Please see the documentation of StudentsT()
for some properties
of the StudentsT distribution, as well as extensive examples
showing to how calculate p-values and confidence intervals.
quantile()
Usage
# S3 method for class 'StudentsT'
quantile(x, probs, drop = TRUE, elementwise = NULL, ...)
Arguments
- x
A
StudentsT
object created by a call toStudentsT()
.- probs
A vector of probabilities.
- drop
logical. Should the result be simplified to a vector if possible?
- elementwise
logical. Should each distribution in
x
be evaluated at all elements ofprobs
(elementwise = FALSE
, yielding a matrix)? Or, ifx
andprobs
have the same length, should the evaluation be done element by element (elementwise = TRUE
, yielding a vector)? The default ofNULL
means thatelementwise = TRUE
is used if the lengths match and otherwiseelementwise = FALSE
is used.- ...
Arguments to be passed to
qt
. Unevaluated arguments will generate a warning to catch mispellings or other possible errors.
Value
In case of a single distribution object, either a numeric
vector of length probs
(if drop = TRUE
, default) or a matrix
with
length(probs)
columns (if drop = FALSE
). In case of a vectorized
distribution object, a matrix with length(probs)
columns containing all
possible combinations.
Details
This function returns the same values that you get from a Z-table. Note
quantile()
is the inverse of cdf()
. Please see the documentation of
StudentsT()
for some properties
of the StudentsT distribution, as well as extensive examples
showing to how calculate p-values and confidence intervals.
See also
Other StudentsT distribution:
cdf.StudentsT()
,
pdf.StudentsT()
,
random.StudentsT()
Examples
set.seed(27)
X <- StudentsT(3)
X
#> [1] "StudentsT(df = 3)"
random(X, 10)
#> [1] 1.4854556 -0.3809239 -1.8376741 0.1105147 0.3005249 0.1558420
#> [7] -1.5135073 -0.6088114 -2.4080689 -1.1878884
pdf(X, 2)
#> [1] 0.06750966
log_pdf(X, 2)
#> [1] -2.695485
cdf(X, 4)
#> [1] 0.9859958
quantile(X, 0.7)
#> [1] 0.5843897
### example: calculating p-values for two-sided T-test
# here the null hypothesis is H_0: mu = 3
# data to test
x <- c(3, 7, 11, 0, 7, 0, 4, 5, 6, 2)
nx <- length(x)
# calculate the T-statistic
t_stat <- (mean(x) - 3) / (sd(x) / sqrt(nx))
t_stat
#> [1] 1.378916
# null distribution of statistic depends on sample size!
T <- StudentsT(df = nx - 1)
# calculate the two-sided p-value
1 - cdf(T, abs(t_stat)) + cdf(T, -abs(t_stat))
#> [1] 0.2012211
# exactly equivalent to the above
2 * cdf(T, -abs(t_stat))
#> [1] 0.2012211
# p-value for one-sided test
# H_0: mu <= 3 vs H_A: mu > 3
1 - cdf(T, t_stat)
#> [1] 0.1006105
# p-value for one-sided test
# H_0: mu >= 3 vs H_A: mu < 3
cdf(T, t_stat)
#> [1] 0.8993895
### example: calculating a 88 percent T CI for a mean
# lower-bound
mean(x) - quantile(T, 1 - 0.12 / 2) * sd(x) / sqrt(nx)
#> [1] 2.631598
# upper-bound
mean(x) + quantile(T, 1 - 0.12 / 2) * sd(x) / sqrt(nx)
#> [1] 6.368402
# equivalent to
mean(x) + c(-1, 1) * quantile(T, 1 - 0.12 / 2) * sd(x) / sqrt(nx)
#> [1] 2.631598 6.368402
# also equivalent to
mean(x) + quantile(T, 0.12 / 2) * sd(x) / sqrt(nx)
#> [1] 2.631598
mean(x) + quantile(T, 1 - 0.12 / 2) * sd(x) / sqrt(nx)
#> [1] 6.368402