Page 104 - Applied Statistics Using SPSS, STATISTICA, MATLAB and R

3.1 Point Estimation and Interval Estimation 83

When estimating a data parameter the point estimate is usually insufficient. In
fact, in all the cases that the point estimator is characterised by a probability
density function the probability that the point estimate actually equals the true
value of the parameter is zero. Using the spring scales analogy, we see that no
matter how accurate and precise the scales are, the probability of obtaining the
exact weight (with arbitrary large number of digits) is zero. We need, therefore, to
attach some measure of the possible error of the estimate to the point estimate. For
that purpose, we attempt to determine an interval, called confidence interval,
containing the true parameter value θ with a given probability 1– α, the so-called
confidence level:

P (t 1 , n (x ) < θ < t 2 , n (x ) )=1 − α , 3.1

where α is a confidence risk.
The endpoints of the interval (also known as confidence limits), depend on the
available sample and are determined taking into account the sampling distribution:

F T (x ) ≡ F n t (X ) (x ) .

We have assumed that the interval endpoints are finite, the so-called two-sided
(or two-tail) interval estimation. Sometimes we will also use one-sided (or one-
tail) interval estimation by setting t 1 , n (x ) = −∞ or t 2 , n (x ) = +∞ .
Let us now apply these ideas to the spring scales example. Imagine that, as
happens with unbiased point estimators, there were no systematic error and
furthermore the measured errors follow a known normal distribution; therefore, the
measurement error is a one-dimensional random variable distributed as N 0,σ , with
known σ. In other words, the distribution function of the random weight variable,
W, is F W (w ) ≡ F (w ) = N ω ,σ (w ) . We are now able to determine the two-sided
95% confidence interval of ω, given a measurement w, by first noticing, from the
normal distribution tables, that the percentile 97.5% (i.e., 100–α/2, with α in
percentage) corresponds to 1.96σ:
Thus:

F (w ) = . 0 975 ⇒ w . 0 975 = . 1 96 σ . 3.2

Given the symmetry of the normal distribution, we have:

ω
σ
w
P ( < ω + . 1 96σ ) = . 0 975 ⇒ P ( − . 1 96 < w < ω + . 1 96σ ) = . 0 95 ,

leading to the following 95% confidence interval:

−
+
ω 1 . 96 σ < w < ω 1 . 96 σ . 3.3

Hence, we expect that in a long run of measurements 95% of them will be inside
the ω ± 1.96σ interval, as shown in Figure 3.2a.
Note that the inequalities 3.3 can also be written as: