Basic Statistics

THE SIGN TEST 191

of successes equals the proportion of failures, or 7r equals 1/2 or .5. The method used
for testing the null hypothesis here is to use the normal approximation to the binomial
distribution. The value of p is 28/(28 + 8) = .78. Since .78 is larger than 7r we
will subtract the correction factor. We will use the test given in Section 10.6.1, where
the data are paired so the test for a single sample is used, but we will include the
correction factor 14212) and subtract the correction factor since .78 is greater than .5.

(p - .5) rir (1/(2 x 36)

dw

Jm

z=

• (.78 - .5) - 1/72


• 
  

.28 - .0139

• 
  
  
  
  • = 3.19
    .0833
  
  
  
  

Here a two-sided test with a = .05 is appropriate, so the null hypothesis will be
rejected and we would conclude that the health program reduced the days off work.
The null hypothesis being tested is Ho : P(+) = P(-).
The effect of using the correction factor can be evaluated simply by noting its size.
If n = 100, then 1/200 is a very small number, but if n, = 5, 1/10 equals .I0 and that
might affect the conclusion if the difference between p and .5 is small. In making
this test we are assuming that the (Xi, Yz) for each worker are independent and that
the data are at least ordinal.

13.1.2 Sign Test When the Sample Size Is Small

When the sample size is small, say less than 20 and 7r = 1/2 or S, the accuracy of
the normal approximation may be questionable and it is recommended that an exact
test be done. In this case the exact probabilities can be obtained using the formula for
the binomial or from binomial tables. An example of using the formula is given here
for a very small sample size, but the simplest thing to do is to use the results from a
binomial table. The probability that X is equal to any particular number greater than
or equal to zero is given by

P(x) = n! px(l -p)(n-X)

X!(n - X)!

where X! is called X factorial and is equal to the product of X(X - 1)(X -

2)...(2)(1). NotethatO!=landl!=l.

For example, suppose that we have five employees and only one of them suc-

ceeds in lowering his number of days off work. So there are four cases where the

employees fail to lower their number of days off work and one where they succeed.