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Chapter 1: Problem 36

The propagation of fatigue cracks in various aircraft parts has been the subject of extensive study in recent years. The accompanying data consists of propagation lives (flight hours \(/ 10^{4}\) ) to reach a given crack size in fastener holes intended for use in military aircraft ("Statistical Crack Propagation in Fastener Holes under Spectrum Loading," J. Aircraft, 1983: 1028-1032): \(\begin{array}{llllllll}.736 & .863 & .865 & .913 & .915 & .937 & .983 & 1.007 \\\ 1.011 & 1.064 & 1.109 & 1.132 & 1.140 & 1.153 & 1.253 & 1.394\end{array}\) a. Compute and compare the values of the sample mean and median. b. By how much could the largest sample observation be decreased without affecting the value of the median?

Short Answer

Expert verified
The sample mean is 1.03, the median is 1.009, and the largest observation can be decreased without affecting the median.

Step by step solution

01

Organize the data

List the given data in ascending order: \[0.736, 0.863, 0.865, 0.913, 0.915, 0.937, 0.983, 1.007, 1.011, 1.064, 1.109, 1.132, 1.140, 1.153, 1.253, 1.394\]
02

Calculate the sample mean

To find the mean, sum all the observations and divide by the number of observations.\[\text{Mean} = \frac{0.736 + 0.863 + 0.865 + 0.913 + 0.915 + 0.937 + 0.983 + 1.007 + 1.011 + 1.064 + 1.109 + 1.132 + 1.140 + 1.153 + 1.253 + 1.394}{16}\]Calculate the total sum first:\[\text{Sum} = 0.736 + 0.863 + 0.865 + 0.913 + 0.915 + 0.937 + 0.983 + 1.007 + 1.011 + 1.064 + 1.109 + 1.132 + 1.140 + 1.153 + 1.253 + 1.394 = 16.475\]Then calculate the mean:\[\text{Mean} = \frac{16.475}{16} = 1.0296875\]Thus, the sample mean is approximately 1.03.
03

Determine the sample median

Locate the middle of the sorted data. Since there are 16 observations, the median will be the average of the 8th and 9th observations.The 8th observation is 1.007, and the 9th is 1.011.Calculate the median:\[\text{Median} = \frac{1.007 + 1.011}{2} = 1.009\]Thus, the sample median is 1.009.
04

Analyze the effect on median of decreasing the largest observation

The largest sample observation is 1.394. To change the median, we would need to lower this observation such that it affects the middle two values (the 8th and 9th positions when sorted). Since lowering 1.394 does not affect the middle two numbers (1.007, 1.011), it can be decreased by any amount without changing the median.

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Key Concepts

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

Sample Mean Calculation
The sample mean is a fundamental concept in descriptive statistics that provides an average value of a data set. To calculate the sample mean, you take the sum of all observations in the dataset and divide it by the number of observations.
For the given problem, we have 16 data points representing the propagation lives in certain aircraft parts. By summing all these values, we find a total of 16.475. To obtain the mean, we divide this sum by 16, which equals approximately 1.03.
  • Formula: The formula for the mean is given by \(\text{Mean} = \frac{\Sigma x_i}{n}\) where \(x_i\) represents each data point, and \(n\) is the number of observations.
  • Interpretation: The sample mean provides a central value of the dataset, indicating the average propagation life. This helps in understanding the general tendency of fatigue crack propagation in the study.
Median Calculation
The median is another crucial measure in descriptive statistics, particularly for understanding the central tendency of a dataset. Unlike the mean, the median is less sensitive to extreme values in the data.
To find the median, especially with 16 data points, you must first ensure the data is ordered. With our sorted data, the median lies between the 8th and 9th observations since we have an even number of values.
  • Finding the Middle: The dataset sorted in ascending order allows us to easily identify the middle by averaging the 8th (1.007) and 9th (1.011) observations. The median hence is \(\frac{1.007 + 1.011}{2} = 1.009\).
  • Significance: The median offers insights into the distribution of the data, and in many cases, can provide a better measure of central tendency than the mean when dealing with skewed data or outliers.
Data Ordering
Arranging data in order is a vital step in descriptive statistics, particularly when calculating the median. Properly ordered data allows for easier identification of key statistics and provides a clearer view of the dataset's distribution.
For the current exercise, organizing the data enabled us to effectively determine the median. The process involves arranging the data from the smallest to largest value, which as presented, changes our data to:
  • 0.736, 0.863, 0.865, 0.913, 0.915, 0.937, 0.983, 1.007,
  • 1.011, 1.064, 1.109, 1.132, 1.140, 1.153, 1.253, 1.394.

  • Benefits of Data Ordering: This task helps in various statistical analyses, reducing potential errors when calculating statistics like the median, percentiles, and detecting any unusual data points.
  • Impact on Analysis: Ordering provides a quick visual check on how evenly distributed the data is, or if it contains clusters or gaps that may indicate further areas for analysis.
Statistical Analysis of Aircraft Fatigue Cracks
Understanding fatigue cracks in aircraft parts involves applying statistical methods to study how cracks evolve over time and under different conditions. This analysis is essential for ensuring safety in aircraft engineering.
For this exercise, the data represents the propagation lives of cracks under specified conditions, measured over several aircraft parts. By using statistical tools like the mean and median, engineers can better estimate how crack propagation behaves across similar circumstances.
  • Significance in Engineering: Knowing the average (mean) and central value (median) of crack propagation gives essential feedback on the overall safety and performance expectations of components used in aircraft.
  • Descriptive Statistics Role: Descriptive statistics provide invaluable first steps in data analysis before more complex inferential statistical methods are applied, which might involve predictions and uncertainty estimations.
  • Data Insights: Recognition of median and mean alongside data ordering may also indicate skewness or possible outliers within the crack life data, causing further investigation or calibration of engineering models and safety standards.

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Most popular questions from this chapter

Consider numerical observations \(x_{1}, \ldots, x_{n}\). It is frequently of interest to know whether the \(x_{t}\) 's are (at least approximately) symmetrically distributed about some value. If \(n\) is at least moderately large, the extent of symmetry can be assessed from a stem-and-leaf display or histogram. However, if \(n\) is not very large, such pictures are not particularly informative. Consider the following alternative. Let \(y_{1}\) denote the smallest \(x_{i}, y_{2}\) the second smallest \(x_{i}\), and so on. Then plot the following pairs as points on a twodimensional coordinate system: \(\left(y_{n}-\bar{x}, \bar{x}-y_{1}\right)\), \(\left(y_{n-1}-\bar{x}, \bar{x}-y_{2}\right),\left(y_{n-2}-\bar{x}, \bar{x}-y_{3}\right), \ldots\). There are \(n / 2\) points when \(n\) is even and \((n-1) / 2\) when \(n\) is odd. a. What does this plot look like when there is perfect symmetry in the data? What does it look like when observations stretch out more above the median than below it (a long upper tail)? b. The accompanying data on rainfall (acre-feet) from 26 seeded clouds is taken from the article "A Bayesian Analysis of a Multiplicative Treatment Effect in Weather Modification" (Technometrics, 1975: 161-166). Construct the plot and comment on the extent of symmetry or nature of departure from symmetry. \(\begin{array}{rrrrrrr}4.1 & 7.7 & 17.5 & 31.4 & 32.7 & 40.6 & 92.4 \\ 115.3 & 118.3 & 119.0 & 129.6 & 198.6 & 200.7 & 242.5 \\ 255.0 & 274.7 & 274.7 & 302.8 & 334.1 & 430.0 & 489.1 \\ 703.4 & 978.0 & 1656.0 & 1697.8 & 2745.6 & & \end{array}\)

a. If a constant \(c\) is added to each \(x_{i}\) in a sample, yielding \(y_{i}=x_{i}+c\), how do the sample mean and median of the \(y_{i}\) 's relate to the mean and median of the \(x_{i}\) 's? Verify your conjectures. b. If each \(x_{i}\) is multiplied by a constant \(c\), yielding \(y_{i}=c x_{i}\), answer the question of part (a). Again, verify your conjectures.

a. Give three different examples of concrete populations and three different examples of hypothetical populations. b. For one each of your concrete and your hypothetical populations, give an example of a probability question and an example of an inferential statistics question.

A study of the relationship between age and various visual functions (such as acuity and depth perception) reported the following observations on area of scleral lamina \(\left(\mathrm{mm}^{2}\right)\) from human optic nerve heads ("Morphometry of Nerve Fiber Bundle Pores in the Optic Nerve Head of the Human," Exper. Eye Res., 1988: 559-568): \(\begin{array}{lllllllll}2.75 & 2.62 & 2.74 & 3.85 & 2.34 & 2.74 & 3.93 & 4.21 & 3.88 \\ 4.33 & 3.46 & 4.52 & 2.43 & 3.65 & 2.78 & 3.56 & 3.01 & \end{array}\) a. Calculate \(\sum x_{i}\) and \(\sum x_{i}^{2}\). b. Use the values calculated in part (a) to compute the sample variance \(s^{2}\) and then the sample standard deviation \(s\).

The article "Snow Cover and Temperature Relationships in North America and Eurasia" (J. Climate Appl. Meteorol., 1983: 460-469) used statistical techniques to relate the amount of snow cover on each continent to average continental temperature. Data presented there included the following ten observations on October snow cover for Eurasia during the years 1970-1979 (in million \(\mathrm{km}^{2}\) ): \(\begin{array}{llllllllll}6.5 & 12.0 & 14.9 & 10.0 & 10.7 & 7.9 & 21.9 & 12.5 & 14.5 & 9.2\end{array}\) What would you report as a representative, or typical, value of October snow cover for this period, and what prompted your choice?
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