Question

Use the range to approximate the value of \(s\). Then calculate the actual value of \(s .\) Is the actual value close to the estimate? $$ n=10 \text { measurements: } 5,2,3,6,1,2,4,5,1,3 $$

Short answer

Answer: In this exercise, the estimated value of \(s\) (Range) is 5, while the actual value of \(s\) (Standard deviation) is approximately 1.646. The difference between the estimated and actual values shows that the range may not be an accurate measure of dispersion, as it only considers the highest and lowest values in a dataset and ignores the variability of all the data points. The standard deviation is generally a better measure of spread because it takes into account the variability of all the data points.

Step-by-step solution

Find the Range of Measurements

To find the range, we will identify the highest and lowest value in the given set of measurements and subtract the lowest value from the highest value. Range = Highest Value - Lowest Value In our case, the measurements are: 5, 2, 3, 6, 1, 2, 4, 5, 1, 3 The highest value is 6, and the lowest value is 1. So, Range = 6 - 1 = 5.

Approximate the Value of \(s\) Using the Range

To approximate the value of \(s\), we will simply consider the range we calculated in the previous step. Approximated value of \(s\) = Range = 5.

Calculate the Actual Value of \(s\)

We are not given any specific method to calculate the actual value of \(s\). However, we can assume that the question is asking us to calculate the standard deviation, which is a commonly used measure of spread or dispersion in statistics. To calculate the standard deviation (\(s\)), follow these steps: 1. Calculate the mean (\(\bar{x}\)) of the measurements. 2. Subtract the mean from each value and square the differences. 3. Calculate the average of these squared differences. 4. Take the square root of this average. Mean (\(\bar{x}\)) = (5+2+3+6+1+2+4+5+1+3)/10 = 32/10 = 3.2 Squared differences: (5 - 3.2)^2 = 3.24 (2 - 3.2)^2 = 1.44 (3 - 3.2)^2 = 0.04 (6 - 3.2)^2 = 7.84 (1 - 3.2)^2 = 4.84 (2 - 3.2)^2 = 1.44 (4 - 3.2)^2 = 0.64 (5 - 3.2)^2 = 3.24 (1 - 3.2)^2 = 4.84 (3 - 3.2)^2 = 0.04 Sum of squared differences = 27.12 Mean of squared differences = 27.12/10 = 2.712 Standard deviation (\(s\)) = \(\sqrt{2.712}\) ≈ 1.646

Compare the Estimated and Actual Values

Now that we have both estimated and actual values of \(s\), let's compare them: Estimated value of \(s\) (Range) = 5 Actual value of \(s\) (Standard deviation) ≈ 1.646 The actual value of \(s\) is not particularly close to the estimated value (range). This shows that the range can sometimes provide a rough approximation of the spread or dispersion in a dataset, but it is not always an accurate measure. The standard deviation is generally a better measure of spread because it takes into account the variability of all the data points.

Key concepts

Range

The range is a simple way to understand how spread out the numbers in a data set are. It helps us identify the difference between the largest and the smallest values. This basic statistical tool gives us an indication of how well-spread or how concentrated the data points are on a number line. To calculate the range, you subtract the smallest number in the data set from the largest.
For example, in our data set of measurements: 5, 2, 3, 6, 1, 2, 4, 5, 1, 3, the largest number is 6 and the smallest is 1, making the range 6 - 1 = 5.
While the range is easy to compute and provides a snapshot of data dispersion, it fails to convey the distribution of numbers in between the extremes.

Mean

The mean is an average that gives us a central value around which the numbers in the data set cluster. It's calculated by adding all the numbers together and dividing by the number of data points. In statistics, the mean is often used because it considers every value in the data set.
With our example, where the data set is 5, 2, 3, 6, 1, 2, 4, 5, 1, 3, the mean is calculated as:
\[ \bar{x} = \frac{5 + 2 + 3 + 6 + 1 + 2 + 4 + 5 + 1 + 3}{10} = 3.2 \]This mean shows us that if each value were evenly distributed, every data point would be close to 3.2. It's a good first step in summarizing data, though it doesn't provide information about dispersion.

Data Dispersion

Data dispersion relates to how spread out the data points are around the mean. When data points diverge widely from the mean, we have high dispersion. Conversely, when they are clustered near the mean, we have low dispersion.
Understanding data dispersion is crucial because two sets of numbers can have the same mean but different dispersions. In simple terms, it provides insight into the consistency of the data.

• High dispersion: larger range of differences among numbers
• Low dispersion: smaller range of differences among numbers

In our example, using the variance and standard deviation gives us a more comprehensive understanding of data dispersion.

Statistical Measurement

Statistical measurement provides tools for summarizing and interpreting quantitative data. It allows us to move beyond just raw numbers to meaningful interpretations. Standard deviation is one such measurement. It tells us how much the data deviates from the mean on average.
To calculate the standard deviation, follow these steps:

• Find the mean of the data.
• Calculate each data point's deviation from the mean, and square it.
• Compute the average of these squared deviations.
• Take the square root of this average.

For instance, our data set's standard deviation is around 1.646. This means that the data points tend, on average, to be 1.646 units away from the mean.
The standard deviation often provides a clearer picture of data dispersion compared to range, as it accounts for every data point's role in the distribution.