17 A random sample of n 7 wheels of cheese yielded sample s

17 A random sample of n = 7 wheels of cheese yielded sample standard derivation, measured in pounds, as s = 7/5. Assume the weight is normally distributed N(Mu, sigma^2). Find a 99.0% confidence interval for sigma

Solution

Sample Size

7

Sample Standard Deviation = 7/5=

1.4

Confidence Level

99%

Degrees of Freedom

6

Sum of Squares =(n-1)*s^2 =6*1.4*1.4=

11.76

Single Tail Area

0.005

Lower Chi-Square Value

0.6757

Upper Chi-Square Value

18.5476

Results

Interval Lower Limit for Variance=11.76/18.5476=

0.6340

Interval Upper Limit for Variance=11.76/0.6757=

17.4035

Interval Lower Limit for Standard Deviation= sqrt(0.634)=

0.7963

Interval Upper Limit for Standard Deviation=sqrt(17.4035)=

4.1717

Sample Size	7
Sample Standard Deviation = 7/5=	1.4
Confidence Level	99%
Degrees of Freedom	6
Sum of Squares =(n-1)*s^2 =6*1.4*1.4=	11.76
Single Tail Area	0.005
Lower Chi-Square Value	0.6757
Upper Chi-Square Value	18.5476
Results
Interval Lower Limit for Variance=11.76/18.5476=	0.6340
Interval Upper Limit for Variance=11.76/0.6757=	17.4035
Interval Lower Limit for Standard Deviation= sqrt(0.634)=	0.7963
Interval Upper Limit for Standard Deviation=sqrt(17.4035)=	4.1717