Question

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 range:(a) Construct a boxplot of the data and comment on any interesting features. (b) Is it plausible that the given sample observations came from a normal distribution? Explain and provide any associated graphs. (c) Calculate a two-sided 95% confidence interval for the true average degree of polymerization. (d) Does the interval suggest that 440 is a plausible value for the true average degree of polymerization? Explain. (e) Does the interval suggest that 450 is a plausible value for the true average degree of polymerization? Explain

Answer 1

Answer:

Step-by-step explanation:

Hello!

You have the data corresponding to 17 observations on the degree of polymerization for paper specimens whose viscosity times concentration fell in a certain range.

a) Boxplot attached.

Box: The 2 quantile (Me) appears to be in the middle of the box, the distance between the 1st quantile and the second quantile appears to be equal than the distance between the 2nd quantile and the 3rd quantile. The mean (black square) is greater than the 2nd quantile. In general, the distribution inside the box appears to be symmetrical (comparing just the Me<X[bar], perhaps slightly right-skewed)

Whiskers: The upper whisker is approximately twice as long as the lower whisker. There are no visible outliers in the data set.

In general, it appears as if the data set is right-skewed.

b) I would be more comfortable with a more symmetrical boxplot, but this one shows enough symmetry to be able to assume that the variable may have a normal distribution.

c) If we assume the normal distribution of the variable, you can use a student t to estimate the true value of the average degree of polymerization of paper:

[X[bar] ± $t_{n-1;1-\alpha /2}$ * $\frac{S}{\sqrt{n} }$]

95% CI: $t_{16;0.975}= 2.120$

X[bar]= 438.29

S= 15.14

[438.29 ± 2.120 * $\frac{15.14}{\sqrt{17} }$]

[430.47;446.03]

With a 95% confidence level, you'd expect the interval [430.47;446.03] to include the true value of the average degree of polymerization of paper.

d) and e)

For these two items, you have to use the calculated CI in c) to decide whether or not there is significant evidence to conclude that two possible values of μ are plausible.

To decide over a hypothesis test using a CI several conditions should be met:

1) The hypothesis has to be two-tailed, and for the same parameter estimated in the interval

2) Both the CI and the Hypothesis test should have complementary levels, this means that if the CI was made using a confidence level of 1 - α= 0.95, then the test should have a significance level of α= 0.05

If these conditions are met, the decision rule is:

If the interval doesn't include the value of the parameter stated in the null hypothesis, then you reject the null hypothesis.

If the interval includes the value of the parameter stated in the null hypothesis, then you do not reject the null hypothesis.

d) μ=440?

Hypothesis:

H₀: μ = 440

H₁: μ ≠ 440

α: 0.05

95% CI: [430.47;446.03]

440 is included in the interval, the decision is to not reject the null hypothesis.

Using a 5% significance level, there is significant evidence to conclude that the population mean of the degree of polymerization of paper is 440.

e) μ=450?

H₀: μ = 450

H₁: μ ≠ 450

α: 0.05

95% CI: [430.47;446.03]

450 is not included in the interval, the decision is to reject the null hypothesis.

At a level of 5%, the decision is to reject the null hypothesis, this means that the population mean of the degree of polymerization of paper is different than 450.

I hope it helps!