3. Oranges sold Price per Orange y (100s) x (cents) 6 10 4 20 5 30...
3.
|
Oranges sold |
Price per Orange |
|
y (100s) |
x (cents) |
|
6 |
10 |
|
4 |
20 |
|
5 |
30 |
|
4 |
40 |
|
3 |
50 |
|
1 |
60 |
Above are six weeks of data on the quantity of oranges sold (y) and price (x):
-
Estimate the parameters of this equation.
-
Predict the quantity of oranges sold if the price is 25 cents per orange.
-
Derive the OLS estimators for 0 and 1 in the simple regression model.
Homework Answers
For the given data on Oranges sold for each given price/Orange, set a regression equation in which the dependent variable is "Oranges sold y(100s)" and the independent variable is "Price per Orange x(cents)".
--------------------------------------------------------------------------------------------------------------------------------------------------------------
The result is:
Thus, the intercept is 6.733333333 and the slope-coefficient is -0.082857143
It shall be noted that the intercept and the slope-coefficient are the parameters of the regression equation
y(100s) = 6.733333333 - 0.082857143 x
--------------------------------------------------------------------------------------------------------------------------------------------------------------
When x=25 cents
y(100s) = 6.733333333 - 0.082857143 * 25
=4.661904762
Hence, the quantity of Oranges sold is 4.661904762*100 = 466 Oranges
--------------------------------------------------------------------------------------------------------------------------------------------------------------
When price is 0 cent, the OLS estimator - intercept is 6.733333333 and expected y(100s) is 6.733333333
When price is 1 cent, the expected y(100s) is:
y(100s) = 6.733333333 - 0.082857143 * 1
=6.65047619
Add Answer to:
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