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

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

[1] 430.508

right

[1] 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.