Question

A computer manager needs to know how efficiency of her new computer program depends on the size of incoming data. Efficiency will be measured by the number of processed requests per hour. Applying the program to data sets of different sizes, she gets the following results:-

Data Size (gigabytes) (x)

6 7 7 8 10 10 15

Processed requests(y)

40 55 50 41 17 26 16

a. Find the equation of regression line

b. Interpret regression co-efficient and y intercept

c. Can we take Data Size as dependent variable if yes then how? If not then why? Explain.

d. Could this problem be studied using correlation if yes then how? If not then why? Explain.

Answer 1

a.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & x & y & xy & x^2 & y^2 \\ \hline & 6 & 40 & 240 & 36 & 1600\\ \hdashline & 7 & 55 &385 & 49 & 3025\\ \hdashline & 7 & 50 & 400 & 49 & 2500\\ \hdashline & 8 & 41 & 328 & 64 & 1681\\ \hdashline & 10 & 17 & 170 & 100 & 289\\ \hdashline & 10 & 26 & 260 & 100 & 676\\ \hdashline & 15 & 16 & 240 & 225 & 256 \\ \hdashline Sum=& 63 & 245 & 1973 & 623 & 10027 \end{array}$$

$$\bar{x}={1\over n}\displaystyle\sum_{i=1}^nx_i={63\over 7}=9$$

$$\bar{y}={1\over n}\displaystyle\sum_{i=1}^ny_i={245\over 7}=35$$

$$SS_{xx}=\displaystyle\sum_{i=1}^nx_i^2-{1\over n}\bigg(\displaystyle\sum_{i=1}^nx_i\bigg)^2 =$$ $$=623-{63^2\over 7}=56$$

$$SS_{yy}=\displaystyle\sum_{i=1}^ny_i^2-{1\over n}\bigg(\displaystyle\sum_{i=1}^ny_i\bigg)^2 =$$ $$=10027-{245^2\over 7}=1452$$

$$SS_{xy}=\displaystyle\sum_{i=1}^nx_iy_i-{1\over n}\bigg(\displaystyle\sum_{i=1}^nx_i\bigg) \bigg(\displaystyle\sum_{i=1}^ny_i\bigg)=$$ $$=1973-{63\cdot 245\over 7}=-232$$

$$m=\dfrac{SS_{xy}}{S_{xx}}=\dfrac{-232}{56}=-\dfrac{29}{7}\approx-4.142857$$

$$b=\bar{y}-m\cdot \bar{x}=35-(-\dfrac{29}{7})\cdot9=$$ $$=\dfrac{506}{7}\approx72.285714$$

The equation of regression line

$$y=-\dfrac{29}{7}x+\dfrac{504}{7}$$

$$Processed\ requests=-\dfrac{29}{7}(Data\ Size)+\dfrac{504}{7}$$

$$y=-4.142857x+72.285714$$

b.

It is obtained that, there is negative linear relationship between “Data size (x)” and “Processed requests (y)”. Thus, as the number of processed requests increased, the data size decreases.

If Data size increases by 1 gigabyte, then the number of processed requests per hour decreases by 4.142857.

The processing request on the size of incoming data of size 0 gigabytes is $72.285714.$

c.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:c} & x & y & xy & x^2 & y^2 \\ \hline & 40 & 6 & 240 & 1600 & 36\\ \hdashline & 55 & 7 & 385 & 3025 & 49\\ \hdashline & 50 & 7 & 400 & 2500 & 49\\ \hdashline & 41 & 8 & 328 & 1681 & 64\\ \hdashline & 17 & 100 & 170 & 289 & 100\\ \hdashline & 26 & 10 & 260 & 676 & 100\\ \hdashline & 16 & 15 & 240 & 256 & 225 \\ \hdashline Sum=& 245 & 63 & 1973 & 10027 & 623 \end{array}$$

$$\bar{x}={1\over n}\displaystyle\sum_{i=1}^nx_i={245\over 7}=35$$

$$\bar{y}={1\over n}\displaystyle\sum_{i=1}^ny_i={63\over 7}=9$$

$$SS_{xx}=\displaystyle\sum_{i=1}^nx_i^2-{1\over n}\bigg(\displaystyle\sum_{i=1}^nx_i\bigg)^2 =$$ $$=10027-{245^2\over 7}=1452$$

$$SS_{yy}=\displaystyle\sum_{i=1}^ny_i^2-{1\over n}\bigg(\displaystyle\sum_{i=1}^ny_i\bigg)^2 =$$ $$=623-{63^2\over 7}=56$$

$$m=\dfrac{SS_{xy}}{S_{xx}}=\dfrac{-232}{1452}=-\dfrac{58}{363}\approx-0.159780$$

$$b=\bar{y}-m\cdot \bar{x}=9-(-\dfrac{58}{363})\cdot35=$$ $$=\dfrac{5297}{363}\approx14.592287$$

The equation of regression line

$$y=-\dfrac{58}{363}x+\dfrac{5297}{363}$$

$$Data\ size=-\dfrac{58}{363}(Processed\ requests)+\dfrac{1263}{121}$$

$$y=-0.159780x+14.592287$$

We can take Data Size as dependent variable.

d.

$$r=\dfrac{SS_{xy}}{\sqrt{SS_{xx}}\sqrt{SS_{yy}}}=\dfrac{-232}{\sqrt{56}\sqrt{1452}}\approx-0.8136$$

The correlation coefficient between "Data size" and " Processed request" is $-0.8136.$

Thus, there is strong negative correlation between the processed request and the size of the incoming data.

Hence this problem could be studied using correlation.