The 1x2 Table tests

The 1x2 Table tests

the_1x2_table_tests(X, n, pi0)

Arguments

X

the number of successes

n

the total number of observations

pi0

a given probability

Value

NULL. This function should be called for its printed output

Examples

# Example: The number of 1st order male births (Singh et al. 2010)
the_1x2_table_tests(singh_2010["1st", "X"], singh_2010["1st", "n"], pi0 = 0.513)
#> H_0: pi = 0.513  vs  H_A: pi ~= 0.513 
#> Estimate of pi: 250/533 = 0.469
#>  
#> Test                P-value  (test statistic) 
#> ------------------------------------------------ 
#> Wald                0.0420   (Z = -2.034) 
#> Wald with CC        0.0466   (Z = 1.990) 
#> Likelihood ratio    0.0423   (T = 4.121, df = 1) 
#> Score               0.0423   (Z = -2.030) 
#> Score with CC       0.0469   (Z = 1.987) 
#>  
#>  
#> Mid-P binomial      0.0425 
#>  
#> ------------------------------------------------ 
#> CC = continuity correction 
# Example: The number of 2nd order male births (Singh et al. 2010)
the_1x2_table_tests(singh_2010["2nd", "X"], singh_2010["2nd", "n"], pi0 = 0.513)
#> H_0: pi = 0.513  vs  H_A: pi ~= 0.513 
#> Estimate of pi: 204/412 = 0.495
#>  
#> Test                P-value  (test statistic) 
#> ------------------------------------------------ 
#> Wald                0.4685   (Z = -0.725) 
#> Wald with CC        0.4993   (Z = 0.676) 
#> Likelihood ratio    0.4685   (T = 0.525, df = 1) 
#> Score               0.4684   (Z = -0.725) 
#> Score with CC       0.4992   (Z = 0.676) 
#>  
#>  
#> Mid-P binomial      0.4689 
#>  
#> ------------------------------------------------ 
#> CC = continuity correction 
# Example: The number of 3rd order male births (Singh et al. 2010)
the_1x2_table_tests(singh_2010["3rd", "X"], singh_2010["3rd", "n"], pi0 = 0.513)
#> H_0: pi = 0.513  vs  H_A: pi ~= 0.513 
#> Estimate of pi: 103/167 = 0.617
#>  
#> Test                P-value  (test statistic) 
#> ------------------------------------------------ 
#> Wald                0.0058   (Z = 2.758) 
#> Wald with CC        0.0074   (Z = 2.679) 
#> Likelihood ratio    0.0070   (T = 7.277, df = 1) 
#> Score               0.0073   (Z = 2.683) 
#> Score with CC       0.0092   (Z = 2.605) 
#>  
#>  
#> Mid-P binomial      0.0072 
#>  
#> ------------------------------------------------ 
#> CC = continuity correction 
# Example: The number of 4th order male births (Singh et al. 2010)
the_1x2_table_tests(singh_2010["4th", "X"], singh_2010["4th", "n"], pi0 = 0.513)
#> H_0: pi = 0.513  vs  H_A: pi ~= 0.513 
#> Estimate of pi: 33/45 = 0.733
#>  
#> Test                P-value  (test statistic) 
#> ------------------------------------------------ 
#> Wald                0.0008   (Z = 3.342) 
#> Wald with CC        0.0015   (Z = 3.174) 
#> Likelihood ratio    0.0025   (T = 9.129, df = 1) 
#> Score               0.0031   (Z = 2.957) 
#> Score with CC       0.0050   (Z = 2.808) 
#>  
#>  
#> Mid-P binomial      0.0029 
#>  
#> ------------------------------------------------ 
#> CC = continuity correction 
# Example: Ligarden et al. (2010)
the_1x2_table_tests(ligarden_2010["X"], ligarden_2010["n"], pi0 = 0.5)
#> H_0: pi = 0.500  vs  H_A: pi ~= 0.500 
#> Estimate of pi: 13/16 = 0.812
#>  
#> Test                P-value  (test statistic) 
#> ------------------------------------------------ 
#> Wald                0.0014   (Z = 3.203) 
#> Wald with CC        0.0039   (Z = 2.882) 
#> Likelihood ratio    0.0094   (T = 6.738, df = 1) 
#> Score               0.0124   (Z = 2.500) 
#> Score with CC       0.0244   (Z = 2.250) 
#>  
#>  
#> Mid-P binomial      0.0127 
#>  
#> ------------------------------------------------ 
#> CC = continuity correction