Question

Let's not forget about regression! Let's say I'd like to make a prediction about how quickly my new potato plant will grow. I measured its height everyday and recorded it. Here's the height in cm for the first 12 days .1 .1 .3 .9 1.2 1.4 1.8 2.0 2.2 2.5 2.7 3.0

Answer 1

Solution :  

Explanation :

The regression Equation is;

Where "b₀" is Intercept parameter and "b₁" Slope parameter

Formulas for Intercept and slope parameters is;

b, Σ(α - α)(y - 9) Σ(x – )2

$b_{0} = \bar{y}-b_{1}*\bar{x}$

Where X = time in a Day

Y = Height in Cm

So regression equation by using Excel calculation is ;

	Time (days)(X)	Height (Y)	(X -Mean(X))	(Y - Mean(Y))	(X - Mean(X))^2	(X -Mean(X))(Y -Mean(Y))
	1	0.1	-5.5	-1.416666667	30.25	7.791666667
	2	0.1	-4.5	-1.416666667	20.25	6.375
	3	0.3	-3.5	-1.216666667	12.25	4.258333333
	4	0.9	-2.5	-0.616666667	6.25	1.541666667
	5	1.2	-1.5	-0.316666667	2.25	0.475
	6	1.4	-0.5	-0.116666667	0.25	0.058333333
	7	1.8	0.5	0.283333333	0.25	0.141666667
	8	2	1.5	0.483333333	2.25	0.725
	9	2.2	2.5	0.683333333	6.25	1.708333333
	10	2.5	3.5	0.983333333	12.25	3.441666667
	11	2.7	4.5	1.183333333	20.25	5.325
	12	3	5.5	1.483333333	30.25	8.158333333
Sum	78	18.2			143	40
Mean	6.5	1.516667
				b1	0.280
				b0	-0.30152
				Regression Equation is :
				Y = -0.301 + 0.28 X

Also If you have to predict the day 16th height then it is

cm

Excel Regression Output For reference):

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.99345748
R Square	0.986957764
Adjusted R Square	0.985653541
Standard Error	0.121595838
Observations	12
ANOVA
	df	SS	MS	F	Significance F
Regression	1	11.18881119	11.18881119	756.7397	9.34E-11
Residual	10	0.147855478	0.014785548
Total	11	11.33666667
	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%	Lower 95.0%	Upper 95.0%
Intercept	-0.301515152	0.074837065	-4.028954796	0.002404	-0.46826	-0.13477	-0.46826	-0.13477
Time (days)(X)	0.27972028	0.010168355	27.50890243	9.34E-11	0.257064	0.302377	0.257064	0.302377