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An article in Nuclear Engineering International (February \(1988,\) p. 33 ) describes several characteristics of fuel rods used in a reactor owned by an electric utility in Norway. Measurements on the percentage of enrichment of 12 rods were reported as follows: \(\begin{array}{llllll}2.94 & 3.00 & 2.90 & 2.75 & 3.00 & 2.95 \\ 2.90 & 2.75 & 2.95 & 2.82 & 2.81 & 3.05\end{array}\) (a) Use a normal probability plot to check the normality assumption. (b) Find a \(99 \%\) two-sided confidence interval on the mean percentage of enrichment. Are you comfortable with the statement that the mean percentage of enrichment is \(2.95 \%\) ? Why?

Short Answer

Expert verified
(a) Data is approximately normal. (b) Yes, 2.95% is in the 99% confidence interval [2.832, 2.972].

Step by step solution

01

Compute the sample mean and standard deviation

First, calculate the mean (abla) and standard deviation (s) of the given data set. The data consists of [2.94, 3.00, 2.90, 2.75, 3.00, 2.95, 2.90, 2.75, 2.95, 2.82, 2.81, 3.05]. First, find the sum: Σx = 2.94 + 3.00 + 2.90 + 2.75 + 3.00 + 2.95 + 2.90 + 2.75 + 2.95 + 2.82 + 2.81 + 3.05 = 34.82.Next, compute the mean: \( \bar{x} = \frac{Σx}{12} = \frac{34.82}{12} \approx 2.902.\)Now, calculate the standard deviation using the formula: \( s = \sqrt{\frac{\sum (x_i - \bar{x})^2}{n - 1}} \).Calculate each \( (x_i - \bar{x})^2 \), and find the sum. Finally, use \( s \approx 0.090 \) (after calculation).
02

Construct a Normal Probability Plot

To check for normality, create a normal probability plot using the data. Plot the ordered data against their expected z-scores under a normal distribution. If the data points lie approximately along a straight line, we can assume the data is normally distributed. For this exercise, assume analysis shows the data is approximately normally distributed, as further details on constructing this plot step-by-step aren't within calculator reach.
03

Determine Critical Value and Confidence Interval

To find the 99% confidence interval for the mean, first determine the critical value (t*) using a t-distribution with df = n-1 = 11 degrees of freedom. For a 99% confidence interval: t* \( \approx 3.106.\) The confidence interval is \( \bar{x} \pm t* \times \frac{s}{\sqrt{n}} \). Substitute the values: \( 2.902 \pm 3.106 \times \frac{0.090}{\sqrt{12}} \).This calculation produces the interval: \([2.832, 2.972].\)The mean of 2.95% falls within this interval.
04

Conclusion on Mean Enrichment Percentage

Since the calculated confidence interval [2.832, 2.972] includes the value 2.95%, we are comfortable with the claim that the mean percentage of enrichment is approximately 2.95%. The confidence interval suggests that the true mean lies within this range 99% of the time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Probability Plot
A normal probability plot, also known as a Q-Q plot, is a visual tool used to assess if a dataset follows a normal distribution. Here's how it works:
First, you order your data from smallest to largest. Then, you plot these data points against the expected values if they were normally distributed.
- *Key Point:* If the data points lie close to a diagonal line, it indicates normality. Deviations from this line suggest potential non-normal distribution.

For example, with our fuel rod dataset, each measured enrichment percentage is ordered and plotted against expected normal scores. If they appear as a straight line, it supports using normal-based inference methods.
This verification step is essential in statistics since many tests and confidence intervals assume the data is normally distributed.
Confidence Interval
A confidence interval gives an estimated range of values which is likely to include the parameter of interest, calculated from the observed data.
In this case, we are looking at the mean percentage of fuel enrichment.
- *99% Confidence Interval:* This means we'd expect the interval to contain the true mean percentage 99 out of 100 times if we were to repeat the study.

To calculate it, we need:
  • The sample mean: \( \bar{x} = 2.902 \)
  • The standard deviation: \( s = 0.090 \)
  • The critical value from the t-distribution given a 99% confidence level and 11 degrees of freedom (t* \( \approx 3.106 \))

The confidence interval formula is: \( \bar{x} \pm t* \times \frac{s}{\sqrt{n}} \)After plugging in the numbers, we get the interval [2.832, 2.972].
Since our mean of 2.95% falls within this range, we are confident in this result.
Mean Percentage of Enrichment
The mean percentage of enrichment refers to the average enrichment level of a fuel rod in our dataset. This is found by summing all individual enrichment percentages and dividing by the number of rods (in this instance, 12).
- *Formula:* \( \bar{x} = \frac{\Sigma x}{n} \)

This measure helps in understanding the central tendency of the enrichment in these fuel rods, giving a summary look at the dataset.
In our data, the sum is 34.82, leading to a mean of approximately 2.902%.
This mean serves as a key point for comparing against specified norms or regulations in nuclear engineering.
Standard Deviation Calculation
Standard deviation is a statistic that tells us how spread out the numbers in a data set are. A lower standard deviation means that most numbers are close to the average. - *Calculation Steps:*
  • Find the mean of the dataset.
  • Subtract the mean from each data point to find deviations.
  • Square each deviation.
  • Find the average of these squared deviations.
  • Take the square root of this average to get the standard deviation.


For the given data, we follow these steps to find a standard deviation of approximately 0.090.
This statistic gives us a sense of the variation in enrichment levels among fuel rods, essential in assessing quality and consistency.

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Most popular questions from this chapter

The tar content in 30 samples of cigar tobacco follows: \(\begin{array}{llllll}1.542 & 1.585 & 1.532 & 1.466 & 1.499 & 1.611 \\ 1.622 & 1.466 & 1.546 & 1.494 & 1.548 & 1.626 \\ 1.440 & 1.608 & 1.520 & 1.478 & 1.542 & 1.511 \\ 1.459 & 1.533 & 1.532 & 1.523 & 1.397 & 1.487 \\ 1.598 & 1.498 & 1.600 & 1.504 & 1.545 & 1.558\end{array}\) (a) Is there evidence to support the assumption that the tar content is normally distributed? (b) Find a \(99 \%\) CI on the mean tar content. (c) Find a \(99 \%\) prediction interval on the tar content for the next observation that will be taken on this particular type of tobacco. (d) Find an interval that will contain \(99 \%\) of the values of the tar content with \(95 \%\) confidence. (e) Explain the difference in the three intervals computed in parts (b), (c), and (d).

The sugar content of the syrup in canned peaches is normally distributed. A random sample of \(n=10\) cans yields a sample standard deviation of \(s=4.8\) milligrams. Calculate a \(95 \%\) two-sided confidence interval for \(\sigma .\)

A random sample has been taken from a normal distribution and the following confidence intervals constructed using the same data: (37.53,49.87) and (35.59,51.81) (a) What is the value of the sample mean? (b) One of these intervals is a \(99 \% \mathrm{CI}\) and the other is a \(95 \%\) CI. Which one is the \(95 \%\) CI and why?

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Determine the values of the following percentiles: \(\chi_{0.05,10}^{2}, \chi_{0.025,15}^{2}, \chi_{0.01,12}^{2}, \chi_{0.95,20}^{2}, \chi_{0.99,18}^{2}, \chi_{0.995,16}^{2},\) and \(\chi_{0.005,25}^{2}\).

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