C.I. of the variance of a normal variable.
[ low, up ] = conint_normvar ( n , v ) [ low, up ] = conint_normvar ( n , v , level ) [ low, up ] = conint_normvar ( n , v , level , twosided )
a 1-by-1 matrix of doubles, positive, integer value, the number of samples
a 1-by-1 matrix of doubles, positive, the variance estimate
a 1-by-1 matrix of doubles, the confidence level (default level = 1.-0.95=0.05). level is expected to be in the range [0.,0.5]
a 1-by-1 matrix of booleans, the side of the interval (default twosided = %t). If twosided is true, then low and up is a two-sided interval [low,up]. If twosided is false, then the intervals are [-inf,up] and [low,inf].
a 1-by-1 matrix of doubles, the estimated lower bound
a 1-by-1 matrix of doubles, the estimated upper bound
Computes a confidence interval of the variance of a normal variable, that is,
computes v
as the variance of x
and
computes confidence intervals for v
.
If twosided
is true, then
low
and up
are such that
the variance var(X)
is such that
P(low < var < up) = 1-level
If twosided
is false, then
low
and up
are such that
the variance var(X)
is such that
P(var(X) < up) = 1-level
and
P(low < var(X)) = 1-level
This function makes the assumption that the
data in x
has a normal distribution with
expectation mu(X)
and
variance var(X)
.
We use a Chi-square distribution to compute the confidence interval.
// Sheldon Ross, Example 7.3h, page 252 // Compute a 90% confidence interval for the // variance of a normal variable. x = [0.123 0.124 0.126 0.120 0.130 ... 0.133 0.125 0.128 0.124 0.126] n = size(x,"*") v = variance(x) [ low, up ] = conint_normvar ( n , v , 1.-0.90 ) // Then P(7.264D-06 <= var(X) <= 36.96D-06) = 0.90 // Give the confidence interval for the standard // deviation [sqrt(low) sqrt(v) sqrt(up)] // Example from Gilbert Saporta, // Section 13.5.3, "Variance d'une loi normale" // Section 13.5.3.2, "m est inconnu" // Get the C.I. for the variance of a normal // variable with n=30 and var(X)=12. x = [6;2;2;8;12;6;6;13;.. 8;2;10;0;7;5;4;4;3;.. 7;4;9;6;2;-3;5;7;2;4;5;2;2]; n = size(x,"*") v = variance(x) [ low, up ] = conint_normvar ( n , v , 1.-0.90 ) // Then P(8.177 < var(X) < 19.65) = 0.90 // The 90% C.I. is (8.177,19.65) // Notice that Saporta gets (8.46,20.33). // This is because Saporta uses n as the denominator // in the variance estimate, while we use (n-1). // Example from // "Probability and Statistics for Engineers and Scientists", // Walpole, Myers, Myers, Ye // Example 9.17: x = [46.4, 46.1, 45.8, 47.0, 46.1, .. 45.9, 45.8, 46.9, 45.2, 46.0]; n = size(x,"*") v = variance(x) [ low, up ] = conint_normvar ( n , v , 1.-0.95 ) // Then (0.135,0.953) is a 95% C.I. for the variance. | ![]() | ![]() |
http://en.wikipedia.org/wiki/Confidence_interval
"Introduction to probability and statistics for engineers and scientists", Sheldon Ross, Third Edition, 2004
"Probabilités, Analyse de Données et Statistique", Gilbert Saporta, 2nd Ed., 2006
// "Probability and Statistics for Engineers and Scientists", Ronald Walpole, Raymond Myers, Sharon Myers, Keying Ye, Eight Edition, 2006