Because estimates of Health expectancies
are
means of random variables assumed to be independent, according to the
central limit theorem, they tend to have normal distributions. So,
the latter normal is often assumed. It permits to test if
differences between two Sullivan indices are caused by random
fluctuations or not.
Therefore, if it is wanted to test the hypothesis of equality of two
health expectancies, 
and 
(for example,
for males and females, or two different times, countries...), it is sufficient
to compute the following Z-score:
If the absolute value of the statistic Z is too large (in comparison with what a normal law should give), the hypothesis is rejected. Then, the two health expectancies can be considered as not equal.
Since, we have always:
where 
and 
are the standard errors of
and . Hence, the sum of the two standard
errors (right hand of the previous equation) is used as denominator of
equation 26. The level of significance for Z depends on the
probability which is accepted for the case of a false conclusion of
inequality.
For example, for p=0.05, the hypothesis is refused for:
In general, see normal law distribution. Table 4 gives an example of such calculation of ratio Z. The differences between the two indicators DFLE does not seem to be the consequences of random fluctuations.
| Age |
|
|
|
|
|
| Z | |
 | 1991 | 1981 |
 |
 | ||||
| 0 | 70.11 | 0.25 | 67.77 | 0.23 | 2.34 | 0.48 | 4.86 | *** |
| 5 | 65.69 | 0.25 | 63.48 | 0.23 | 2.30 | 0.48 | 4.54 | *** |
 | | | | |
| | | |
|
| | | | |
| | | |
|
| | | | |
| | | |
| 65 | 12.28 | 0.22 | 9.88 | 0.21 | 3.40 | 0.42 | 5.67 | *** |
| 70 | 8.39 | 0.21 | 6.86 | 0.20 | 1.93 | 0.41 | 4.73 | *** |
| 75 | 5.83 | 0.20 | 4.25 | 0.18 | 1.04 | 0.38 | 4.17 | *** |
| 80 | 3.39 | 0.18 | 2.34 | 0.16 | 1.96 | 0.34 | 3.04 | *** |
| 85 | 1.75 | 0.17 | 1.51 | 0.15 | 1.48 | 0.32 | 0.69 | *** |
| ***= probability of equality of the two indicators < 0.001. | ||||||||
