Problem 6 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

Consider two datasets: \(1,5,9\) and \(2,4,6,8 .\) a. Denote the sample means of the two datasets by \(\bar{x}\) and \(\bar{y}\). Is it true that the average \((\bar{x}+\bar{y}) / 2\) of \(\bar{x}\) and \(\bar{y}\) is equal to the sample mean of the combined dataset with 7 elements? b. Suppose we have two other datasets: one of size \(n\) with sample mean \(\bar{x}_{n}\) and another dataset of size \(m\) with sample mean \(\bar{y}_{m} .\) Is it always true that the average \(\left(\bar{x}_{n}+\bar{y}_{m}\right) / 2\) of \(\bar{x}_{n}\) and \(\bar{y}_{m}\) is equal to the sample mean of the combined dataset with \(n+m\) elements? If no, then provide a counterexample. If yes, then explain this. c. If \(m=n\), is \(\left(\bar{x}_{n}+\bar{y}_{m}\right) / 2\) equal to the sample mean of the combined dataset with \(n+m\) elements?

Step-by-Step Solution

Answer

a. Yes, it's equal. b. No, except when sizes are equal. c. Yes, if sizes are equal.

Step 1: Calculate Sample Mean of First Dataset

For the first dataset, the numbers are 1, 5, and 9. The sample mean is calculated as follows:\[ \bar{x} = \frac{1 + 5 + 9}{3} = \frac{15}{3} = 5. \]

Step 2: Calculate Sample Mean of Second Dataset

For the second dataset, the numbers are 2, 4, 6, and 8. The sample mean is calculated as:\[ \bar{y} = \frac{2 + 4 + 6 + 8}{4} = \frac{20}{4} = 5. \]

Step 3: Calculate Average of Sample Means

The average of the two sample means \( \bar{x} = 5 \) and \( \bar{y} = 5 \) is:\[ \frac{\bar{x} + \bar{y}}{2} = \frac{5 + 5}{2} = 5. \]

Step 4: Calculate Sample Mean of Combined Dataset

Combine both datasets: 1, 5, 9, 2, 4, 6, 8. The combined sample mean is:\[ \bar{z} = \frac{1 + 5 + 9 + 2 + 4 + 6 + 8}{7} = \frac{35}{7} = 5. \]

Step 5: Analyze General Case for Different Dataset Sizes

The sample mean of combined datasets depends on the size and values of each. The formula \( \frac{\bar{x}_{n} + \bar{y}_{m}}{2} \) only gives the mean if \(n = m\). Otherwise, values should be weighted by the number of elements.

Step 6: Provide Counterexample for Different Sizes

Example: Suppose \( \bar{x}_{n} = 2 \) for set size 2 and \( \bar{y}_{m} = 8 \) for set size 3. Using \( (2+8)/2 = 5 \) for combined mean is incorrect as actual mean is:\[ \bar{z} = \frac{2\times2 + 3\times8}{5} = \frac{28}{5} = 5.6. \]

Step 7: Verify for Equal Sizes

If sizes are the same \( m = n \), then indeed the average \( \frac{\bar{x}_{n} + \bar{y}_{m}}{2} \) matches the sample mean of the combined dataset because each mean is equally weighted.

Key Concepts

Understanding Datasets

When discussing statistics, we often work with datasets, which are collections of data points or numbers organized for analysis. In the original exercise, we used two datasets, which are simply groups of numerical values. The first dataset had three numbers: 1, 5, and 9. The second dataset included four numbers: 2, 4, 6, and 8. Each dataset provides a set of values from which statistical analyses, like finding the sample mean, can be performed.

The sample mean is a measure of average value in a dataset. It provides a single, representative value that summarizes the entire dataset. To calculate it, add all data points together and divide by the total number of points. Each dataset can have its unique characteristics based on the data it contains, such as average value (mean), spread (variance), and more.

In our example, calculating the sample means for these datasets helped us understand the central tendency of each set of numbers. This concept is a fundamental building block in statistics, helping to simplify and summarize data collected for analysis.

The Role of Weighted Averages

Weighted averages are used when combining datasets of different sizes. If each dataset has a different number of data points, their means should contribute proportionally based on their sizes to the overall average. This ensures that each value's influence is relative to how many times it appears in total.

In the exercise, the regular mean of two datasets \((\bar{x} + \bar{y}) / 2\) assumes equal contribution, regardless of size. However, when datasets are of different sizes, use a weighted average:

• Multiply each dataset's mean by its respective size.
• Add these products together.
• Divide by the total number of data points from all combined datasets.

This method accurately reflects the average taking into account the number of items in each dataset, preventing over- or under-representation of any single dataset.

Consider an example where one dataset has only two data points with a mean of 2 and another has three data points with a mean of 8. The true mean using a weighted approach is more accurate because it accounts for the additional data's increased influence.

Exploring Counterexamples

In mathematics, a counterexample is a specific case for which a general statement does not hold true. These examples help to clarify rules by demonstrating exceptions or misapplications. In the task at hand, the idea that the arithmetic average of two sample means \(\frac{\bar{x}_n + \bar{y}_m}{2}\) is the same as the sample mean of the combined dataset can be examined using counterexamples.

Consider two sets: one with a small sample mean and small size, another with a large sample mean and large size. For instance, a dataset of size 2 with a mean of 2 and another of size 3 with a mean of 8 won't balance out simply by averaging the two means. Using the weighted average method, as calculated earlier, gives a more accurate picture of the actual central tendency. In our counterexample:

• The mean of the smaller dataset misrepresents its significance in the combined set if only the simple average is used.
• The larger dataset's mean rightfully exerts more influence due to its greater size.

Thus, the arithmetic mean of two sample means only equals the combined mean when both datasets have the same size, highlighting the limitation and guidance given by counterexamples.