Skip to contents

Evaluate the probability mass function of a StudentsT distribution

Source: R/StudentsT.R
pdf.StudentsT.Rd

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.

Usage

# S3 method for class 'StudentsT'
pdf(d, x, drop = TRUE, elementwise = NULL, ...)

# S3 method for class 'StudentsT'
log_pdf(d, x, drop = TRUE, elementwise = NULL, ...)

Arguments

d

A StudentsT object created by a call to StudentsT().

x

A vector of elements whose probabilities you would like to determine given the distribution d.

drop

logical. Should the result be simplified to a vector if possible?

elementwise

logical. Should each distribution in d be evaluated at all elements of x (elementwise = FALSE, yielding a matrix)? Or, if d and x have the same length, should the evaluation be done element by element (elementwise = TRUE, yielding a vector)? The default of NULL means that elementwise = TRUE is used if the lengths match and otherwise elementwise = FALSE is used.

...

Arguments to be passed to dt. 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(x) columns (if drop = FALSE). In case of a vectorized distribution object, a matrix with length(x) columns containing all possible combinations.

See also

Other StudentsT distribution: cdf.StudentsT(), quantile.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