# Solved:The article “Measuring and understanding the Aging of Kraft Insulating Paper in Power Transformers”

#### ByRaju Chaudhari

Aug 22, 2020

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:

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

a. Construct a boxplot of the data and comment on any interesting features.
b. Is it plausible that the given sample observations were selected from a normal distribution?
c. Calculate a two-sided 95% confidence interval for true average degree of polymerization (as did the authors of the article). Does the interval suggest that 440 is a plausible value for true average degree if polymerization? What about 450?

#### Solution

Let $X$ denote degree of polymerization for paper specimens.

x<-c(418,421,421,422,425,427,431,434,437,439,446,447,448,453,454,463,465)

(a) Boxplot of the data is as follows:

boxplot(x,horizontal = TRUE,col="lightblue",
main="degree of polymerization")

It appears that the distribution of degree of polymerization is positively skewed as the box-plot has a longer tail towards right side.

(b) Normality Check

qqnorm(x,col="blue")
qqline(x,distribution = qnorm,col="red")

It is observed from the normality plot that the distribution of degree of polymerization is slightely positively skewed but still we can assume that it is approximately normally distributed.

(c) Two-sided 95% confidence interval for true average degree of polymarization

xbar<-mean(x)
s<-sd(x)
n<-length(x)
alpha<-0.05
error<-qt(0.975,n-1)*s/sqrt(n)
left<-round(xbar-error,3)
right<-round(xbar+error,3)
left
 430.508
right
 446.081

The $95$ percent confidence interval for true average degree of polymerization 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(438.2941176-2.12\times \frac{15.1441602}{\sqrt{17}},438.2941176+2.12\times \frac{15.1441602}{\sqrt{17}}\big)\\ &= \big(430.508,446.081\big) \end{aligned}

As the value 440 lies inside of the $95$ percent confidence interval $(430.508, 446.081)$, 440 is a plausible value for the true average degree of polymerization.

As the value 450 lies outside of the $95$ percent confidence interval $(430.508, 446.081)$, 450 is not a a plausible value for the true average degree of polymerization.