Explained: "Fit the curve of the form y=abx to..."

Lloyd Allen

Answered question

2021-11-29

Fit the curve of the form

y = a b^{x}

to the following data, then find the standard deviation.

\begin{array}{cc} x & 3 & 4 & 5 & 6 & 7 \\ y & 1.8 & 2.5 & 3.6 & 4.7 & 6.6 \end{array}

Answer & Explanation

Ancessitere

Beginner2021-11-30Added 17 answers

$\begin{array}{cc} x & \log y & (x - \overset{―}{x}) & (x - \overset{―}{x})^{2} & (y - \overset{―}{y}) & (y - \overset{―}{y})^{2} & (x - \overset{―}{x}) \cdot (y - \overset{―}{y}) \\ 3 & 1.09861 & - 2 & 4 & - 0.46779 & 0.21883 & 0.935581095 \\ 4 & 1.38629 & - 1 & 1 & - 0.18011 & 0.03244 & 0.180108475 \\ 5 & 1.60944 & 0 & 0 & 0.04304 & 0.00185 & 0 \\ 6 & 1.79176 & 1 & 1 & 0.22536 & 0.05079 & 0.225356633 \\ 7 & 1.94591 & 2 & 4 & 0.37951 & 0.14403 & 0.759014626 \end{array}$
$\log (y) = \log (a) + \log (b) x$
The mean is given by
$\overset{―}{x} = \frac{1}{n} \sum x$
$\overset{―}{x} = \frac{25}{5} = 5$
$\frac{\log Y}{\log (y)} = \frac{7.832}{5} = 1.5664$
Answer:
The slope is given by
$β_{1} = \frac{\sum (x - \overset{―}{x}) (\log (y) - \overset{―}{\log (y)})}{\sum (x - {\overset{―}{x}}^{2}}}$
$β_{1} = \frac{2.1001}{10} = 0.21001$
The intercept is given by
$β_{0} = \overset{―}{\log (y)} - β_{1} \overset{―}{x}$
$β_{0} = 1.5664 - 0.21001 \times 5 = 0.5164$
The regression line is given by
$\log (y) = 0.5164 + 0.21001 x$
$y = 1.676 {(8.1735)}^{x}$

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