/*! 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 13 The following data on distilled ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The following data on distilled alcohol content (\%) for a sample of 35 port wines was extracted from the article "A Method for the Estimation of Alcohol in Fortified Wines Using Hydrometer Baumé and Refractometer Brix" (Amer. J. Enol. Vitic., 2006: \(486-490\) ). Each value is an average of two duplicate measurements. $$ \begin{array}{lllllllll} 16.35 & 18.85 & 16.20 & 17.75 & 19.58 & 17.73 & 22.75 & 23.78 & 23.25 \\ 19.08 & 19.62 & 19.20 & 20.05 & 17.85 & 19.17 & 19.48 & 20.00 & 19.97 \\ 17.48 & 17.15 & 19.07 & 19.90 & 18.68 & 18.82 & 19.03 & 19.45 & 19.37 \\ 19.20 & 18.00 & 19.60 & 19.33 & 21.22 & 19.50 & 15.30 & 22.25 & \end{array} $$ Use methods from this chapter, including a boxplot that shows outliers, to describe and summarize the data.

Short Answer

Expert verified
Mean is 19.46%. Median is 19.20%. Outliers: 15.30%, 22.75%, 23.25%, 23.78%.

Step by step solution

01

Organize the Data

First, list all the alcohol content values in ascending order for better visualization of the distribution. Sorted data: 15.30, 16.20, 16.35, 17.15, 17.48, 17.73, 17.75, 17.85, 18.00, 18.68, 18.82, 18.85, 19.03, 19.07, 19.08, 19.17, 19.20, 19.20, 19.33, 19.37, 19.45, 19.48, 19.50, 19.58, 19.60, 19.62, 19.90, 19.97, 20.00, 20.05, 21.22, 22.25, 22.75, 23.25, 23.78.
02

Calculate Descriptive Statistics

Compute the mean, median, variance, and standard deviation. - **Mean** is calculated as the sum of all values divided by the number of values. - **Median** is the middle value in the sorted list. For the provided data:- Mean: \( \text{Mean} = \frac{\text{Sum of all values}}{35} = \frac{681.00}{35} = 19.46 \%\)- Median: Since there are 35 numbers, the median is the 18th value, which is 19.20%.
03

Identify the Quartiles and Interquartile Range (IQR)

To find the quartiles:- **Q1 (First Quartile)** is the middle value of the first half of the data (8th value): 17.75%.- **Q3 (Third Quartile)** is the middle value of the second half of the data (27th value): 20.00%.The **IQR** is calculated as:\( \text{IQR} = Q3 - Q1 = 20.00\% - 17.75\% = 2.25\% \) .
04

Determine and Visualize Outliers Using a Boxplot

Calculate the potential outliers using 1.5 * IQR below Q1 and above Q3: - **Lower bound** = Q1 - 1.5 * IQR = 17.75% - 1.5 * 2.25% = 14.375% - **Upper bound** = Q3 + 1.5 * IQR = 20.00% + 1.5 * 2.25% = 23.375% Numbers outside these bounds are considered outliers. Thus, 15.30%, 22.75%, 23.25%, and 23.78% are outliers. Use these values to construct a boxplot highlighting the outliers.
05

Summarize the Results

The data has a mean alcohol content of 19.46%, with a median of 19.20%. The distribution is slightly skewed with an IQR of 2.25%. The outliers identified suggest some extreme variations in alcohol content. The majority of values fall within a typical range, but a few notable exceptions exist.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Boxplot
A boxplot, also known as a box-and-whisker plot, is a valuable graphical representation of data that helps us gain a quick understanding of the distribution and variability. It displays the five-number summary: minimum, first quartile (Q1), median, third quartile (Q3), and maximum.

The central box shows the interquartile range, which is the middle 50% of the data. The line inside the box represents the median. "Whiskers" extend from the box to the smallest and largest values, excluding any outliers. Outliers are denoted by individual points outside the whiskers.

In this data set, a boxplot helps visualize the general trends and emphasizes the outliers, such as 15.30%, 22.75%, 23.25%, and 23.78%, making it easier to interpret the spread and central tendencies of the wine alcohol content.
Interquartile Range (IQR)
The Interquartile Range (IQR) is a measure that indicates the spread of the middle 50% of data points. It is calculated as the difference between the third quartile (Q3) and the first quartile (Q1). Interquartile Range provides a robust measure of variability that isn't influenced by outliers or extreme values.

In our wine data example, the first quartile (Q1) is 17.75% and the third quartile (Q3) is 20.00%. Thus, the IQR is calculated as: \[ \text{IQR} = Q3 - Q1 = 20.00\% - 17.75\% = 2.25\% \] An IQR of 2.25% indicates that the central half of the alcohol content figures are spread within 2.25% of each other. This helps in understanding the consistency of the alcohol content across the different wine samples.
Outliers
Identifying outliers is crucial as they can significantly affect statistical analysis. An outlier is a data point that falls outside the range of what is expected and can skew the results.

To find outliers, we use the IQR to calculate the bounds. If a value is more than 1.5 * IQR below Q1 or above Q3, it's considered an outlier. In our dataset, the outlier boundaries are:
  • Lower bound: 17.75% - 1.5 * 2.25% = 14.375%
  • Upper bound: 20.00% + 1.5 * 2.25% = 23.375%
Values beyond these, such as 15.30%, 22.75%, 23.25%, and 23.78%, are outliers. Recognizing these outliers helps in refining the analysis and understanding variations in the data.
Mean and Median
Mean and median are fundamental measures in descriptive statistics. They help in understanding the central tendency of a dataset.

The **mean** is calculated by dividing the sum of all data points by the number of observations. For this wine data, the mean is: \[ \text{Mean} = \frac{681.00}{35} = 19.46\% \] The **median** is the middle value when the data is sorted. Here, with 35 values, the median is the 18th data point, 19.20%.

Comparing the mean and median provides insights into the data distribution. A close mean and median suggest a symmetrical distribution, while a large difference could indicate skewness. In this case, the mean is slightly higher, hinting at a right-skewed distribution due to some high outliers.
Quantitative Data Analysis
Quantitative data analysis involves applying mathematical and statistical methods to numerical data. It allows us to draw conclusions and gain insights from the figures. This process includes calculating various statistics such as the mean, median, standard deviation, and interquartile range.

The analysis of the wine data involved organizing values in ascending order, simplified visualization with a boxplot, and summarizing with descriptive statistics. These steps assist in understanding overall trends, variability, and identifying peculiarities in the data.

By understanding these quantitative measures, one can make informed judgments about data tendencies, predict future patterns, and spot irregularities, providing a comprehensive understanding of datasets like alcohol content in wines.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article "Effects of Short-Term Warming on Low and High Latitude Forest Ant Communities" (Ecoshpere, May 2011, Article 62) described an experiment in which observations on various characteristics were made using minichambers of three different types: (1) cooler (PVC frames covered with shade cloth), (2) control (PVC frames only), and (3) warmer (PVC frames covered with plastic). One of the article's authors kindly supplied the accompanying data on the difference between air and soil temperatures \(\left({ }^{\circ} \mathrm{C}\right)\). \(\begin{array}{ccc}\text { Cooler } & \text { Control } & \text { Warmer } \\\ 1.59 & 1.92 & 2.57 \\ 1.43 & 2.00 & 2.60 \\ 1.88 & 2.19 & 1.93 \\ 1.26 & 1.12 & 1.58 \\ 1.91 & 1.78 & 2.30 \\ 1.86 & 1.84 & 0.84 \\ 1.90 & 2.45 & 2.65 \\ 1.57 & 2.03 & 0.12 \\ 1.79 & 1.52 & 2.74 \\ 1.72 & 0.53 & 2.53 \\\ 2.41 & 1.90 & 2.13 \\ 2.34 & & 2.86 \\ 0.83 & & 2.31 \\ 1.34 & & 1.91 \\\ 1.76 & & \end{array}\) a. Compare measures of center for the three different samples. b. Calculate, interpret, and compare the standard deviations for the three different samples. c. Do the fourth spreads for the three samples convey the same message as do the standard deviations about relative variability? d. Construct a comparative boxplot (which was included in the cited article) and comment on any interesting features.

A Pareto diagram is a variation of a histogram for categorical data resulting from a quality control study. Each category represents a different type of product nonconformity or production problem. The categories are ordered so that the one with the largest frequency appears on the far left, then the category with the second largest frequency, and so on. Suppose the following information on nonconformities in circuit packs is obtained: failed component, 126; incorrect component, 210 ; insufficient solder, 67; excess solder, 54 ; missing component, 131. Construct a Pareto diagram.

Automated electron backscattered diffraction is now being used in the study of fracture phenomena. The following information on misorientation angle (degrees) was extracted from the article "Hbservations on the Faceted Initiation Site in the Dwell-Fatigue Tested Ti-6242 Alloy: Crystallographic Orientation and Size Effects" (Metallurgical and Materials Trans., 2006: 1507-1518). \(\begin{array}{lcccc}\text { Class: } & 0-<5 & 5-<10 & 10-<15 & 15-<20 \\\ \text { Rel freq: } & .177 & .166 & .175 & .136 \\ \text { Class: } & 20-<30 & 30-<40 & 40-<60 & 60-<90 \\ \text { Rel freq: } & .194 & .078 & .044 & .030\end{array}\) a. Is it true that more than \(50 \%\) of the sampled angles are smaller than \(15^{\circ}\), as asserted in the paper? b. What proportion of the sampled angles are at least \(30^{\circ} ?\) c. Roughly what proportion of angles are between \(10^{\circ}\) and \(25^{\circ} ?\) d. Construct a histogram and comment on any interesting features.

For each of the following hypothetical populations, give a plausible sample of size 4 : a. All distances that might result when you throw a football b. Page lengths of books published 5 years from now c. All possible earthquake-strength measurements (Richter scale) that might be recorded in California during the next year d. All possible yields (in grams) from a certain chemical reaction carried out in a laboratory

Blood pressure values are often reported to the nearest \(5 \mathrm{mmHg}\) ( \(100,105,110\), etc.). Suppose the actual blood pressure values for nine randomly selected individuals are \(\begin{array}{lllllll}118.6 & 127.4 & 138.4 & 130.0 & 113.7 & 122.0 & 108.3 \\\ 131.5 & 133.2 & & & & & \end{array}\) a. What is the median of the reported blood pressure values? b. Suppose the blood pressure of the second individual is \(127.6\) rather than \(127.4\) (a small change in a single value). How does this affect the median of the reported values? What does this say about the sensitivity of the median to rounding or grouping in the data?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.