Description Usage Arguments Details Value Author(s) References Examples

Use empirical likelihood ratio to test the hypothesis Ho: (1-b0)th quantile of sample 1 = (1-t0)th quantile of sample 2. This is the same as testing Ho: R(t0)= b0, where R(.) is the ROC curve.

The log empirical likelihood been maximized is

* ∑_{d1=1} \log Δ F_1(t1_i) + ∑_{d1=0} \log [1-F_1(t1_i)]
+ ∑_{d2=1} \log Δ F_2(t2_j) + ∑_{d2=0} \log [1-F_2(t2_j)] .*

This empirical likelihood ratio has a chi square limit under Ho.

1 | ```
ROCnp(t1, d1, t2, d2, b0, t0)
``` |

`t1` |
a vector of length n. Observed times, may be right censored. |

`d1` |
a vector of length n, censoring status. d=1 means t is uncensored; d=0 means t is right censored. |

`t2` |
a vector of length m. Observed times, may be right censored. |

`d2` |
a vector of length m, censoring status. |

`b0` |
a scalar between 0 and 1. |

`t0` |
a scalar, betwenn 0 and 1. |

Basically, we first test (1-b0)th quantile of sample 1 = c and also test (1-t0)th quantile of sample 2 = c. This way we obtain two log likelihood ratios.

Then we minimize the sum of the two log likelihood ratio over c.

See the tech report below for details on a similar setting.

A list with the following components:

`"-2LLR"` |
the -2 loglikelihood ratio; have approximate chisq
distribution under |

`cstar` |
the estimated common quantile. |

Mai Zhou.

Zhou, M. and Liang, H (2008).
Empirical Likelihood for Hybrid Two Sample Problem with
Censored Data. *Univ. Kentucky Tech. Report.*

Su, H., Zhou, M. and Liang, H. (2011). Semi-parametric
hybrid empirical likelihood inference for two-sample comparison with censored data. *Lifetime Data Analysis*,
**17**, 533-551.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 | ```
#### An example of testing the equality of two medians. No censoring.
ROCnp(t1=rexp(100), d1=rep(1,100), t2=rexp(120), d2=rep(1,120), b0=0.5, t0=0.5)
##########################################################################
#### Next, an example of finding 90 percent confidence interval of R(0.5)
#### Note: We are finding confidence interval for R(0.5). So we are testing
#### R(0.5)= 0.35, 0.36, 0.37, 0.38, etc. try to find values so that
#### testing R(0.5) = L , U has p-value of 0.10, then [L, U] is the 90 percent
#### confidence interval for R(0.5).
#set.seed(123)
#t1 <- rexp(200)
#t2 <- rexp(200)
#ROCnp( t1=t1, d1=rep(1, 200), t2=t2, d2=rep(1, 200), b0=0.5, t0=0.5)$"-2LLR"
#### since the -2LLR value is less than 2.705543 = qchisq(0.9, df=1), so the
#### confidence interval contains 0.5.
#gridpoints <- 35:65/100
#ELvalues <- gridpoints
#for( i in 1:31 ) ELvalues[i] <- ROCnp(t1=t1, d1=rep(1, 200),
# t2=t2, d2=rep(1, 200), b0=gridpoints[i], t0=0.5)$"-2LLR"
#myfun1 <- approxfun(x=gridpoints, y=ELvalues)
#uniroot( f= function(x){myfun1(x)-2.705543}, interval= c(0.35, 0.5) )
#uniroot( f= function(x){myfun1(x)-2.705543}, interval= c(0.5, 0.65) )
#### So, taking the two roots, we see the 90 percent confidence interval for R(0.5)
#### in this case is [0.4478081, 0.5889425].
``` |

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