Let’s, e.g., take data series
data={#,#+(2Mod[#,2]-1)/2//N}&/@Range[10]
{{1,1.5},{2,1.5},{3,3.5},{4,3.5},{5,5.5},{6,5.5},{7,7.5},{8,7.5},{9,9.5},{10,9.5}}
The confidence level that correspond to one sigma is
CLOneSigma=CDF[NormalDistribution[0,1],1]-CDF[NormalDistribution[0,1],-1]//N
0.682689
With this CL the LinearModelFit for this series is
fit=LinearModelFit[data,{1,x},x,ConfidenceLevelCLOneSigma]
FittedModel
0.166667+0.969697x

The SinglePredictionErrors are
spe1=fit["SinglePredictionErrors"]
{0.638524,0.615084,0.5969,0.584464,0.578145,0.578145,0.584464,0.5969,0.615084,0.638524}
The half of interval between SinglePredictionBands is
spe2=Table
#[[2]]-#[[1]]
2
&[fit["SinglePredictionBands"]/.xdata[[i,1]]],{i,Length[data]}
{0.681004,0.656005,0.636611,0.623347,0.616608,0.616608,0.623347,0.636611,0.656005,0.681004}
I believed that the halves of SinglePredictionBands (“the confidence bands based on single observations”, as it is said in docs) must be equal to SinglePredictionErrors (“the standard errors for the predicted response of single observations”). However
spe2/spe1
{1.06653,1.06653,1.06653,1.06653,1.06653,1.06653,1.06653,1.06653,1.06653,1.06653}
The ratio is not one, but
InverseCDF[StudentTDistribution[Length[data]-2],CDF[NormalDistribution[0,2],2]]//N
1.06653
Why?