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Chapter 12: Problem 2

Statistical Literacy If two or more data values are the same, how is the rank of each of the tied data computed?

Short Answer

Expert verified
Compute the average of the ranks tied values would occupy and assign it to each of the tied values.

Step by step solution

01

Understanding Ranks for Unique Values

Before addressing ties, we must understand that the rank of a data value is its position in an ordered list. If all values are unique, each value receives a distinct rank according to its order.
02

Identifying Tied Data Values

When two or more data values are identical, they are said to be 'tied.' For instance, if two data values are the same and occur at positions 4 and 5 in the sorted list, these positions are relevant for computing their rank.
03

Computing Mean Rank for Tied Values

To assign ranks to tied values, compute the average (mean) of the ranks they would occupy if they were distinct. Continuing with the previous example, the values in positions 4 and 5 would both receive a rank of \(\frac{4 + 5}{2} = 4.5\).
04

Assigning the Computed Rank

Once the mean rank is calculated, assign this average rank to all tied values instead of individual positions to ensure fairness across the data set.

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Key Concepts

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

Rank Calculation
Understanding rank calculation is essential for analyzing data sets. Ranking essentially involves assigning a position to each data point within an ordered list. This order is determined by the data values themselves, usually from lowest to highest. Being able to calculate ranks helps make sense of where each data point stands relative to others. For unique values, this is straightforward since each value occupies a distinct position.

However, complexities arise when data values are not unique. Nonetheless, the concept of ranking is crucial for many statistical methods, especially when trying to understand distributions and trends within data sets.
Tied Data Values
Tied data values occur when two or more data points have the same value within a data set. In a perfect world, each data point would be unique, but it often happens that some values tie. For instance, we might see this in test scores where two students achieve the exact same mark.

In statistics, tied values can be problematic if ranks are assigned naively, as doing so could undermine the integrity of the data analysis. Identifying these ties is the first step; once identified, a fair and consistent method for ranking them is necessary. This usually means determining their average position in the ordered set before assigning ranks.
Mean Rank
Calculating the mean rank for tied values is a common solution in statistics, ensuring fairness and uniformity. When data values are tied, rather than assigning different ranks which could skew results, the mean rank method calculates an average rank for the tied positions.

To compute a mean rank, add up the ranks the tied values would occupy if they were each unique, then divide by the number of tied values. For example, if there are ties at positions 3, 4, and 5, the mean rank is calculated as \( \frac{3 + 4 + 5}{3} \). This mean rank is then uniformly assigned to each of the tied values.
Data Sorting
Sorting data is one of the preliminary steps in rank calculation and is crucial for proper analysis. Data sorting involves organizing values in a manner that allows for easier comparison and ranking.

Typically, sorting is done in ascending or descending order depending on the needs of the analysis. This step simplifies the identification of unique and tied values, thereby directly guiding the rank assignment process. Well-sorted data lays a foundation for clear and effective interpretation of the statistical results.
Statistical Methods
Statistical methods are systematic approaches that utilize mathematical techniques to analyze data sets. Understanding these methods is integral to comprehending concepts like rank calculation and tied data values.

Choosing the appropriate method depends on the data type and the analysis objectives. For example, when dealing with tied ranks, the method of mean rank ensures that data interpretation remains fair and justifiable. These methods not only inform how we calculate and interpret rankings but also support deeper insights into data trends and patterns across various fields.

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Most popular questions from this chapter

Doctor's Degree: Years of Study Is the average length of time to earn a doctorate different from one field to another? Independent random samples from large graduate schools gave the following averages for length of registered time (in years) from bachelor's degree to doctorate. Sample A was taken from the humanities field, and sample \(\mathrm{B}\) from the social sciences field. (Reference: Education Statistics, U.S. Department of Education.) \begin{tabular}{l|cccccccccccc} \hline Field \(\mathrm{A}\) & \(8.9\) & \(8.3\) & \(7.2\) & \(6.4\) & \(8.0\) & \(7.5\) & \(7.1\) & \(6.0\) & \(9.2\) & \(8.7\) & \(7.5\) & \\ \hline Field B & \(7.6\) & \(7.9\) & \(6.2\) & \(5.8\) & \(7.8\) & \(8.3\) & \(8.5\) & \(7.0\) & \(6.3\) & \(5.4\) & \(5.9\) & \(7.7\) \\ \hline \end{tabular} Use a \(1 \%\) level of significance to test the claim that there is no difference in the distributions of time to complete a doctorate for the two fields.

Horse Trainer: Jumps A horse trainer teaches horses to jump by using two methods of instruction. Horses being taught by method \(\mathrm{A}\) have a lead horse that accompanies each jump. Horses being taught by method \(\mathrm{B}\) have no lead horse. The table shows the number of training sessions required before each horse performed the jumps properly. \begin{tabular}{l|cccccccccccc} \hline Method A & 28 & 35 & 19 & 41 & 37 & 31 & 38 & 40 & 25 & 27 & 36 & 43 \\\ \hline Method B & 42 & 33 & 26 & 24 & 44 & 46 & 34 & 20 & 48 & 39 & 45 & \\ \hline \end{tabular} Use a \(5 \%\) level of significance to test the claim that there is no difference between the training session distributions.

Psychology: Testing An army psychologist gave a random sample of seven soldiers a test to measure sense of humor and another test to measure aggressiveness. Higher scores mean greater sense of humor or more aggressiveness. \begin{tabular}{l|ccccccc} \hline Soldier & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline Score on humor test & 60 & 85 & 78 & 90 & 93 & 45 & 51 \\ Score on aggressiveness test & 78 & 42 & 68 & 53 & 62 & 50 & 76 \\ \hline \end{tabular} (i) Ranking the data with rank 1 for highest score on a test, make a table of ranks to be used in a Spearman rank correlation test. (ii) Using a \(0.05\) level of significance, test the claim that rank in humor has a monotone-decreasing relation to rank in aggressiveness.

Statistical Literacy Consider the Spearman rank correlation coefficient \(r_{s}\) for data pairs \((x, y) .\) What is the monotone relationship, if any, between \(x\) and \(y\) implied by a value of (a) \(r_{s}=0 ?\) (b) \(r_{s}\) close to \(1 ?\) (c) \(r_{s}\) close to \(-1 ?\)

Economics: Stocks As an economics class project, Debbie studied a random sample of 14 stocks. For each of these stocks, she found the cost per share (in dollars) and ranked each of the stocks according to cost. After 3 months, she found the earnings per share for each stock (in dollars). Again, Debbie ranked each of the stocks according to earnings. The way Debbie ranked, higher ranks mean higher cost and higher earnings. The results follow, where \(x\) is the rank in cost and \(y\) is the rank in earnings. \begin{tabular}{l|rrrrrrrrrrrrrr} \hline tock & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline rank & 5 & 2 & 4 & 7 & 11 & 8 & 12 & 3 & 13 & 14 & 10 & 1 & 9 & 6 \\ rank & 5 & 13 & 1 & 10 & 7 & 3 & 14 & 6 & 4 & 12 & 8 & 2 & 11 & 9 \\ \hline \end{tabular} Using a \(0.01\) level of significance, test the claim that there is a monotone relation, either way, between the ranks of cost and earnings.
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