Question

Use the given information to find the number of degrees of​ freedom, the critical values chi Subscript Upper L Superscript 2 and chi Subscript Upper R Superscript 2​, and the confidence interval estimate of sigma. It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution. Platelet Counts of Women 99​% ​confidence; nequals23​, sequals65.4.

Answer 1

Solution:

n=23

s=65.4

confidence level=0.99

alpha=0.01

alpha/2=0.01/2=0.005

1-alpha/2=1-0.005=0.995

df = n- 1=23-1=22

99% confidence interval for standard deviation is

sqrt(n*s^2/chisqupper<sigma<sqrt(n*s^2/chisqlower)

chisqupper in excel

=CHIINV(0.005,22)

=  42.79565

chilower in excel is

=CHIINV(0.995,22)

= 8.642716

99% confidence interval for standard deviation is

sqrt{(23-1)*65.4^2/42.79565)<sigma<sqrt{(23-1)*65.4^2/8.642716)

46.891<sigma<104.343

99% lower limit for standard deviation is 46.891

99% upper limit for standard deviation is 104.343