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 95​% ​confidence; nequals22​, sequals65.6.

Answer 1

Solution :

Given that,

c = 0.95

s = 65.6

n = 22

At 95% confidence level the is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

_/2,df = _0.025,21 = 35.48

and

_{1- /2,df} = _0.975,21 = 10.28

²_L = ²_/2,df = 35.48

²_R = ²_{1 - /2,df} = 10.28

The 95% confidence interval for is,

s (n-1) / _/2,df < < s (n-1) / _{1- /2,df}

65.6 ( 22 - 1 ) / 35.48 < < 65.6 ( 22 - 1 ) / 10.28

50.47 < < 93.75

( 50.47 , 93.75)