Question

X= weight of copper

Y=Actual Price of Copper in (USD)

x : 0.3 0.4 0.5 0.5 1.0 0.7

y : 510 1151 1343 1410 5669 2277

(a) Find the regression equation for the data points given.

(b) Determine the percentage of variation in price of Copper y that is explained by the weight x.

(c) Is it reasonable to predict the price of a 0.8 grams of copper using this model?

(d) Is it reasonable to predict the price of a 1.5 grams of copper using this model?

This is strictly to check answers.

Answer 1

Here we have given that

X : weight of copper

Y: Actual Price of Copper in (USD)

X	Y
0.3	510
0.4	1151
0.5	1343
0.5	1410
1	5669
0.7	2277

(A)

Now we want to find the least square regression line

we know that

where Y is response variable and X is predictor variable

is intercept

is slope

We can solve this question using excel.

First enter data into excel.

Click on Data -------> Data Analysis --------> Regression ------->

Input

Input Y Range : select y values

Input X Range :select values of x.

Output Range : select any empty cell

we get this output

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.97
R Square	0.94
Adjusted R Square	0.92
Standard Error	523.85
Observations	6
	Coefficients	Standard Error	t Stat	P-value
Intercept	-2006.98	571.80	-3.51	0.02
X	7177.02	935.84	7.67	0.00

we get the least square regression line is

=-2006.98 + 7177.02 ( X)

where

= - 2006.98

= 7177.02

(B)

In the above output we get the R-square= 0.94 =94%

The percentage of variation in price of copper y that is explained by the weight x is 94%

(C)

it is reasonable to predict the price of 1.5 grams of copper using this model

x=weight of copper in gram=1.5

the least square regression line is

=-2006.98 + 7177.02 ( X)

=-2006.98 + 7177.02 ( 1.5)

=8759

The price of 1.5 grams of copper using this model is 8759 USD