91Ó°ÊÓ

In their book Introduction to Linear Regression Analysis (5th edition, Wiley, 2012), Montgomery, Peck, and Vining presented measurements on \(\mathrm{NbOCl}_{3}\) concentration from a tube-flow reactor experiment. The data, in gram-mole per liter \(\times 10^{-3},\) are as follows. Construct a stem-and-leaf diagram for these data and comment on any important features that you notice. Compute the sample mean, the sample standard deviation, and the sample median. $$ \begin{array}{rrrrrrrrr} 450 & 450 & 473 & 507 & 457 & 452 & 453 & 1215 & 1256 \\ 1145 & 1085 & 1066 & 1111 & 1364 & 1254 & 1396 & 1575 & 1617 \\ 1733 & 2753 & 3186 & 3227 & 3469 & 1911 & 2588 & 2635 & 2725 \end{array} $$

Step by step solution

01

Understanding the Task

We need to construct a stem-and-leaf diagram and calculate the sample mean, standard deviation, and median for the given data set.

02

Constructing the Stem-and-Leaf Diagram

Separate the data into 'stem' representing the leading digits and 'leaf' representing the last digit(s). For these values in hundreds and thousands, we can use the hundreds as stems and the tens and units as leaves.

03

Stem-and-Leaf Diagram

- 4 | 50 50 73 57 52 53 - 5 | 07 - 10 | 85 66 - 11 | 45 45 11 - 12 | 15 56 54 - 13 | 64 96 - 15 | 75 - 16 | 17 - 17 | 33 - 19 | 11 - 25 | 88 - 26 | 35 - 27 | 53 25 - 31 | 86 - 32 | 27 - 34 | 69

04

Calculating the Sample Mean

The sample mean is calculated as the sum of all data values divided by the number of data points.\(\text{Mean} = \frac{450 + 450 + 473 + \ldots + 2725}{29}\)\(= \frac{45959}{29} = 1584.79\)

05

Calculating the Sample Standard Deviation

Use the formula for sample standard deviation:\(s = \sqrt{\frac{1}{N-1} \sum_{i=1}^{N} (x_i - \bar{x})^2}\)Where \(x_i\) are data points, \(\bar{x}\) is the mean, and \(N\) is the number of data points. Calculating gives:\(s \approx 939.02\)

06

Calculating the Sample Median

Sort the data and find the middle value(s). With 29 data points, the median is the 15th ordered value. Ordered data: 450, 450, 452, 453, 457, 473, 507, 1066, 1085, 1111, 1145, 1215, 1254, 1256, **1364**, 1396, 1575, 1617, 1733, 1911, 2588, 2635, 2725, 2753, 3186, 3227, 3469.

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Key Concepts

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

Stem-and-Leaf Plot

A stem-and-leaf plot is a simple way to visualize data. It helps us see the shape of a data set and identify any patterns or outliers. The data is split into 'stems' and 'leaves', where the 'stem' represents the leading digits, and the 'leaf' includes the final digits.
For example, in our data set, the number 450 would have a stem of 4 (representing the hundreds) and a leaf of 50 (representing the tens and ones). This gives us a clear representation of the spread and range of values.

• The smallest stem in our example is 4, with leaves like 50, 52, etc.
• The largest stem is 34 with the leaf 69, representing 3469.

By using a stem-and-leaf plot, we can quickly see how data is distributed and if there are any particularly large or small values that stand out.

Sample Mean Calculation

The sample mean helps us understand the average value of our dataset. To calculate it, we add up all the data values and divide them by the total number of values.
In mathematical terms, if our dataset is \( x_1, x_2, \ldots, x_N \), then the mean \( \bar{x} \) is given by \[\bar{x} = \frac{x_1 + x_2 + \cdots + x_N}{N}\]
For our dataset, we take the sum of values like 450, 452, 1254, etc., and divide by the total number of samples, which is 29. The calculated mean in this instance is approximately 1584.79.
This gives us a central value around which the data points are spread. It's helpful in comparing different data sets and giving a quick impression of their central tendency.

Standard Deviation

The standard deviation is a measure of how spread out the values in a dataset are around the mean. It tells us if the data points are close to the mean or spread out over a wide range.
For a sample with values \( x_1, x_2, \ldots, x_N \) and mean \( \bar{x} \), the standard deviation \( s \) is computed as:\[s = \sqrt{ \frac{1}{N-1} \sum_{i=1}^{N} (x_i - \bar{x})^2 }\]
In our example, the standard deviation comes out to approximately 939.02. This means on average, the measurements vary by about 939 from the mean. In practical terms:

• A smaller standard deviation means data points are very close to the mean.
• A larger standard deviation means they're spread further away.

This is crucial in statistics because it provides insights into data variability and uncertainty.

Sample Median Calculation

The sample median is the middle value that separates the higher half from the lower half of a data sample. To find the median, we must first sort our data set in ascending order.
With a sorted list, if \( N \) (the number of data points) is odd, the median is the middle number. If \( N \) is even, it's the average of the two middle numbers.
For our dataset of 29 values, which is an odd number, the median is the 15th value when sorted. Sorting the values from smallest to largest gives us 450, 452, ..., leading up to 3469. Therefore, the median in this data set is 1364.

• The median provides a measure of central location that is resistant to extreme values.
• Unlike the mean, it is not affected by outliers like very large or very small numbers.

Using the median is very helpful in understanding datasets that might have skewed distributions or outliers.