## Info

Xs (critical) = 15.51

### Analysis of Inclination-Only Data

There are many situations in which declination information is not available. The common case is that of deep-sea cores, but bore cores drilled on land are often not oriented or may be available only as a series of fragmentary pieces.

If I0 is the inclination of the true mean direction then a simple average of the observed sample inclinations will give an estimate of 70 that is too shallow (except for /0 = 0). Suppose that the true mean inclination /0 = 90°. If the directions are symmetrically distributed about this mean (as expected for a Fisher distribution), virtually all of the observed inclinations will be <90°. Their arithmetic mean will therefore always be less than the true value. If the true mean has (D0 = 0°, IQ = 80°), an observation at (180°, 85°) is actually 15° from the true mean, but without the declination information it is "folded back" to being only 5° from the mean. If the foldback problem is small, it is possible to obtain an approximately unbiased estimate of both I0 and the precision k. This has been examined by Briden and Ward (1966), Kono (1980b), McFadden and Reid (1982), and Cox and Gordon (1984).

Following the most general analysis given by McFadden and Reid (1982), consider a set of observations /b /2, ..., IN drawn from a Fisher distribution whose true mean direction has an inclination /0. Define 0O as^ the complement of IQ (i.e., 0O = 90° - /0), 9, as the complement of /„ and 0O as the maximum likelihood estimate (which in this instance is approximately unbiased) for 0O. Let where the summations are for i from 1 to N. The value of 0ft is then a solution of which may be obtained by any simple iterative method. The correct solution is obvious as it is close to the average of the 0,. The best estimate I0 for I0 is then

C = 2cos(9o-0,) = cos0o£cos0, +sin0oEsin9, S = E sin(0o - 0,) = sin 0OE cos 0, - cos 0OE sin 0, ,

N cos0o +(sin2 0O - cos2 0o)Zcos0, -2sin0o cos0o£sin0, = 0, (3.2.31)
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