x | 1 | 2 | 3 | 4 |

y | 1.65 | 2.70 | 4.50 | 7.35 |

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We know that exponential model is given by

y = ae^{bx}

Taking log on both side, We get

log y = log a + bx

This equation is similar in form to the linear equation y = a + bx. Thus we evaluate parameters a and b by using the equation of linear regression model as below:

\(b = \frac{n\sum_{i=1}^{n} x_i log y_i – \sum_{i=1}^{n}x_i \sum_{i=1}^{n} log y_i}{n\sum_{i=1}^{n}x_i^2 – (\sum_{i=1}^{n}x_i)^2}\)

And

\(loga = \frac{\sum_{i=1}^n log y_i}{n}-b\frac{\sum_{i=1}^n x_i}{n}\)

Now we calculate required summations as below:

iteration | x_{i} |
y_{i} |
log y_{i} |
x_{i }log y_{i} |
x_{i}^{2} |

1 | 1 | 1.65 | 0.5 | 0.825 | 1 |

2 | 2 | 2.7 | 0.99 | 2.673 | 4 |

3 | 3 | 4.5 | 1.504 | 6.768 | 9 |

4 | 4 | 7.35 | 1.99 | 14.6265 | 16 |

n =4 | \(\sum x_i = 10\) | \(\sum log y_i = 4.984\) | \(\sum x_i \enspace logy_i = 24.8925\) | \(\sum x_i^2 = 30\) |

Now,

\(b = \frac{4 \times (24.892) – 10\times (4.984)}{4 \times 30 – 10^2}\)

b = 2.4865

\(log a = \frac{4.984}{4} – (2.4865)\frac{10}{4}\)

log a = -4.97025

a = e^{-4.97025} = 0.00694

Thus regression formula is

y = 0.00694 x e^{-2.4865x}

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