C.I. of the mean of a normal variable.
[ low, up ] = conint_normmu ( n , me , v ) [ low, up ] = conint_normmu ( n , me , v , level ) [ low, up ] = conint_normmu ( n , me , v , level , twosided ) [ low, up ] = conint_normmu ( n , me , v , level , twosided , known )
a n-by-1 or 1-by-n matrix of doubles, integer value, positive, the number of samples
a 1-by-1 matrix of doubles, the mean 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 doubles, the variance. By default, the variance is assumed to be exact (i.e. by default known=%t). If the variance v
is estimated, then set the known
option to %f.
a 1-by-1 matrix of booleans, the side of the interval (default twosided = %t). If twosided
is %t, then low
and up
make a two-sided interval [low,up]
. If twosided
is %f, then the intervals are [-inf,up]
and [low,inf]
.
a 1-by-1 matrix of booleans, %t if the variance v
is exact (default known = %t). If known
is %t, then v
is the exact variance. If known is %f, then v
is the estimate of the variance.
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 mean of a normal variable, that is,
computes m
as the mean of x
and
computes confidence intervals for mu(X)
.
If twosided
is true, then
low
and up
are such that
the expectation mu(X)
is such that
P(low < mu(X) < up) = 1-level
If twosided
is false, then
low
and up
are such that
the expectation mu
is such that
P(mu(X) < up) = 1-level
and
P(low < mu(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)
.
If the variance is exact (i.e. known=%t
)
then we use a Normal distribution to compute the confidence interval.
If the variance is estimated (i.e. if known=%f
),
then we use Student's T distribution to compute the confidence interval.
// Sheldon Ross, Example 7.3a, page 241 x = [5, 8.5, 12, 15, 7, 9, 7.5, 6.5, 10.5] n = size(x,"*") me = mean(x) // By default, we compute a two-sided 95% C.I. // with exact variance: v = 4. [ low, up ] = conint_normmu ( n , me , v ) low_expected = 7.69 up_expected = 10.31 // Then P(7.69 <= mu <= 10.31) = 0.95 // Get one-sided 95% C.I. // Example 7.3b, p. 242 low_expected = 7.903 up_expected = 10.097 twosided = %f [ low, up ] = conint_normmu ( n , me , v , [] , twosided ) // Then : P(7.903 <= mu) = 0.95 // P(mu <= 10.097) = 0.95 // Example 7.3c, p. 244 // Two-sided 99% interval : (7.28, 10.72) [ low, up ] = conint_normmu ( n , me , v , 1.-0.99 ) // One-sided 99% intervals : (7.447,inf), (−inf, 10.553) [ low, up ] = conint_normmu ( n , me , v , 1.-0.99 , %f ) // EXAMPLE 7.3e, p. 247 // The variance is unknown: use Student's T random variable v = variance(x) [ low, up ] = conint_normmu ( n , me , v , [] , [] , %f ) // Then P(6.63<= mu <= 11.37) = 0.95 // EXAMPLE 7.3f, p.249 x = [54, 63, 58, 72, 49, 92, 70, 73, .. 69, 104, 48, 66, 80, 64, 77]; n = size(x,"*") me = mean(x) v = variance(x) // Two-sided : [ low, up ] = conint_normmu ( n , me , v , 1.-0.95 , [] , %f ) // Then P(60.865<= mu <=77.6683) = 0.95 // One-sided : [ low, up ] = conint_normmu ( n , me , v , 1.-0.95 , %f , %f ) // Then : P (mu <= 76.16) = 0.95 // Then : P (62.368 <= mu) = 0.95 // EXAMPLE 7.3g, Monte-Carlo simulation U = grand(100,1,"def"); x = sqrt(1-U.^2); n = size(x,"*") me = mean(x) v = variance(x) [ low, up ] = conint_normmu ( n , me , v , [] , [] , %f ) // Exact integral is : %pi/4 | ![]() | ![]() |
http://en.wikipedia.org/wiki/Confidence_interval
"Introduction to probability and statistics for engineers and scientists", Sheldon Ross, Third Edition, 2004