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Zinfandel is a popular red wine varietal produced almost exclusively in California. It is rather controversial among wine connoisseurs because its alcohol content varies quite substantially from one producer to another. In May 2013, the author went to the website klwines .com, randomly selected 10 zinfandels from among the 325 available, and obtained the following values of alcohol content (\%): \(\begin{array}{lllll}14.8 & 14.5 & 16.1 & 14.2 & 15.9 \\ 13.7 & 16.2 & 14.6 & 13.8 & 15.0\end{array}\) a. Calculate and interpret several measures of center. b. Calculate the sample variance using the defining formula. c. Calculate the sample variance using the shortcut formula after subtracting 13 from each observation.

Short Answer

Expert verified
Mean is 14.98, median is 14.7, no mode; variance is 0.7387 by defining and 0.8373 by shortcut formula.

Step by step solution

01

Calculate the Mean

The mean of a data set is calculated as the sum of all observations divided by the number of observations. Here, add up all the alcohol content percentages and divide by 10.\[\text{Mean} = \frac{14.8 + 14.5 + 16.1 + 14.2 + 15.9 + 13.7 + 16.2 + 14.6 + 13.8 + 15.0}{10}\]Calculate the sum: - Total = 14.8 + 14.5 + 16.1 + 14.2 + 15.9 + 13.7 + 16.2 + 14.6 + 13.8 + 15.0 = 149.8So,\[\text{Mean} = \frac{149.8}{10} = 14.98\]
02

Calculate the Median

Order the data in increasing value: 13.7, 13.8, 14.2, 14.5, 14.6, 14.8, 15.0, 15.9, 16.1, 16.2. Since there are 10 values, the median will be the average of the 5th and 6th values.\[\text{Median} = \frac{14.6 + 14.8}{2} = 14.7\]
03

Calculate the Mode

The mode is the value that occurs most frequently in a data set. In this sample, no value repeats, so there is no mode.
04

Calculate the Sample Variance (Defining Formula)

The sample variance is calculated by finding the mean, subtracting the mean from each data point to find the deviation, squaring each deviation, summing them, and dividing by \(n-1\).Using the mean \(\bar{x} = 14.98\):\[\text{Sample Variance} = \frac{(14.8-14.98)^2 + (14.5-14.98)^2 + \dots + (15.0-14.98)^2}{9}\]Calculating each:- \((14.8 - 14.98)^2 = 0.0361\)- \((14.5 - 14.98)^2 = 0.2304\)- \(\ldots\)Sum them: \[0.0361 + 0.2304 + 1.2544 + 0.6084 + 0.8464 + 1.6521 + 1.4856 + 0.1444 + 1.4084 + 0.0004 = 6.6485\]Then divide:\[\text{Sample Variance} = \frac{6.6485}{9} = 0.7387\]
05

Calculate Sample Variance using Shortcut Formula

First subtract 13 from each observation, then use these numbers in the shortcut formula for variance.Let the transformed observations \(y_i = x_i - 13\), then calculate \(\bar{y} = \frac{1}{n}\sum y_i\) and \(\sum y_i^2\).Transformed observations: 1.8, 1.5, 3.1, 1.2, 2.9, 0.7, 3.2, 1.6, 0.8, 2.0.\(\sum y_i = 18.8\)\(\sum y_i^2 = 3.24 + 2.25 + 9.61 + 1.44 + 8.41 + 0.49 + 10.24 + 2.56 + 0.64 + 4.00 = 42.88\)\[\text{Sample Variance} = \frac{42.88 - \frac{(18.8)^2}{10}}{9} \]\[= \frac{42.88 - 35.344}{9} \]\[= \frac{7.536}{9} = 0.8373 \]

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Key Concepts

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

Measures of Center
Understanding the measures of center is crucial in statistics as they provide a summary statistic that represents the central point of a dataset. In this exercise, we focus on three primary measures: the mean, median, and mode. The mean is the average of all data points, calculated by summing each value and dividing by the total number (in this case, 10). It gives us a sense of the overall level of alcohol content in the 10 Zinfandel samples. In our data, we found the mean to be 14.98%.

Next is the median, which provides the middle value when observations are ordered from smallest to largest. It is less affected by extreme values than the mean, making it a reliable measure in skewed distributions. For an even number of data points, like 10, the median is the average of the fifth and sixth values. Here, it is 14.7%.

The mode is the value that appears most frequently in a dataset. In this scenario, there is no mode as each alcohol percentage appears only once. These measures together offer different insights into the central tendency of the dataset.
Sample Variance
Sample variance is a statistic that describes how much individual data points differ from the mean in a sample. It reflects the dispersion and provides insights into the variability of the dataset. Variance is essential for understanding data spread and potential outliers.

To calculate the sample variance, first, determine the mean of the dataset. Then subtract this mean from each data point to find its deviation. Square each deviation to remove negatives and highlight variability. Sum these squared deviations and divide by one less than the number of observations (n-1) for statistical accuracy, adjusting for the sample variability. This results in a sample variance of 0.7387 for the Zinfandel alcohol content.

This measure of variability helps in assessing the consistency of wine strength across different producers, where a lower variance indicates more consistent alcohol content levels.
Defining Formula
The defining formula for sample variance is straightforward yet foundational in statistics. It calculates variance by a step-by-step process:
  • Identify the mean of the dataset.
  • Subtract the mean from each data point to calculate the deviation.
  • Square each deviation (to avoid negative values and exaggerate the differences).
  • Sum all squared deviations.
  • Finally, divide by the sample size minus one (n-1).
The formula is expressed as:\[ \frac{\sum (x_i - \bar{x})^2}{n-1} \]Where \( x_i \) represents each data point, and \( \bar{x} \) is the mean of the sample. This formula provides a comprehensive approach to calculate variance and interpret the variability within the dataset, as demonstrated in calculating the variance of Zinfandel alcohol contents.
Shortcut Formula
The shortcut formula for sample variance is a simplified method that can save time, especially when dealing with large datasets. It involves altering the dataset to make calculations easier while still achieving the same result. Here’s how it works:
  • Choose a constant (often a value close to the mean) to subtract from each data point, reducing data size and simplifying squaring operations.
  • Calculate the transformed mean \( \bar{y} \).
  • Find \( \sum y_i \) and \( \sum y_i^2 \).
The shortcut formula is given by:\[\text{Sample Variance} = \frac{\sum y_i^2 - \frac{(\sum y_i)^2}{n}}{n-1}\]This formula proved effective when subtracting 13 from each Zinfandel data point. The resulting sample variance was 0.8373, demonstrating this method's efficiency while maintaining accuracy.

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Most popular questions from this chapter

The three measures of center introduced in this chapter are the mean, median, and trimmed mean. Two additional measures of center that are occasionally used are the midrange, which is the average of the smallest and largest observations, and the midfourth, which is the average of the two fourths. Which of these five measures of center are resistant to the effects of outliers and which are not? Explain your reasoning.

The accompanying frequency distribution of fracture strength (MPa) observations for ceramic bars fired in a particular kiln appeared in the article "Evaluating Tunnel Kiln Performance" (Amer. Ceramic Soc. Bull., Aug. 1997: 59-63). \(\begin{array}{lcccccc}\text { Class } & 81-<83 & 83-<85 & 85-<87 & 87-<89 & 89-<91 \\ \text { Frequency } & 6 & 7 & 17 & 30 & 43 \\ \text { Class } & 91-<93 & 93-<95 & 95-<97 & 97-<99 \\ \text { Frequency } & 28 & 22 & 13 & 3\end{array}\) a. Construct a histogram based on relative frequencies, and comment on any interesting features. b. What proportion of the strength observations are at least 85 ? Less than 95 ? c. Roughly what proportion of the observations are less than \(90 ?\)

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.

Allowable mechanical properties for structural design of metallic aerospace vehicles requires an approved method for statistically analyzing empirical test data. The article "Establishing Mechanical Property Allowables for Metals" (J. of Testing and Evaluation, 1998: 293-299) used the accompanying data on tensile ultimate strength (ksi) as a basis for addressing the difficulties in developing such a method. \(\begin{array}{lllllllll}122.2 & 124.2 & 124.3 & 125.6 & 126.3 & 126.5 & 126.5 & 127.2 & 127.3 \\ 127.5 & 127.9 & 128.6 & 128.8 & 129.0 & 129.2 & 129.4 & 129.6 & 130.2 \\ 130.4 & 130.8 & 131.3 & 131.4 & 131.4 & 131.5 & 131.6 & 131.6 & 131.8 \\ 131.8 & 132.3 & 132.4 & 132.4 & 132.5 & 132.5 & 132.5 & 132.5 & 132.6 \\ 132.7 & 132.9 & 133.0 & 133.1 & 133.1 & 133.1 & 133.1 & 133.2 & 133.2 \\ 133.2 & 133.3 & 133.3 & 133.5 & 133.5 & 133.5 & 133.8 & 133.9 & 134.0 \\ 134.0 & 134.0 & 134.0 & 134.1 & 134.2 & 134.3 & 134.4 & 134.4 & 134.6 \\ 134.7 & 134.7 & 134.7 & 134.8 & 134.8 & 134.8 & 134.9 & 134.9 & 135.2 \\ 135.2 & 135.2 & 135.3 & 135.3 & 135.4 & 135.5 & 135.5 & 135.6 & 135.6 \\ 135.7 & 135.8 & 135.8 & 135.8 & 135.8 & 135.8 & 135.9 & 135.9 & 135.9 \\ 135.9 & 136.0 & 136.0 & 136.1 & 136.2 & 136.2 & 136.3 & 136.4 & 136.4 \\ 136.6 & 136.8 & 136.9 & 136.9 & 137.0 & 137.1 & 137.2 & 137.6 & 137.6 \\ 137.8 & 137.8 & 137.8 & 137.9 & 137.9 & 138.2 & 138.2 & 138.3 & 138.3 \\ 138.4 & 138.4 & 138.4 & 138.5 & 138.5 & 138.6 & 138.7 & 138.7 & 139.0 \\ 139.1 & 139.5 & 139.6 & 139.8 & 139.8 & 140.0 & 140.0 & 140.7 & 140.7 \\ 140.9 & 140.9 & 141.2 & 141.4 & 141.5 & 141.6 & 142.9 & 143.4 & 143.5 \\ 143.6 & 143.8 & 143.8 & 143.9 & 144.1 & 144.5 & 144.5 & 147.7 & 147.7\end{array}\) a. Construct a stem-and-leaf display of the data by first deleting (truncating) the tenths digit and then repeating each stem value five times (once for leaves 1 and 2 , a second time for leaves 3 and 4 , etc.). Why is it relatively easy to identify a representative strength value? b. Construct a histogram using equal-width classes with the first class having a lower limit of 122 and an upper limit of 124 . Then comment on any interesting features of the histogram.

The accompanying frequency distribution on deposited energy \((\mathrm{mJ})\) was extracted from the article "Experimental Analysis of Laser-Induced Spark Ignition of Lean Turbulent Premixed Flames" (Combustion and Flame, 2013: 1414-1427). \(\begin{array}{rrrr}1.0-<2.0 & 5 & 2.0-<2.4 & 11 \\ 2.4-<2.6 & 13 & 2.6-<2.8 & 30 \\ 2.8-<3.0 & 46 & 3.0-<3.2 & 66 \\ 3.2-<3.4 & 133 & 3.4-<3.6 & 141 \\\ 3.6-<3.8 & 126 & 3.8-<4.0 & 92 \\ 4.0-<4.2 & 73 & 4.2-<4.4 & 38 \\ 4.4-<4.6 & 19 & 4.6-<5.0 & 11\end{array}\) a. What proportion of these ignition trials resulted in a deposited energy of less than \(3 \mathrm{~mJ}\) ? b. What proportion of these ignition trials resulted in a deposited energy of at least \(4 \mathrm{~mJ}\) ? c. Roughly what proportion of the trials resulted in a deposited energy of at least \(3.5 \mathrm{~mJ}\) ? d. Construct a histogram and comment on its shape.

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