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Chapter 2: Problem 17

An article in Archaeometry involved an analysis of 26 samples of Romano- British pottery found at four different kiln sites in the United Kingdom. \({ }^{7}\) The samples were analyzed to determine their chemical composition. The percentage of iron oxide in each of five samples collected at the Island Thorns site was: \(\begin{array}{llll}1.28, & 2.39, & 1.50, & 1.88, & 1.51\end{array}\) a. Calculate the range. b. Calculate the sample variance and the standard deviation using the computing formula. c. Compare the range and the standard deviation. The range is approximately how many standard deviations?

Short Answer

Expert verified
Answer: There are approximately 2.54 standard deviations in the range of the dataset.

Step by step solution

01

Dataset given for the percentage of iron oxide

The dataset containing the percentage of iron oxide in five samples of Romano-British pottery collected at the Island Thorns site is given as: \(\begin{array}{llll}1.28, & 2.39, & 1.50, & 1.88, & 1.51\end{array}\)
02

Calculate the Range

To calculate the range, we need to find the difference between the highest and the lowest values in the dataset. The highest value is 2.39, and the lowest value is 1.28. Range = Highest value - Lowest value Range = \(2.39 - 1.28 = 1.11\) So, the range of the data is 1.11.
03

Calculate the Sample Mean

To calculate the sample variance and standard deviation, we need to find the sample mean first. The sample mean can be found by summing all the values in the dataset and dividing it by the number of samples: Mean = \(\frac{\sum_{i=1}^n x_i}{n}\) Mean = \(\frac{1.28 + 2.39 + 1.50 + 1.88 + 1.51}{5} = \frac{8.56}{5} = 1.712\) Thus, the sample mean is 1.712.
04

Calculate the Sample Variance

Now, we can calculate the sample variance using the computing formula. Computing formula for sample variance is given by: \(s^2 = \frac{\sum_{i=1}^n (x_i - \bar{x})^2}{n-1}\) First, let's find the sum of squares for the dataset \((x_i - \bar{x})^2\): Sum of squares = \((1.28-1.712)^2 + (2.39-1.712)^2 + (1.50-1.712)^2 + (1.88-1.712)^2 + (1.51-1.712)^2\) Sum of squares = \((0.1924 + 0.4581 + 0.0449 + 0.0283 + 0.0406)\) Sum of squares = \(0.7643\) Now, let's use this value to find the sample variance: Sample Variance = \(\frac{0.7643}{5-1} = \frac{0.7643}{4} = 0.1911\) So, the sample variance of the data is 0.1911.
05

Calculate the Standard Deviation

The standard deviation is the square root of the sample variance. Standard Deviation = \(\sqrt{Sample\ Variance} = \sqrt{0.1911} = 0.437\) Thus, the standard deviation of the data is approximately 0.437.
06

Compare the Range and the Standard Deviation

To find how many standard deviations are there in the range, we need to compare the range (1.11) and the standard deviation (0.437). So, we'll divide the range by the standard deviation: \(\frac{Range}{Standard\ Deviation} = \frac{1.11}{0.437} \approx 2.54\) So, the range is approximately 2.54 standard deviations.

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Key Concepts

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

Range Calculation
Understanding the range in statistical analysis is fundamental for grasping the dispersion of data points. Range represents the difference between the highest and lowest values within a dataset. In our exercise involving the analysis of iron oxide in pottery samples, the range is calculated by subtracting the smallest percentage (1.28) from the largest (2.39), yielding a range of 1.11.

The significance of the range lies in its ability to provide a snapshot of the data's spread; however, it is sensitive to outliers and does not reflect the distribution of all the data points. Despite its limitations, range is a quick and straightforward measure of variability, especially useful in non-complex data sets.
Sample Variance
Sample variance provides a numerical value that represents how data points in a set are spread out from their mean. It is an estimation of the variance within an entire population. To compute sample variance, one calculates the squared deviations of each data point from the sample mean and then averages these squared deviations.

In our Archaeometry article example, the sample variance is found by squaring each sample's deviation from the mean, summing these squares, and finally dividing by the number of samples minus one (to account for the sample as an estimator of the population). This process yielded a sample variance of 0.1911, which measures the iron oxide content variability among the pottery samples.
Standard Deviation
The standard deviation is a step further in the dissection of our data's spread鈥攊t's the square root of the sample variance, offering a measure of dispersion that is in the same unit as the original data.

It tells us, on average, how far each data point is from the mean. For the pottery sample analysis, we calculated a standard deviation of 0.437. Compared to the range, the standard deviation provides a more robust picture of variability because it takes into account all data points, not just the extreme values.

Understanding Variability

Standard deviation is widely accepted as a vital tool since it helps to understand the amount of variety or dispersion for a set of data. In the context of the exercise, it reflected how much variation in the percentage of iron oxide could be expected within samples from the same site. Thus, the standard deviation is more informative than the range, illustrating variability within the context of each piece of data included in the sample.

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Most popular questions from this chapter

The number of television viewing hours per household and the prime viewing times are two factors that affect television advertising income. A random sample of 25 households in a particular viewing area produced the following estimates of viewing hours per household: $$ \begin{array}{rrrrr} 3.0 & 6.0 & 7.5 & 15.0 & 12.0 \\ 6.5 & 8.0 & 4.0 & 5.5 & 6.0 \\ 5.0 & 12.0 & 1.0 & 3.5 & 3.0 \\ 7.5 & 5.0 & 10.0 & 8.0 & 3.5 \\ 9.0 & 2.0 & 6.5 & 1.0 & 5.0 \end{array} $$ a. Scan the data and use the range to find an approximate value for \(s\). Use this value to check your calculations in part \(\mathrm{b}\). b. Calculate the sample mean \(\bar{x}\) and the sample standard deviation \(s\). Compare \(s\) with the approximate value obtained in part a. c. Find the percentage of the viewing hours per household that falls into the interval \(\bar{x} \pm 2 s\). Compare with the corresponding percentage given by the Empirical Rule.

A favorite summer pastime for many Americans is camping. In fact, camping has become so popular at the California beaches that reservations must sometimes be made months in advance! Data from a USA Today Snapshot is shown below. \({ }^{13}\) The Snapshot also reports that men go camping 2.9 times a year, women go 1.7 times a year; and men are more likely than women to want to camp more often. What does the magazine mean when they talk about 2.9 or 1.7 times a year?

Find the five-number summary and the IQR for these data: $$ 19,12,16,0,14,9,6,1,12,13,10,19,7,5,8 $$

A set of \(n=10\) measurements consists of the values \(5,2,3,6,1,2,4,5,1,3 .\) a. Use the range approximation to estimate the value of \(s\) for this set. (HINT: Use the table at the end of Section \(2.5 .)\) b. Use your calculator to find the actual value of \(s\). Is the actual value close to your estimate in part a? c. Draw a dotplot of this data set. Are the data moundshaped? d. Can you use Tchebysheff's Theorem to describe this data set? Why or why not? e. Can you use the Empirical Rule to describe this data set? Why or why not?

Here are a few facts reported as Snapshots in USA Today. \- The median hourly pay for salespeople in the building supply industry is \(\$ 10.41 .^{15}\) \- Sixty-nine percent of U.S. workers ages 16 and older work at least 40 hours per week. \({ }^{16}\) \- Seventy-five percent of all Associate Professors of Mathematics in the U.S. earn \(\$ 91,823\) or less. \(^{17}\) Identify the variable \(x\) being measured, and any percentiles you can determine from this information.
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