Question

05) 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: 418 422 446 434 425 447 427 448 454 Is it plausible that the given sample observations were selected from a normal distribution? Calculate a two-sided 90% confidence interval for true average degree of polymerization Suppose that we need to predict the limit stress for 18th observation. Calculate the 95% prediction interval for this single observation.

Answer 1

(i)

Following is the output of descriptive statistics and box plot generated by excel:

Descriptive statistics Data 17 count mean sample standard deviation sample variance minimum maximum range 438.29 15.14 229.35 418 465 47 BoxPlot 410 420 430 440 450 460 470 Data

Box plot shows that data is approximately symmetric.There is no outlier. So we can assume that data come from normal population.

(b)

Here we have =438.29, S =15.14, n =17 Since populaiton SD is unknown so t critical value will be used. Level of significance: a =0.1 Critical value: t =1.746 Degree of freedom: df =n-1=16 The requried confidence interval is: S iht =438.29 + (15.14*1.746/sqrt(17)) =438.29 + 6.4113 =(431.88,444.7) Excel function for critical value: TINV(0.1,16) Hence, the required confidence interval is (431.88,444.7).

iii)

The critical value for 95% confidence interval is 2.120.

Excel function for critical value: ETINV(0.05,16)

The 95% prediction interval is

$\bar{x}\pm t_{\alpha/2,df}s\cdot \sqrt{1+\frac{1}{n}}=438.29\pm 2.120\cdot 15.14\cdot \sqrt{1+\frac{1}{17}} =438.29\pm 33.03=(405.26, 471.32)$

Hence,

$405.26\leq x_{18}\leq 471.32$