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Lesson 15 - 6. Inferences Between Two Variables. Objectives. Perform Spearman’s rank-correlation test. Vocabulary. Rank-correlation test -- nonparametric procedure used to test claims regarding association between two variables.
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Lesson 15 - 6 Inferences Between Two Variables
Objectives • Perform Spearman’s rank-correlation test
Vocabulary • Rank-correlation test -- nonparametric procedure used to test claims regarding association between two variables. • Spearman’s rank-correlation coefficient -- test statistic, rs 6Σdi² rs = 1 – -------------- n(n²- 1)
Association • Parametric test for correlation: • Assumption of bivariate normal is difficult to verify • Used regression instead to test whether the slope is significantly different from 0 • Nonparametric case for association: • Compare the relationship between two variables without assuming that they are bivariate normal • Perform a nonparametric test of whether the association is 0
Tale of Two Associations Similar to our previous hypothesis tests, we can have a two-tailed, a left-tailed, or a right-tailed alternate hypothesis • A two-tailed alternative hypothesis corresponds to a test of association • A left-tailed alternative hypothesis corresponds to a test of negative association • A right-tailed alternative hypothesis corresponds to a test of positive association
Test Statistic for Spearman’s Rank-Correlation Test Small Sample Case: (n ≤ 100) The test statistic will depend on the size of the sample, n, and on the sum of the squared differences (di²). 6Σdi² rs = 1 – -------------- n(n²- 1) where di = the difference in the ranks of the two observations (Yi – Xi) in the ith ordered pair. Spearman’s rank-correlation coefficient, rs, is our test statistic z0 = rs √n – 1 Large Sample Case: (n > 100)
Critical Value for Spearman’s Rank-Correlation Test Small Sample Case: (n ≤ 100) Using αas the level of significance, the critical value(s) is (are) obtained from Table XIII in Appendix A. For a two-tailed test, be sure to divide the level of significance, α, by 2. Large Sample Case: (n > 100)
Hypothesis Tests Using Spearman’s Rank-Correlation Test Step 0 Requirements: 1. The data are a random sample of n ordered pairs. 2. Each pair of observations is two measurements taken on the same individual Step 1 Hypotheses:(claim is made regarding relationship between two variables, X and Y) H0: see below H1: see below Step 2 Ranks:Rank the X-values, and rank the Y-values. Compute the differences between ranks and then square these differences. Compute the sum of the squared differences. Step 3 Level of Significance:(level of significance determines the critical value) Table XIII in Appendix A. (see below) Step 4 Compute Test Statistic: Step 5 Critical Value Comparison: 6Σdi² rs = 1 – -------------- n(n²- 1)
Expectations • If X and Y were positively associated, then • Small ranks of X would tend to correspond to small ranks of Y • Large ranks of X would tend to correspond to large ranks of Y • The differences would tend to be small positive and small negative values • The squared differences would tend to be small numbers • If X and Y were negatively associated, then • Small ranks of X would tend to correspond to large ranks of Y • Large ranks of X would tend to correspond to small ranks of Y • The differences would tend to be large positive and large negative values • The squared differences would tend to be large numbers
Example 1 from 15.6 Calculations:
Example 1 Continued • Hypothesis: H0: X and Y are not associated Ha: X and Y are associated • Test Statistic: 6 Σdi² 6 (6) 36rs = 1 - ----------- = 1 – ------------- = 1 - -------- = 0.929 n(n² - 1) 8(64 - 1) 8(63) • Critical Value: 0.738 (from table XIII) • Conclusion: Since rs > CV, we reject H0; therefore there is a relationship between club-head speed and distance.
Summary and Homework • Summary • The Spearman rank-correlation test is a nonparametric test for testing the association of two variables • Test is a comparison of the ranks of the paired data values • Critical values for small samples are given in tables • Critical values for large samples can be approximated by a calculation with the normal distribution • Homework • problems 3, 6, 7, 10 from the CD