/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 54 Consider the following observati... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider the following observations on shear strength (MPa) of a joint bonded in a particular manner (from a graph in the article "Diffusion of Silicon Nitride to Austenitic Stainless Steel without Interlayers," Metallurgical Trans., 1993: 1835-1843). \(\begin{array}{rrrrrr}22.2 & 40.4 & 16.4 & 73.7 & 36.6 & 109.9 \\ 30.0 & 4.4 & 33.1 & 66.7 & 81.5 & \end{array}\) a. What are the values of the fourths, and what is the value of \(f_{s}\) ? b. Construct a boxplot based on the five-number summary, and comment on its features. c. How large or small does an observation have to be to qualify as an outlier? As an extreme outlier? d. By how much could the largest observation be decreased without affecting \(f_{s}\) ?

Short Answer

Expert verified
a. Q1: 22.2, Q3: 73.7, \( f_{s} = 51.5 \). b. Boxplot shows a right-skewed distribution. c. No outliers or extreme outliers. d. Largest value can decrease to 73.7.

Step by step solution

01

Arrange Data in Ascending Order

First, arrange the given data in ascending order: 4.4, 16.4, 22.2, 30.0, 33.1, 36.6, 40.4, 66.7, 73.7, 81.5, 109.9.
02

Determine the Fourths

The fourths correspond to the 25th and 75th percentiles. For 11 data points: - The 1st quartile (Q1) is the 3rd data point: 22.2. - The 3rd quartile (Q3) is the 9th data point: 73.7.
03

Calculate the Spread of the Fourths (fs)

Calculate the spread of the fourths, which is the interquartile range (IQR):\[ f_{s} = Q3 - Q1 = 73.7 - 22.2 = 51.5 \]
04

Construct the Five-Number Summary

The five-number summary consists of the minimum, Q1, median, Q3, and maximum: - Minimum: 4.4 - Q1: 22.2 - Median: 36.6 - Q3: 73.7 - Maximum: 109.9
05

Sketch the Boxplot

Draw a boxplot with the above five-number summary: - The box spans from Q1 (22.2) to Q3 (73.7). - The line at 36.6 represents the median. - Whiskers extend to the minimum (4.4) and maximum (109.9). Comment: The boxplot shows a right-skewed distribution.
06

Determine Outlier and Extreme Outlier Ranges

Outliers lie outside the range calculated by:- Lower boundary: \( Q1 - 1.5 \times IQR = 22.2 - 1.5 \times 51.5 = -55 \)- Upper boundary: \( Q3 + 1.5 \times IQR = 73.7 + 1.5 \times 51.5 = 151 \)Extreme outliers lie outside the range calculated by:- Lower boundary: \( Q1 - 3 \times IQR = 22.2 - 3 \times 51.5 = -133.3 \)- Upper boundary: \( Q3 + 3 \times IQR = 73.7 + 3 \times 51.5 = 229.2 \) Thus, no values are outliers or extreme outliers according to this.
07

Determine Reduction in Largest Observation without Affecting fs

To keep the value of \( f_{s} \) (51.5) the same, the reduction should not affect the values of Q1, Q3, or any central/other observations that determine the quartiles. Therefore, the largest data point can be decreased as much as possible without becoming less than 73.7 (Q3), as this would change \( Q3 \).

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Key Concepts

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

Shear Strength
Shear strength is an important property that measures how well a material can withstand sliding forces along its planes. In the context of this problem, we are looking at shear strength data from joints bonded in a specific way. Understanding shear strength is crucial in fields such as civil engineering, materials science, and mechanical engineering. Strong shear strength is vital for ensuring that structural components like beams, joints, and connectors can safely support loads without failing. To determine the shear strength of a material, tests often include applying force until the material fails or deforms beyond its limits. Values provided, in this case, are measured in Megapascals (MPa), which is a unit of pressure or stress. By analyzing the shear strength values using statistical techniques, we can gain insights into how consistent the material's strength is and identify any variability in the data.
Interquartile Range
The interquartile range (IQR) is a measure of statistical dispersion, capturing the range within which the central 50% of data lies. It is particularly useful for understanding the spread and central tendency of data while minimizing the effect of outliers.In this problem, the IQR is calculated as the difference between the third quartile (Q3) and the first quartile (Q1). With the given data:
  • Q1 = 22.2
  • Q3 = 73.7
The IQR is calculated as:\[IQR = Q3 - Q1 = 73.7 - 22.2 = 51.5\]This value tells us that the middle 50 percent of the shear strength observations lie between 22.2 MPa and 73.7 MPa. The IQR helps identify variability and can be key in assessing overall stability and reliability of the material's shear strength.
Outliers
Outliers are data points that differ significantly from other values. Identifying outliers is crucial because they can distort statistical analyses, sometimes leading to misleading conclusions.To ascertain whether there are outliers within the data, we use the IQR. Outliers are generally defined as points lying outside of:
  • Lower boundary: \( Q_1 - 1.5 \times IQR \)
  • Upper boundary: \( Q_3 + 1.5 \times IQR \)
In our case:
  • Lower boundary: \( 22.2 - 1.5 \times 51.5 = -55 \)
  • Upper boundary: \( 73.7 + 1.5 \times 51.5 = 151 \)
Since all data points are within this range, we find that there are no outliers in this data set. Additionally, extreme outliers are more severe and lie outside:
  • Lower boundary: \( Q_1 - 3 \times IQR = -133.3 \)
  • Upper boundary: \( Q_3 + 3 \times IQR = 229.2 \)
Thus, this evaluation further confirms no extreme outliers are present in this dataset.
Five-Number Summary
The five-number summary provides a compact overview of a dataset, highlighting key data points: the minimum, first quartile ( Q1 ), median, third quartile ( Q3 ), and maximum. These elements help in summarizing the dataset's distribution efficiently. For the given shear strength values, the five-number summary is:
  • Minimum: 4.4
  • Q1: 22.2
  • Median: 36.6
  • Q3: 73.7
  • Maximum: 109.9
This summary is graphically represented using a boxplot, where the box spans from Q1 to Q3 , with a median line inside it, and whiskers extending to the minimum and maximum values. The five-number summary and the boxplot provide a clear picture of how data is spread, its central tendency, and whether any skewness is present. In our case, the boxplot reveals a right-skewed distribution, suggesting variability in higher shear strength values.

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