Fitting an exponential curve

The {x,y} coordinates represent {time(sec), temperature ((^o)C)} .

data0={{0,88},{30,56},{60,44},{90,38},{120,33},{150,30},{180,27},{210,26},{240,25}};

data={{30,56},{60,44},{90,38},{120,33},{150,30},{180,27},{210,26},{240,25}};

data=data0;

ListPlot[data,PlotStyle->PointSize[0.020],PlotRange->{0,100},AxesOrigin->{20,0}]

f1=FindFit[data,a (E^(-t/tau)) +b,{{a,60},tau,{b,25}},t]

FindFit assumes that each data point has the same error bar which, in this case, it does (~1 (C^o)).

Plot[a (E^(-t/tau)) +b/.f1,{t,0,250},PlotRange->{0,100},Epilog->Prepend[Point/@data,PointSize[0.015]]]

Using the fitted parameters from FindFit, calculate the (χ^2).

chisqmin=(∑_(i=1)^(Length[data]))((((data[[i,2]]-(a (E^(-data[[i,1]]/tau)) +b/.f1)))^2)/(1.0^2))

chisq[τ_]:=(∑_(i=1)^(Length[data]))((((data[[i,2]]-(a (E^(-data[[i,1]]/τ)) +b/.f1)))^2)/(1.0^2))

Plot the (χ^2) as a function τ to determine the uncertainty in the time constant (δτ).

Plot[chisq[τ],{τ,(tau/.f1)-2.,(tau/.f1)+2.}]

Find the uncertainty in the time constant (δτ).

f2=FindRoot[chisq[τ]==chisqmin+1,{τ,(tau/.f1)+1}]

δτ=Abs[(tau/.f1) -( τ/.f2)]

Print["τ = ",(tau/.f1)," ± ",δτ]
Print["(χ^2)/dof = ",chisqmin," / ",Length[data]-3]

So, the final result is τ = (49.0 ± 1.3) sec.  with a (χ^2)/dof  = 23/6.
The (χ^2)/dof is a little high and probably reflects a situation where the actual "setting error," or uncertainty in the temperature (δT), is probably underestimated, and is more likely somewhere between 1 (C^o) and  2 (C^o).

Question:  What happens if you increase the setting error in the (χ^2)'s from 1(C^o) to  2(C^o)?    More specifically, how does the (χ^2)/dof change?  And, how does the uncertainty in the time constant (δτ) change?   Does this make sense?