quantile.Normal: Determine quantiles of a Normal distribution

In distributions3: Probability Distributions as S3 Objects

Determine quantiles of a Normal distribution

Description

Please see the documentation of Normal() for some properties of the Normal distribution, as well as extensive examples showing to how calculate p-values and confidence intervals. quantile()

Usage

## S3 method for class 'Normal'
quantile(x, probs, drop = TRUE, elementwise = NULL, ...)

Arguments

x	A Normal object created by a call to Normal().
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 of probs (elementwise = FALSE, yielding a matrix)? Or, if x and probs 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 qnorm. Unevaluated arguments will generate a warning to catch mispellings or other possible errors.

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 Normal() for some properties of the Normal distribution, as well as extensive examples showing to how calculate p-values and confidence intervals.

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.

See Also

Other Normal distribution: cdf.Normal(), fit_mle.Normal(), pdf.Normal()

Examples

set.seed(27)

X <- Normal(5, 2)
X

mean(X)
variance(X)
skewness(X)
kurtosis(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 Z-test

# here the null hypothesis is H_0: mu = 3
# and we assume sigma = 2

# exactly the same as: Z <- Normal(0, 1)
Z <- Normal()

# data to test
x <- c(3, 7, 11, 0, 7, 0, 4, 5, 6, 2)
nx <- length(x)

# calculate the z-statistic
z_stat <- (mean(x) - 3) / (2 / sqrt(nx))
z_stat

# calculate the two-sided p-value
1 - cdf(Z, abs(z_stat)) + cdf(Z, -abs(z_stat))

# exactly equivalent to the above
2 * cdf(Z, -abs(z_stat))

# p-value for one-sided test
# H_0: mu <= 3   vs   H_A: mu > 3
1 - cdf(Z, z_stat)

# p-value for one-sided test
# H_0: mu >= 3   vs   H_A: mu < 3
cdf(Z, z_stat)

### example: calculating a 88 percent Z CI for a mean

# same `x` as before, still assume `sigma = 2`

# lower-bound
mean(x) - quantile(Z, 1 - 0.12 / 2) * 2 / sqrt(nx)

# upper-bound
mean(x) + quantile(Z, 1 - 0.12 / 2) * 2 / sqrt(nx)

# equivalent to
mean(x) + c(-1, 1) * quantile(Z, 1 - 0.12 / 2) * 2 / sqrt(nx)

# also equivalent to
mean(x) + quantile(Z, 0.12 / 2) * 2 / sqrt(nx)
mean(x) + quantile(Z, 1 - 0.12 / 2) * 2 / sqrt(nx)

### generating random samples and plugging in ks.test()

set.seed(27)

# generate a random sample
ns <- random(Normal(3, 7), 26)

# test if sample is Normal(3, 7)
ks.test(ns, pnorm, mean = 3, sd = 7)

# test if sample is gamma(8, 3) using base R pgamma()
ks.test(ns, pgamma, shape = 8, rate = 3)

### MISC

# note that the cdf() and quantile() functions are inverses
cdf(X, quantile(X, 0.7))
quantile(X, cdf(X, 7))

distributions3 documentation built on Sept. 30, 2024, 9:37 a.m.