Question

Suppose the population variance is unknown and the sample standard deviation is 1.9. Compute the correct quantile for each confidence interval: (Use 3 decimal places)

(a) A 95% confidence interval from a sample of size 18.

(c) A 80% confidence interval from a sample of size 3.

Answer 1

a)

	here n =	18
	s2=	3.610
Critical value of chi square distribution for n-1=17 df and 95 % CI
Lower critical value χ2L=		7.564
Upper critical valueχ2U=		30.191

for Confidence interval of standard deviation:
Lower bound =√((n-1)s2/χ2U)=sqrt((18-1)*(3.61/30.191))=		1.426
Upper bound =√((n-1)s2/χ2L)=sqrt((18-1)*(3.61/7.564))=		2.848
from above 95% confidence interval for population standard deviation =(1.426<σ<2.848)

c)

Critical value of chi square distribution for n-1=2 df and 80 % CI
Lower critical value χ2L=		0.211
Upper critical valueχ2U=		4.605

for Confidence interval of standard deviation:
Lower bound =√((n-1)s2/χ2U)=		1.252
Upper bound =√((n-1)s2/χ2L)=		5.850
80% confidence interval for population standard deviation =(1.252<σ<5.85)