How to compute "SinglePredictionConfidenceIntervalTable"-Values on validation data with NonlinearModeFit?

When I perform a NonlinearModelfit I get a FittedModel object and I can see the values of the property "SinglePredictionConfidenceIntervalTable". Now I apply the FittedModel object on a new data case (a validation data case) and get the estimated model value for this case. But I need additionally the confidence interval for this single predicted response for the validation case like I can see it for each data case in the training data. Is it possible to calculate this?

Quoting Wikipedia

prediction intervals predict the distribution of individual future points, whereas confidence intervals and credible intervals of parameters predict the distribution of estimates of the true population mean or other quantity of interest that cannot be observed.

You can obtain the single prediction intervals for a new data point as shown below. sp contains the 1, 2 & 3 standard deviation single prediction intervals based on the sample data.

data = {{0, 1}, {1, 0}, {3, 2}, {5, 4}, {6, 4}, {7, 5}};

nlm = NonlinearModelFit[data, Log[a + b x^2], {a, b}, x];

{sd1, sd2, sd3} =
2 (CDF[NormalDistribution[0, 1], #] - 0.5) & /@ {1, 2, 3};

{bands68[x_], bands95[x_], bands99[x_]} = Table[
nlm["SinglePredictionBands",
ConfidenceLevel -> cl], {cl, {sd1, sd2, sd3}}];

(* new data *)
x0 = 4;

sp = Flatten[{nlm[#], bands68[#], bands95[#], bands99[#]} &@x0];
sp2 = {x0, #} & /@ sp;

Show[
Plot[{nlm[x], bands68[x], bands95[x], bands99[x]}, {x, -1, 8},
Filling -> {2 -> {1}, 3 -> {2}, 4 -> {3}}],
ListPlot[data, PlotStyle -> Directive[Green, PointSize[Large]]],
ListPlot[sp2, PlotStyle -> Directive[Blue, PointSize[Large]]],
AxesOrigin -> {-1, -9}, PlotRange -> {{-1, 8}, {-9, 13}},
Frame -> True, PlotRangePadding -> 0] TableForm[Partition[Rest[sp], 2],
TableHeadings -> {{"σ=1", "σ=2", "σ=3"}, None}]
σ=1  |  2.16509      4.21813
σ=2  |  0.611601     5.77162
σ=3  |  -2.76111     9.14433
• The factor 2 is the rule of thumb for 1.96, correct? Should one use the Student rather than the Normal distribution, because the number of data points is known even though small? Apr 13 '17 at 8:36
• @Volker The SinglePredictionBands function uses Student's t-distribution under the hood. You can see the calculation for a linear case demonstrated here: mathematica.stackexchange.com/a/69819/363 Apr 13 '17 at 9:15
• @Volker The factor 2 is two standard deviations, for which the confidence level is 0.9545. I.e. sd = 2; cl = 2 (CDF[NormalDistribution[0, 1], sd] - 0.5) and in reverse: sd = Quantile[NormalDistribution[0, 1], 1 - (0.5 - cl/2)]. 1.96 is the number of standard deviations representing the 95% confidence interval, i.e. sd = Quantile[NormalDistribution[0, 1], 1 - (0.5 - 0.95/2)]. They are used interchangeably, but it's more precise to state which is being used: std dev or confidence level. Apr 13 '17 at 9:46