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Compute p value.

Source: R/bruceR-stats_1_basic.R
p.Rd

Compute p value.

Usage

p(
  z = NULL,
  t = NULL,
  f = NULL,
  r = NULL,
  chi2 = NULL,
  n = NULL,
  df = NULL,
  df1 = NULL,
  df2 = NULL,
  digits = 2
)

p.z(z)

p.t(t, df)

p.f(f, df1, df2)

p.r(r, n)

p.chi2(chi2, df)

Arguments

z, t, f, r, chi2

\(z\), \(t\), \(F\), \(r\), \(\chi^2\) value.

n, df, df1, df2

Sample size or degree of freedom.

digits

Number of decimal places of output. Defaults to 2.

Value

p value statistics.

Functions

  • p.z(): Two-tailed p value of \(z\).

  • p.t(): Two-tailed p value of \(t\).

  • p.f(): One-tailed p value of \(F\). (Note: \(F\) test is one-tailed only.)

  • p.r(): Two-tailed p value of \(r\).

  • p.chi2(): One-tailed p value of \(\chi^2\). (Note: \(\chi^2\) test is one-tailed only.)

Examples

p.z(1.96)
#> [1] 0.04999579
p.t(2, 100)
#> [1] 0.04821218
p.f(4, 1, 100)
#> [1] 0.04821218
p.r(0.2, 100)
#> [1] 0.04603629
p.chi2(3.84, 1)
#> [1] 0.05004352

p(z=1.96)
#> z = 1.96, p = 0.050 *  
p(t=2, df=100)
#> t(100) = 2.00, p = 0.048 *  
p(f=4, df1=1, df2=100)
#> F(1, 100) = 4.00, p = 0.048 *  
p(r=0.2, n=100)
#> r(98) = 0.20, p = 0.046 *  
p(chi2=3.84, df=1)
#> χ²(1) = 3.84, p = 0.050 .