Question

In the statistical treatment of data one often needs to compute the quantities $$ \bar{x}=\frac{1}{n} \sum_{i=1}^{n} x_{i}, \quad s^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}, $$ where \(x_{1}, x_{2}, \ldots, x_{n}\) are the given data. Assume that \(n\) is large, say, \(n=10,000\). It is easy to see that \(s^{2}\) can also be written as $$ s^{2}=\frac{1}{n} \sum_{i=1}^{n} x_{i}^{2}-\bar{x}^{2} $$ (a) Which of the two methods to calculate \(s^{2}\) is cheaper in terms of overall computational cost? Assume \(\bar{x}\) has already been calculated and give the operation counts for these two options. (b) Which of the two methods is expected to give more accurate results for \(s^{2}\) in general? (c) Give a small example, using a decimal system with precision \(t=2\) and numbers of your choice, to validate your claims.

Short answer

Answer: Method 1, which calculates the variance using deviations from the mean, is more accurate in general because there is a lower risk of loss of significance or round-off errors compared to Method 2, which calculates the variance using the mean of squares and is more prone to these errors when dealing with large numbers and subtraction of large values.

Step-by-step solution

Method 1: Variance using the deviation from the mean

We are given the following formula for this method: $$ s^{2}=\frac{1}{n}\sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} $$ Assuming we have already calculated the mean \(\bar{x}\), we need the following operations for each \(x_i\): subtraction, squaring, and then addition to the sum. This needs to be done n (10,000) times. So, the total number of operations required for this method is approximately \(3n\).

Method 2: Variance using the mean of squares

The second formula to calculate the variance is given as: $$ s^{2}=\frac{1}{n} \sum_{i=1}^{n} x_{i}^{2}-\bar{x}^{2} $$ For this method, we first need to calculate the square of each \(x_i\) and then add them up. Finally, we subtract the square of the mean \(\bar{x}\). So, the total number of operations required for this method is approximately \(2n + 1\) (squaring n times, addition n-1 times, and subtraction once).

Comparison: Computational cost

Comparing both methods, we can see that Method 2 is cheaper in terms of computational cost as it requires fewer operations \((2n+1) < 3n\).

Comparison: Accuracy

Method 1 calculates the variance directly using the deviations from the mean, so there is a lower risk of loss of significance or round-off errors. In contrast, Method 2 is more prone to such errors, especially when calculating the squares of large numbers and then subtracting large values. Therefore, Method 1 is expected to give more accurate results in general.

Example with a decimal system of t=2

Let's take the following data points with t=2 (2 decimal places): [1.25, 1.66, 1.33]. 1. Calculate mean: \(\bar{x} = \frac{1}{3}(1.25 + 1.66 + 1.33) \approx 1.41\). For Method 1: 2a. Calculate deviations from mean: [1.25 - 1.41, 1.66 - 1.41, 1.33 - 1.41] ≈ [-0.16, 0.25, -0.08]. 2b. Calculate squares of deviations: [0.03, 0.06, 0.01]. 2c. Calculate variance: \(s^{2} = \frac{1}{3}(0.03 + 0.06 + 0.01) \approx 0.03\). For Method 2: 3a. Calculate squares of data points: [1.56, 2.75, 1.77]. 3b. Calculate mean of squares: \(\frac{1}{3}(1.56 + 2.75 + 1.77) \approx 2.03\). 3c. Calculate the square of the mean (1.41)^2 ≈ 1.99. 3d. Calculate variance: \(s^2 = 2.03 - 1.99 \approx 0.04\). In this example, using t=2 precision, Method 1 gives us a more accurate result for the variance since the value for Method 2 is slightly off due to the loss of significance in the arithmetic operations.

Key concepts

Statistical Treatment of Data

Understanding the statistical treatment of data involves mastering several core techniques that allow us to summarize and interpret sets of numbers. The ultimate goal is to transform raw data into meaningful insights, which can then be used to make informed decisions.

In statistics, the concept of variance is integral to understanding data dispersion. Variance helps us appreciate just how much individual data points in a set differ from the average value. It's a critical concept, not just in theoretical mathematics but also in practical domains like engineering, social science, and almost any field that relies on data analysis.

Executing such statistical calculations with a large dataset can be challenging. Computing the mean, variance, and other statistics, must be done with a balance between computational efficiency and the precision of results. This is especially relevant when dealing with large numbers where computation errors can accumulate, impacting the integrity of the findings.

Variance Formula

The variance of a dataset is essentially an average of the squared deviations from the mean. There are two primary methods to calculate variance denoted by the formula, the first being variance using the deviation from the mean \(s^2 = \frac{1}{n}\sum_{i=1}^{n}(x_{i}-\bar{x})^2\), and the second using the mean of the squares \(s^2 = \frac{1}{n} \sum_{i=1}^{n} x_{i}^2 - \bar{x}^2\).

Understanding the difference between these two variance formulas is crucial for students not only to perform calculations but also to choose the methodology that suits their computational resources and accuracy requirements.

Method Comparison

Even though logically, these methods should yield the same result, in practice, due to limited numerical precision (e.g., computational limits, rounding off), they can produce slightly different results. When dealing with a vast array of data, the choice between these formulas can influence both the computational cost and the susceptibility to errors.

Numerical Precision and Accuracy

Numerical precision and accuracy are central to statistics and scientific computations. They refer to how closely a calculated or measured value agrees with the true value and the extent to which these values are expressed with a fine degree of detail.

Impact on Variance Calculation

When it comes to calculating variance, the numerical precision matters because the operations involved may amplify any round-off errors. For large datasets, these errors could significantly affect the final result. A precise t-value determines the number of significant digits with which the computation is carried out, and selecting an appropriate t-value is vital for maintaining the integrity of the result.

Method 1, involving direct deviations from the mean, generally conserves numerical accuracy better than Method 2, which is more prone to round-off errors because of the way large numbers are squared and then combined. This feature makes Method 1 preferable in situations where numerical precision is non-negotiable, despite its higher computational cost.