58862

ANSWER:

MARK

21.028