The confidence interval for the mean μof the population uses the computed
variance
s^2 =^1
n− 1
∑n
j=1
(xj−x)^2 (11 .82)
to produce the γ= 1 −αconfidence level
CON F {x−
√s
ntα/^2 ,n−^1 ≤μ≤x+
√s
ntα/^2 ,n−^1 } (11 .83)
where nis the sample size.
Confidence interval for the variance σ^2
The confidence interval for the variance σ^2 of the population having a normal
distribution is based upon the fact that the variable Y = (n−1)s^2 /σ^2 follows a chi-
square distribution with n− 1 degrees of freedom , where again nrepresents the sample
size.
First select a level of confidence γ= 1 −αand then from
a chi-square distribution table with n− 1 degrees of free-
dom determine the χ^2 α/ 2 ,n − 1 and χ^21 −α/ 2 ,n− 1 values which
represent the points corresponding to the tail areas of the
chi-square probability density function as illustrated.
Secondly, one must calculate the variance squared s^2 using equation (11.82), then
construct the confidence interval for the variance of a normal distribution given by
CON F {(n−1) s
2
χ^21 −α/ 2 ,n− 1
≤σ^2 ≤(n−1) s
2
χ^2 α/ 2 ,n− 1
} (11 .84)
Least Squares Curve Fitting
A set of data points (x 1 , y 1 ),(x 2 , y 2 ),(x 3 , y 3 ),... ,(xi, yi),... , (xn− 1 , y n− 1 ),(xn, y n) can
be plotted on ordinary graph paper and then a line y=β 0 +β 1 xcan also be plotted
to obtain a figure such as illustrated in the figure 11-14.
Assume that the data points are normally distributed about the straight line
and that errors e 1 , e 2 ,... , en occur in the y-variable, where the errors are defined as
the differences between the y-value on the line and the y-value of the data point.
What would be the “best” straight line to represent the given data points? There
