POSSIBLE SOLUTION(S)

The data does not appear to be going to 0, but rather leveling off at a positive number, say b.

We seek to find the function of the form y = f(x) = Ae-kx + b which fits this data best in the sense that it minimizes the sum of the squares of the differences between the y values of the actual data, yi, and y values of the theoretical model, f(xi).

Let us try our hand at guessing a good A, k, and b and then try it out to see how well it fits. Enter the function below and to try your values for A, k, and b. Simply type in a command, e.g., try[10,.4,5]] will plot the function f(x) = 10 Exp[-.4 x] + 5 and the data.

Input := 

try[A_,k_,b_] := Show[p1, 
 	Plot[A Exp[-k (x-1962)] + b,{x,1962,1996},
 		PlotRange->{0,10},
 		DisplayFunction->Identity] ]

Input := 

try[80,.12,5]

Output =

-Graphics-

We enter our theoretical model, based on obervations or previous theory. We note there are three parameters, A, k, and b which need to be determined.

Input := 

f[x_] :=  A Exp[-k (x - 1962)] + b

We form the sum of the squares of the difference between the y values of the actual data,yi, and y values of the theoretical model, f(xi). Here SS[A,k,b] is a function of the two parameters, A , k, and b. TimeList[[i]][[1]] picks off the first element of each data pair, i.e. the xi, while TimeList[[i]][[2]] picks off the second element of each data pair, i.e. yi. Then f[TimeList[[i]][[1]]] actually computes the value of the theoretical function at xi. Finally, SS[A,k,b] computes the differences in these y values, squares them, and adds the sum of the square differences for all data in the TimeList set.

Input := 

SS[A_,k_,b_] = Sum[
		(f[ TimeList[[i]][[1]]] - TimeList[[i]][[2]])^2,
     		{i,1,  Length[TimeList]}]; 

We seek the values of A, k, and b] which minimize this sum of squares, SS[A, k, b].

Input := 

vals = FindMinimum[SS[A,k,b],{A,80},{k,.12},{b,5}]

Output =

{4.91117, {A -> 82.0849, k -> 0.143559, b -> 6.8807}}

Hence we have values for A, k, and b.

And our theoretical model appears to be:

Input := 

ft[x_] := A Exp[-k (x - 1962)] + b/.vals[[2]]

Input := 

fPlot = Plot[ft[x],{x,1962,1996}]

Output =

-Graphics-

And now to see how good our fit is, we compare the theoretical plot, fPlot with the actual plot of the data

Input := 

Show[p1,fPlot]

Output =

-Graphics-

Not too shabby......

What will the average sound bite be in 1996?

Input := 

ft[1996]

Output =

7.50368

Thus we expect to get about an average of 7.5 seconds of uninterrupted time for Presidential candidates in the evening news in 1996.

Think on that!!!!

This model would predict in the long run that the average length of a Presidental sound bite woulc approach b=6.8807, within the range of the article's claim.