Download - Lesson 15 - 6

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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

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

Small Sample Case: (n ≤ 100)

Large Sample Case: (n > 100)

Critical Value for Spearman’s Rank-Correlation Test

Left-Tailed Two-Tailed Right-Tailed

Significance α α/2 α

Decision Rule

Reject if rs < -CV

Reject if rs < -CV or rs > CV

Reject if rs > CV

Using α as 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.

Small Sample Case: (n ≤ 100)

Large Sample Case: (n > 100)

Hypothesis Tests Using Spearman’s Rank-Correlation TestStep 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: Left-Tailed Two-Tailed Right-Tailed

Significance α α/2 α

H0 not associated not associated not associated

H1 negatively associated associated positively associated

Decision Rule

Reject if rs < -CVReject if

rs < -CV or rs > CVReject if rs > CV

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

S D S-Rank D-Rank d = X - Y d²

100 257 2.5 1 1.5 2.25

102 264 5 4 1 1

103 274 6 6 0 0

101 266 4 5 -1 1

105 277 7.5 8 -0.5 0.25

100 263 2.5 3 -0.5 0.25

99 258 1 2 -1 1

105 275 7.5 7 0.5 0.25

102 267 Ave Sum 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) 36 rs = 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.

Example done in Excel

Club-Head Speed Distance Rank - X Rank - Y difference d²

100 257 2.5 1 1.5 2.25102 264 5 4 1 1103 274 6 6 0 0101 266 4 5 -1 1105 277 7.5 8 -0.5 0.25100 263 2.5 3 -0.5 0.2599 258 1 2 -1 1105 275 7.5 7 0.5 0.25n = 8

Club-Head

Speed Distance 6 sumClub-Head Speed 1 rs = 0.928571Distance 0.938695838 1 rc = 0.738

Rank - X Rank - YRank - X 1Rank - Y 0.927778183 1

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

Homework Problem 3

Problem 3

X Y Rank - X Rank - Y difference d22 1.4 1 1 0 04 1.8 2 2 0 08 2.1 3.5 3 0.5 0.258 2.3 3.5 4 -0.5 0.259 2.6 5 5 0 0

n = 5 0.5 sum Rank - X Rank - Y rs = 0.975

Rank - X 1 rc = 1Rank - Y 0.974679434 1 FTR

Homework Problem 6

Problem 6

X Y Rank - X Rank - Y difference d20 0.8 1 1 0 0

0.5 2.3 2 3 -1 11.4 1.9 3.5 2 1.5 2.251.4 2.5 3.5 4 -0.5 0.253.9 5 5 5 0 04.6 6.8 6 6 0 0n = 6

3.5 sumrs = 0.9

Rank - X Rank - Y rc = 0.886Rank - X 1 RejectRank - Y 0.898645105 1

Homework Problem 7Problem 7

BS % Income Rank - X Rank - Y difference d217.4 24289 1 1 0 029.8 33749 9 9 0 024.6 29043 4 4 0 022.3 26100 2 2 0 023.7 28831 3 3 0 026.8 30758 8 8 0 025 29944 6 7 -1 1

26.4 29340 7 5 2 424.7 29372 5 6 -1 1n = 9

6 sumrs = 0.95

Rank - X Rank - Y rc = 0.6Rank - X 1 RejectRank - Y 0.95 1