# Solved:The alternating current (AC) breakdown voltage of an insulating liquid indicates its dielectric strength. The article “Testing Practices for the AC Breakdown Voltage

#### ByDr. Raju Chaudhari

Aug 22, 2020

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, 1995:21-26) 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 52 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. Construct a boxplot of the data and comment on interesting features.
b. Calculate and interpret a 95% CI for true average breakdown voltage $\mu$. Does it appear that $\mu$ has been precisely estimated? Explain.
c. Suppose the investigator believes that virtually all values of breakdown voltage are between 40 and 70. What sample size would be appropriate for the 95% CI to have a width of 2 kV (so that $\mu$ is estimated to within 1kV with 95% confidence)?

#### Solution

Let $X$ denote the breakdown voltage (kV) of a particular circuit under certain conditions.

x<-c(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) Blelow is the Boxplot of breakdown voltage

boxplot(x,main="")

The distribution of breakdown voltage is approximately normal with one outlier on either side.

(b) 95% Confidence interval for true average breakdown voltage $\mu$

xbar<-round(mean(x),4)
s<-round(sd(x),4)
n<-length(x)
alpha<-0.05
SE<-s/sqrt(n)
E<-qt(0.975,n-1)*SE
left<-round(xbar-E,3)
right<-round(xbar+E,3)
left
[1] 53.189
right
[1] 56.227

The $95$ percent confidence interval for true average breakdown voltage $\mu$ is

\begin{aligned} &\big(\overline{X}-t_{\alpha/2,n-1}\times \frac{s}{\sqrt{n}},\overline{X}+t_{\alpha/2,n-1}\times \frac{s}{\sqrt{n}}\big)\\ & = \big(54.7083-2.012\times \frac{5.2307}{\sqrt{48}},54.7083+2.012\times \frac{5.2307}{\sqrt{48}}\big)\\ &= \big(53.189,56.227\big) \end{aligned}
It appears that $\mu$ has been precisely estimated.

(c) Sample size estimation:

alpha<-0.05
s<-round(sd(x),4)
E<-1
z<-round(qnorm(1-alpha/2),3)
n<-z^2*s^2/(E^2)
n
[1] 105.107

\begin{aligned} n &= \frac{z^2_{\alpha/2}s^2}{E^2}\\ &= \frac{1.96^2\times 5.2307^2}{1^2}\\ &= 105.1070307\\ &\approx 105 \end{aligned}

Suppose the investigator believes that virtually all values of breakdown voltage are between 40 and 70, the appropriate sample size for the 95% CI to have a width of 2 kV would be $n\approx 105$