Question

The article “Measuring and Understanding the Aging of Kraft Insulating Paper in Power Transformers” (IEEE Electrical Insul. Mag., 1996: 28–34) contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range:

418, 434, 454, 421, 437, 463, 421, 439, 465, 422, 446, 425, 447, 427, 448, 431, 453

(a) Calculate a 99% confidence interval for the true average degree of polymerization.

(b) Does the confidence interval suggest that µ = 435 is a plausible value for the true average degree of polymerization? (c) Does the confidence interval suggest that µ = 460 is a plausible value for the true average degree of polymerization?

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	418	411.8505
	434	18.4393
	454	246.6753
	421	299.0859
	437	1.6747
	463	610.3815
	421	299.0859
	439	0.4983
	465	713.2051
	422	265.4977
	446	59.3809
	425	176.7331
	447	75.7927
	427	127.5567
	448	94.2045
	431	53.2039
	453	216.2635
Total	7451	3669.5295

Mean X̅ = Σ Xi / n
X̅ = 7451 / 17 = 438.2941
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 3669.5295 / 17 -1 ) = 15.1442

Confidence Interval
X̅ ± t(α/2, n-1) S/√(n)
t(α/2, n-1) = t(0.01 /2, 17- 1 ) = 2.921
438.2941 ± t(0.01/2, 17 -1) * 15.1442/√(17)
Lower Limit = 438.2941 - t(0.01/2, 17 -1) 15.1442/√(17)
Lower Limit = 427.5652
Upper Limit = 438.2941 + t(0.01/2, 17 -1) 15.1442/√(17)
Upper Limit = 449.023
99% Confidence interval is ( 427.5652 , 449.023 )

Part b)

µ = 435

Since the value µ = 435 lies in the interval, hence we can suggest that µ = 435 is a plausible value for the true average degree of polymerization.

Part c)

µ = 460

Since the value µ = 460 does not lies in the interval, hence we can not suggest that µ = 460 is a plausible value for the true average degree of polymerization.