The Total Deviation Index (TDI) describes a boundary, \(\kappa\), such that \(100p\%\) of observations are less than \(\kappa\).
The Coverage Probability (CP) can be considered the inverse of the TDI. It is the proportion, \(p\), of observations that are less than a pre-specified boundary \(\kappa\).
In order to use the statistics for assessing agreement, acceptance criteria is required. The acceptance criteria for the TDI is determined a priori based on biologically and analytically relevant criteria.
Parameter \(p\) is usually chosen to be 0.90 and can be used as an input in the TDI calculation to calculate the boundary \(\kappa\) which can then be compared to the acceptance criteria. Usually, if both \(\kappa\) and it’s upper 95% confidence limit is less than the acceptance criteria, agreement can be concluded.
The acceptance criteria can also be used as an input in the calculation of CP to calculate \(p\). If the value of \(p\) and it’s lower 95% confidence limit is greater than 0.90, agreement can be concluded.
Note that, one can choose to calculate either the TDI or CP to assess agreement as if one concludes agreement, the other will too. Although it seems common to report both statistics.
Agreement = function(df, p, k, alpha=0.05, space = 'arithmetic'){
# Calculate the differences
diffs = df[,2] - df[,1]
if (space == 'Percentage'){
diffs = 100*(diffs/((df[,1] + df[,2])/2))
}
n = length(diffs)
e2 = sum(diffs^2) / (n - 1)
tdi = qnorm(1 - ((1 - p) / 2)) * sqrt(e2)
if (e2 != 0){
w = log(e2)
}else{
return(c(tdi, NA, NA, NA))
}
means = df %>%
summarise(across(everything(), mean))
mu_d = means[,1] - means[,2]
d2 = mu_d^2
if (n < 4){
return(c(tdi, NA, NA, NA))
}
sd2 = (sum(diffs^2) / n - d2) * (n / (n - 3))
d2_s2 = (d2 / sd2)
s_w = sqrt(2 * (1 - ((d2^2) / (e2^2))) / (n - 2))
w_u = w + qnorm(1 - alpha) * s_w
e2_u = exp(w_u)
tdi_u = qnorm(1 - ((1 - p) / 2)) * sqrt(e2_u)
cp = pchisq((k ** 2) / sd2, 1, d2_s2)
sd = sqrt(sd2)
kpm = (k + mu_d) / sd
kmm = (k - mu_d) / sd
s_t = (
(
((dnorm(-kpm) - dnorm(kmm))^2) +
(((kmm * dnorm(kmm)) + (kpm * dnorm(-kpm)))^2) / 2
) /
((n - 3) * (cp^2) * ((1 - cp)^2))
)
s_t = sqrt(s_t)
t = log(cp / (1 - cp))
t_l = t - qnorm(1 - alpha) * s_t
cp_l = exp(t_l) / (1 + exp(t_l))
return(c(tdi, tdi_u, cp, cp_l))
}The industry standard for measuring liver inflammation is performed using medical device \(A\). A new medical device, \(B\), has been created which also measures liver inflammation, but measures it in a more cost effective way. In order for the medical device to be accepted into a clinical care pathway, it needs to measure liver inflammation as well as the industry standard. You choose to show that medical device \(B\) is equivalent to medical device \(A\) at measuring liver inflammation.
It is decided that equivalence between devices will be concluded if the TDI is less than 30ms.
A total of 100 individuals have had their liver inflammation measured using both medical device \(A\) and \(B\). To test for equivalence between the medical devices, an acceptance criteria of 30ms on the TDI is used.
set.seed(1)
# Create some dummy data
A = rnorm(100, 800, 50)
B = A + rnorm(100, 0, 10)
# Create a data frame containing the data
df = data.frame(A, B)
# Calculate the TDI and CP
TDI = Agreement(df, 0.9, 30)[1]
TDI_upper = Agreement(df, 0.9, 30)[2]
CP = Agreement(df, 0.9, 30)[3]
CP_lower = Agreement(df, 0.9, 30)[4]Running the above code gives:
Thus as both the TDI and upper 95% confidence limit of the TDI is less than 30, agreement can be concluded. Equally because both the CP and lower 95% confidence limit of the CP is greater than 0.90, agreement can be concluded.