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Basic Analysis and Graphing - SAS

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212 Performing Contingency <strong>Analysis</strong> Chapter 6<br />

Statistical Details for the Contingency Platform<br />

For the purposes of this test, change size to an ordinal variable:<br />

2. In the Columns panel, right-click on the icon next to size <strong>and</strong> select Ordinal.<br />

3. Select Analyze > Fit Y by X.<br />

4. Select sex <strong>and</strong> click Y, Response.<br />

5. Select size <strong>and</strong> click X, Factor.<br />

6. Click OK.<br />

7. From the red triangle menu next to Contingency <strong>Analysis</strong>, select Cochran Armitage Trend Test.<br />

Figure 6.21 Example of the Cochran Armitage Trend Test Report<br />

The two-sided p-value (0.7094) is large. From this, you can conclude that there is no trend in the<br />

proportion of male <strong>and</strong> females that purchase different sizes of cars.<br />

Statistical Details for the Contingency Platform<br />

This section contains statistical details for selected options <strong>and</strong> reports in the Contingency platform.<br />

Statistical Details for the Agreement Statistic Option<br />

Viewing the two response variables as two independent ratings of the n subjects, the Kappa coefficient<br />

equals +1 when there is complete agreement of the raters. When the observed agreement exceeds chance<br />

agreement, the Kappa coefficient is positive, with its magnitude reflecting the strength of agreement.<br />

Although unusual in practice, Kappa is negative when the observed agreement is less than chance<br />

agreement. The minimum value of Kappa is between -1 <strong>and</strong> 0, depending on the marginal proportions.<br />

Quantities associated with the Kappa statistic are computed as follows:<br />

P 0<br />

– P e<br />

κˆ = ---------------- where P <strong>and</strong><br />

1 – P = 0 p P ii<br />

= e p p i. .i<br />

e<br />

i<br />

i<br />

The asymptotic variance of the simple kappa coefficient is estimated by the following:<br />

var<br />

A + B – C<br />

= ------------------------ where , <strong>and</strong><br />

( 1 – P e<br />

) 2 A = p ii<br />

[ 1 – ( p i.<br />

+ p .i<br />

)( 1 – κˆ<br />

)] 2 B ( 1 – κˆ<br />

) 2 p ij<br />

( p .i<br />

+ p j.<br />

) 2<br />

= <br />

n<br />

i≠<br />

j<br />

i

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