Question

Problem 2 The following data can be well-represented by an equation of the form: Y = aebx Xi 2 4 6 8 10 12 yi 12 34 102 256 727 1877 Calculate the following: 1. Standard deviation 2. Standard error of estimate 3. Coefficient of determination (r?)

Answer 1

Taking natural logarithm to both sides of the equation we get

$y=ae^{bx}$

or, In y = In ae

$or,~ln~y=ln~a+ln~e^{bx}$

or, In y = ln a + bx

or, 2 = c+b.

where z = ln y and c = ln a

x	y	z = lny
2	12	2.485
4	34	3.526
6	102	4.625
8	256	5.545
10	727	6.589
12	1877	7.537

In our case, we have N = 6 datasets. Using standard equations, the value of b and c can be determined as follows

$b=\frac{\sum_{i=1}^N z_ix_i -\frac{\sum_{i=1}^Nx_i\sum_{i=1}^Nz_i}{N} }{\sum_{i=1}^Nx_i^2-\frac{(\sum_{i=1}^Nx_i)^2}{N}}$

Using Excel to calculate the terms in this equation, we get

$\sum_{i=1}^N z_ix_i =247.518$

$\sum_{i=1}^Nx_i =42$

$\sum_{i=1}^Nz_i =30.307$

$\sum_{i=1}^Nx_i^2 =364$

Putting these values in the equation we get

$b=\frac{\sum_{i=1}^N z_ix_i -\frac{\sum_{i=1}^Nx_i\sum_{i=1}^Nz_i}{N} }{\sum_{i=1}^Nx_i^2-\frac{(\sum_{i=1}^Nx_i)^2}{N}}$

$or,~b=\frac{247.518 -\frac{42*30.307}{6} }{364-\frac{(42)^2}{6}}$

$or,~b=\frac{247.518 -212.17}{364-294}$

$or,~b=0.505$

Using this the value of c can be caluclated as

$c=\frac{\sum_{i=1}^Nz_i-b\sum_{i=1}^Nx_i}{N}$

$or,~c=\frac{30.307-0.505*42}{6}$

$or,~c=1.516$

Thus the linear fit is

$z=1.516+0.505x$

From our prvious assumption,

$c=ln~a$

$or,~a=e^c=e^{1.516}=4.554$

Thus the exponential equation fit to the data is

$y=4.554e^{0.505x}$

In the subsequent steps I have used Excel for the calculations to avoid human error. All formulae used are written in the headings.

А B с D E F G H J 1 2 3 4 х у sum of (y-y')^2 No. of points N 5789.144 6 2 5 6 7 12 34 4 6 y' = 4.554"en (0.505x) 12.50346702 34.32953175 94.25519721 258.7871651 710.5263033 1950.821739 Y-y' (y-y')^2 -0.503467 0.25347904 -0.329532 0.10859117 7.7448028 59.9819702 -2.787165 7.7682895 16.473697 271.382684 -73.82174 5449.64922 102 256 727 1877 Standarad deviation = (H4/H5-1)^0.5 34.02688 Standard error of estimate = S.D./(H5)^0.5 13.89142 8 8 9 10 12 10 11 12

For the calculation of the R^2 value we will use the linear fit data

- А B С D E F G H J K L M 1 2 3 х Y 12 z = Iny 2.485 z'=1.516+0.505*x 2.526 z-zm -2.566166667 (z'-zm)^2 6.376467 4 2 5 4 34 3.526 3.536 2.29573 R^2 = 111/G11 0.998357 6 6 102 4.546 (z-zm)^2 6.58521136 2.32613336 0.18161803 0.24387136 2.36493136 6.17936736 0.255193 z'-zm -2.525166667 -1.515166667 -0.505166667 0.504833333 1.514833333 2.524833333 7 4.625 5.545 6.589 8 256 -1.525166667 -0.426166667 0.493833333 1.537833333 2.485833333 5.556 8 10 727 6.566 7.576 0.254857 2.29472 6.374783 9 12 1877 7.537 10 11 Mean of z=zm = 5.051167 sum (z-zm)^2 = 17.8811328 sum (z'-zm)^2 = 17.85175 12 13

Thus from here we get

• Standard deviation of the data is 34.02
• The standard estimate of error is 13.89
• The value of
  R
  is 0.998

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