cdf.StudentsT | R Documentation |
Evaluate the cumulative distribution function of a StudentsT distribution
## S3 method for class 'StudentsT'
cdf(d, x, drop = TRUE, elementwise = NULL, ...)
d |
A |
x |
A vector of elements whose cumulative probabilities you would
like to determine given the distribution |
drop |
logical. Should the result be simplified to a vector if possible? |
elementwise |
logical. Should each distribution in |
... |
Arguments to be passed to |
In case of a single distribution object, either a numeric
vector of length probs
(if drop = TRUE
, default) or a matrix
with
length(x)
columns (if drop = FALSE
). In case of a vectorized distribution
object, a matrix with length(x)
columns containing all possible combinations.
Other StudentsT distribution:
pdf.StudentsT()
,
quantile.StudentsT()
,
random.StudentsT()
set.seed(27)
X <- StudentsT(3)
X
random(X, 10)
pdf(X, 2)
log_pdf(X, 2)
cdf(X, 4)
quantile(X, 0.7)
### 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
# 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))
# exactly equivalent to the above
2 * cdf(T, -abs(t_stat))
# p-value for one-sided test
# H_0: mu <= 3 vs H_A: mu > 3
1 - cdf(T, t_stat)
# p-value for one-sided test
# H_0: mu >= 3 vs H_A: mu < 3
cdf(T, t_stat)
### example: calculating a 88 percent T CI for a mean
# lower-bound
mean(x) - quantile(T, 1 - 0.12 / 2) * sd(x) / sqrt(nx)
# upper-bound
mean(x) + quantile(T, 1 - 0.12 / 2) * sd(x) / sqrt(nx)
# equivalent to
mean(x) + c(-1, 1) * quantile(T, 1 - 0.12 / 2) * sd(x) / sqrt(nx)
# also equivalent to
mean(x) + quantile(T, 0.12 / 2) * sd(x) / sqrt(nx)
mean(x) + quantile(T, 1 - 0.12 / 2) * sd(x) / sqrt(nx)
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