Question

An article in Cancer Research “Analyses of littermatched time-to-response data, with modifications for recovery of interlitter information,” (1977, Vol. 37, pp. 3863–3868) tested the tumorigenesis of a drug. A total of 41 rats were randomly selected from litters and given the drug. The times of tumor appearance were recorded as follows: 101, 104, 104, 77, 89, 88, 104, 96, 82, 70, 89, 91, 39, 103, 93, 85, 104, 104, 81, 67, 104, 104, 104, 87, 104, 89, 78, 104, 86, 76, 103, 102, 80, 45, 94, 104, 104, 76, 80, 72, 73 Calculate a 95% confidence interval on the standard deviation of time until a tumor appearance. Assume population is approximately normally distributed. Round your answers to 2 decimal places.

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	101	149.3162
	104	231.6332
	104	231.6332
	77	138.7802
	89	0.0482
	88	0.6092
	104	231.6332
	96	52.1212
	82	45.9752
	70	352.7072
	89	0.0482
	91	4.9262
	39	2478.0982
	103	202.1942
	93	17.8042
	85	14.2922
	104	231.6332
	104	231.6332
	81	60.5362
	67	474.3902
	104	231.6332
	104	231.6332
	104	231.6332
	87	3.1702
	104	231.6332
	89	0.0482
	78	116.2192
	104	231.6332
	86	7.7312
	76	163.3412
	103	202.1942
	102	174.7552
	80	77.0972
	45	1916.7322
	94	27.2432
	104	231.6332
	104	231.6332
	76	163.3412
	80	77.0972
	72	281.5852
	73	249.0242
Total	3640	10231.0252

Mean X̅ = Σ Xi / n
X̅ = 3640 / 41 = 88.7805
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 10231.0252 / 41 -1 ) = 15.993

χ² (0.05/2, 41 - 1) = 59.3417
χ² (1 - 0.05/2) , 41 - 1 ) = 24.433
Lower Limit = (( 41-1 ) 255.776 / χ² (0.05/2) ) = 172.4089
Upper Limit = (( 41-1 ) 255.776 / χ² (0.05/2) ) = 418.7386
95% Confidence interval is ( 172.4089 , 418.7386 )

95% Confidence interval for standard deviation is  ( 13.13 < σ < 20.46 ).