The alternating current (AC) breakdown voltage of an insulating liquid indicates
ID: 3314315 • Letter: T
Question
The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article Testing Practices for the AC Breakdown Voltage Testing of Insulation Liquids (IEEE Electrical Insulation Magazine, 1995: 2126) gave the accompanying sample observations on breakdown voltage (kV) of a particular circuit under certain conditions 62 50 53 57 41 53 55 61 59 64 50 53 64 62 50 68 54 55 57 50 55 50 56 55 46 55 53 54 52 47 47 55 57 48 63 57 57 55 53 59 53 52 50 55 60 50 56 58 (a) Calculate a 87%- 92% , 95%-and 99% confidence interval for true average breakdown voltage. (b) At each of the above levels, determine the smallest sample size needed so that the confidence interval width is 2 kVExplanation / Answer
(a) Here sample mean breakdown voltage x = 54.7083 kV
sample standard deviation s = 5.23 kV
sample size n = 48
Standard erorr of sample mean se = s/ sqrt(n) = 5.23/ sqrt(48) = 0.755 kV
The (1 - )% confidence interval = x +- t(1-)/2. dF * se0
87 % confidence interval = x +- t0.065,47 * se0
54.7083 +- 1.5411 * 0.755
= (53.5448, 55.8718)
92 % confidence interval = x +- t0.04,47 * se0
54.7083 +- 1.7894 * 0.755
= (53.3574, 56.0593)
95 % confidence interval = x +- t0.025,47 * se0
54.7083 +- 2.0117 * 0.755
= (53.1895, 56.2272)
99 % confidence interval = x +- t0.025,47 * se0
54.7083 +- 2.6846 * 0.755
= (52.6815, 56.7351)
(b) Here to calculate the smallest sample size we will use Z statistic here.
Where SE0 = standard error of sample mean = 5.23/ sqrt(n) = 5.23/ n
Here confidence interval width = 2kV = 2 * Z(1-)/2 * SE0
1 = Z(1-)/2 * 5.23/ n
n = Z(1-)/2 * 5.23
HEre for 87% confidence interval
n = Z(0.013 * 5.23
n = (5.23 * 1.5141)2 = 62.72= 63
HEre for 92% confidence interval
n = Z(0.04) * 5.23
n = (5.23 * 1.7507)2 =83.85= 84
HEre for 95% confidence interval
n = Z(0.013 * 5.23
n = (5.23 * 1.96)2 = 105.10= 106
HEre for 99% confidence interval
n = Z(0.01) * 5.23
n = (5.23 * 2.575)2 = 181.53= 182
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