Statistical Methods for Psychology

Kendall defined

It is well known that the number of pairs of Nobjects is given by N(N 2 1) 2.

For our data

Thus, as a measure of the agreement between rankings on Alcohol and Tobacco, Kendall’s

t5.345.

The interpretation of is more straightforward than would be the interpretation of

calculated on the same data (0.37). If t5.345, we can state that if a pair of objects is sam-

pled at random, the probability that the two regions will be ranked in the same order is .345

higher than the probability that they will be ranked in the reverse order.

When there are tied rankings, the calculation of must be modified. For the appropri-

ate correction for ties, see Hays (1981, p. 602 ff).

Significance oft

Unlike Spearman’s rs, there is an accepted method for estimation of the standard error of

Kendall’s t.

Moreover, tis approximately normally distributed for N$10. This allows us to approxi-

mate the sampling distribution of Kendall’s tusing the normal approximation.

For a two-tailed test p 5 .139, which is not statistically significant.

With a standard error of 0.2335, the confidence limits on Kendall’s t, assuming nor-

mality of t,would be

For our example this would produce confidence limits of 2 .11 #t#.80.

Kendall’s thas generally been given preference of Spearman’s rSbecause it is a better

estimate of the corresponding population parameter, and its standard error is known.

Although there is evidence that Kendall’s holds up better than Pearson’s rto nonnor-

mality in the data, that seems to be true only at quite extreme levels. In general, Pearson’s r

on the raw data has been, and remains, the coefficient of choice. (For this data set the Pearson

correlation between the original cost values is r 5 .22, p 5 .509.)

10.4 Analysis of Contingency Tables with Ordered Variables

In Chapter 6 on chi-square, I referred to the problem that arises when the independent vari-

ables are ordinal variables. The traditional chi-square analysis does not take this ordering

into account, but it is important for a proper analysis. As I said in Chapter 6, this section

t

CI=t 6 1.96st=t 6 1.96 ¢

B

2(2N 1 5)

9 N(N 2 1)

≤=t 6 1.96(.2335)

z=

t

st =

t

B

2(2N 1 5)

9 N(N 2 1)

=

.345

B

2(27)

9(11)(10)

=

.345

.2335

=1.48

st=

B

2(2N 1 5)

9 N(N 2 1)

t

t rs

t= 12

2(Number of inversions)

Number of pairs of objects

= 12

2(18)

55

=.345

>

t= 12

2(Number of inversions)

Number of pairs of objects

or

2 S

N(N 2 1)

306 Chapter 10 Alternative Correlational Techniques