Question

a simple random sample of 17 containers showed a mean volume of 1.0183 liters and a standard deviation .003 liters. construct the 99% confidence interval estimate of the standard deviation if the volume for all such containers

a. state the two values

b. show the formula and computations used in computing the confidence interval. state the confidence interval rounding to three decimals

Answer 1

Solution :

Given that,

c = 0.99

s = 0.003

n = 17

a)

At 99% confidence level the $\chi ^{2}$ is ,

$\alpha$ = 1 - 99% = 1 - 0.99= 0.01

$\alpha$ / 2 = 0.01/ 2 = 0.005

$\chi ^{2}$_{$\alpha$ /2,df} = $\chi ^{2}$ _0.005,16 = 34.27

and

$\chi ^{2}$_{1-$\alpha$ /2,df} = $\chi ^{2}$ _0.995,16 = 5.14

$\chi$²_L = $\chi$ ²_{$\alpha$/2,df} = 34.27

$\chi$²_R = $\chi$ ²_{1 - $\alpha$ /2,df} = 5.14

b)

The 99% confidence interval for $\sigma$ is,

s $\sqrt{}$ (n-1) / $\chi ^{2}$ _{$\alpha$ /2,df} < $\sigma$ < s $\sqrt{}$ (n-1) / $\chi ^{2}$ _{1-$\alpha$ /2,df}

0.003$\sqrt{}$( 17 - 1 ) / 34.27< $\sigma$ < 0.003$\sqrt{}$( 17- 1 ) / 5.14

0.002 < $\sigma$ < 0.005

( 0.002 , 0.005)