Question

A market trader sells ball-point pens on his stall. He sells the pens for a different
fixed price, x pens, in each of six weeks. He notes the number of pens, y that he
sells in each of these six weeks. The results are shown in following table.
Week 1 2 3 4 5 6
Pens 10 15 20 25 30 35
Sells 68 60 55 48 38 32
a. Find the correlation coefficient between pens and sells and test the hypothesis
that there is no relation between these variables.
b. Find the regression equation of sells to pens
c. Estimate the weekly sells for 20 pens.
d. Estimate the Error.

Answer 1

Part a)

Let pens be X and sells be Y. The Pearson correlation coefficient between X and Y is obtained as:

$$r = \frac{\sum(x_i - \bar x) (y_i - \bar y)}{\sum(x_i - \bar x)^2 \sum(y_i - \bar y)^2}$$

$\sum(x_i - \bar x) (y_i - \bar y) = -632.5$ $\sum(x_i - \bar x)^2=437.5$ $\sum(y_i - \bar y)^2=920.8333$

Therefore

$$r=\frac{ -632.5}{\sqrt{437.5*920.3333}} = -0.99651$$

Null hypothesis, $H_0: r = 0$

Alternative hypothesis, $H_1: r \not =0$

Since we have 12 data points, the degrees of freedom for the hypothesis test are:

$$12-2=10$$

The critical value associated with 10 degrees of freedom is $\underset{-}{+} 0.576$

Since the correlation is greater than the critical value i.e $-0.99651 >0.576$ , we can conclude that the correlation between pens and sells is significant

Part b)

Regression line for sells on pens

Using Excel, click on the $Data\ Analysis\ tab$ , select $regression$

In the pop-up window, fill $Input\ X\ Range$

With the cell ranges containing pens and

$Input\ Y\ Range$ with cell ranges containing sells.

Excel returns an

$$Intercept\ coefficient=82.6952$$

And

$$Slope\ coefficient=-1.4457$$

The regression equation is therefore

$$Sells=82.6952-1.4457 \ pens$$

Part c)

Weekly sells for 20 pens is

$$=82.6952-1.4457*20$$ $$=53.78095$$

Part d)

The mean square error (MSE) gives an average of the error from the model

$$MSE=\frac{1}{n}\sum(Actual_i - Predicted_i)^2$$

$$=\frac{6.41904}{6}=1.06984$$