3. (8 pts) Text prob X4.5, p. 173 (edited) Assume inventory is growing by day. C

Question

3. (8 pts) Text prob X4.5, p. 173 (edited) Assume inventory is growing by day. Chart the data and use "Add Trendline" to fit 2 trends: Power and Polynomial (order=3). Which one fits the data best and why? Hint: Use coefficients of determination. Day Inventory 1 134 2 267 3 456 4 390 5 777 6 978 7 1510 8 1800 9 2560 10 4250 3. (8 pts) Text prob X4.5, p. 173 (edited) Assume inventory is growing by day. Chart the data and use "Add Trendline" to fit 2 trends: Power and Polynomial (order=3). Which one fits the data best and why? Hint: Use coefficients of determination. Day Inventory 1 134 2 267 3 456 4 390 5 777 6 978 7 1510 8 1800 9 2560 10 4250

Explanation / Answer

Let us consider I as inventory and D as day.

We need to fit a appropriate regression equation of I on D with suitable degree which can be decided upon the value of coefficient of determination R2 , gives the proportion of variance explained by the corresponding regression equation.That is the fitted regression equation can explain the R2 amount of variation of the original data.

Now , first we fit a linear trend as

I = a + b (D) ,

By OLS method we have the estimates of a = -800.33 and b= 384.10 and R2 = 0.8154 , which is itself quite high.

Now we intend to fit a quadratic reression equation of I on D as

I = a+ b(d) + c(D2) .

The estimates are a= 614.83 , b= -323.49 , c = 64.33 , R2 = 0.9617 , which is very high .

Now similarly we take a cubic equation of I on D as,

I = a+ b(d) + c(D2) + d(D3 )

and get a= -328.467 , b= 513.170 ,c= -117.078 ,d= 10.994 with R2 = 0.9868 ,

which is aso a substantial increase in R2 , so we choose the cubic to be the best amonst these three, but if one wishes to work with a simple one he/she can work better with the quadratic one.

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