## The medians test and the examination of the math test scores in Section reduce every observation of a coin toss. Specifically, the medians test judges every observation as being either above or below the pooled sample median. The actual magnitude of every observation is lost. It does not matter how far above or below the median an observation is. Does this seem like a tremendous loss of information? Rank methods meet this loss halfway: Instead of reducing all observations to binary above\below status, rank methods replace the actual observations with their order when the data is sorted.                                                                                                              To illustrate this method, consider the plant growth data from Table . All 20 of the pooled sample values are sorted in Table  from smallest to largest and identified as belonging to either the control or the treated groups. Each observation is also identified with its rank or order number, from 1 to 20 in terms of the pooled sampled. So, for example, the four smallest observation (ranked 1-4) are associated with the control group, and then the next two smallest observations (ranked 5 and 6) are in the treated group.

In rank methods, we replace the observations by their ranks. A large observations achieves a high rank, unlike the medians test where all observations larger than the median are treated equally.   In Table , if  both the treated and the control plants had roughly the same means, then we would also expect these two samples to have roughly the same average rank values.                                           We will return to this example and the SAS program for the rank test in Section . Before we get to that, let us again illustrate the use of ranked data and show that this is a frequently used approach to describing data that is sometimes qualitative, rather than quantitative, in nature.

### Intersection of subring

How to prove the intersection of two subrings is a subring? Let S1 and S2  be two subrings of a ring R. Then S1∩S2 is not empty since ...