Question #116464
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 obtains the
following results,
Data size (gigabytes) 6 7 7 8 10 10 15
Processed requests 40 55 50 41 17 26 16
i. Draw the scatterplot for the data. Be sure to label your axes.
ii. Is there any correlation between the processing request and the size of incoming data?
What is the correlation coefficient?
iii. By what percentage is the processing time dependent on the size of incoming data?
iv. Compute a least square regression line for regressing processing request on the size of
incoming data.
v. Use your regression equation to predict the processing request for an incoming data
of size 17.0 gigabytes
vi. Is the slope statistically significant at α = 5% ?
1
Expert's answer
2020-05-25T20:30:34-0400
Data science megabytes,xProcessed requests,y640755750841101710261516\def\arraystretch{1.5} \begin{array}{c:c} Data \ science\ megabytes,x & Processed \ requests,y \\ \hline 6 & 40 \\ 7 & 55 \\ 7 & 50 \\ 8 & 41 \\ 10 & 17 \\ 10 & 26 \\ 15 & 16 \end{array}

i. The response variable here is the number of processed requests (y),(y), and we attempt to predict it from the size of a data set (x).(x).


xyx2xyy2640362401600755493853025750493502500841643281681101710017028910261002606761516225240324Sum=63245623197310027\def\arraystretch{1.5} \begin{array}{c} & x & y & x^2 & xy & y^2\\ \hline &6 & 40 & 36 & 240 & 1600 \\ & 7 & 55 & 49 & 385 & 3025 \\ & 7 & 50 & 49 & 350 & 2500 \\ & 8 & 41 & 64 & 328 & 1681 \\ & 10 & 17 & 100 & 170 & 289 \\ & 10 & 26 & 100 & 260 & 676 \\ & 15 & 16 & 225 & 240 & 324 \\ Sum=& 63 & 245 & 623 & 1973 & 10027 \end{array}

xˉ=ixin=637=9,yˉ=iyin=2457=35\bar{x}=\dfrac{\sum_ix_i}{n}=\dfrac{63}{7}=9,\bar{y}=\dfrac{\sum_iy_i}{n}=\dfrac{245}{7}=35

Sxx=ixi2nxˉ2=6237(9)2=56S_{xx}=\sum_ix_i^2-n\cdot\bar{x}^2=623-7\cdot(9)^2=56

Sxy=ixiyinxˉyˉ=19737(9)(35)=232S_{xy}=\sum_ix_iy_i-n\cdot\bar{x}\cdot\bar{y}=1973-7\cdot(9)(35)=-232

Syy=iyi2nyˉ2=100277(35)2=1452S_{yy}=\sum_iy_i^2-n\cdot\bar{y}^2=10027-7\cdot(35)^2=1452

Therefore, based on the above calculations, the regression coefficients (the slope m,m, and the yy-

intercept bb ) are obtained as follows:


m=SxySxx=23256=2974.142857m=\dfrac{S_{xy}}{S_{xx}}=\dfrac{-232}{56}=-\dfrac{29}{7}\approx−4.142857

b=yˉmxˉ=35(297)(9)=506772.285714b=\bar{y}-m\bar{x}=35-(-\dfrac{29}{7})(9)=\dfrac{506}{7}\approx72.285714

Therefore, we find that the regression equation is:


Y=72.2857144.142857XY=72.285714-4.142857X


ii. Is there any correlation between the processing request and the size of incoming data?

What is the correlation coefficient?

Correlation cofficient


r=SxySxxSyy=2325614520.8136r=\dfrac{S_{xy}}{\sqrt{S_{xx}}\sqrt{S_{yy}}}=\dfrac{-232}{\sqrt{56}\sqrt{1452}}\approx−0.8136

Strong correlation

iii. By what percentage is the processing time dependent on the size of incoming data?

The coefficient of determination


r2=(232561452)20.6619r^2=(\dfrac{-232}{\sqrt{56}\sqrt{1452}})^2\approx0.6619

66.19 %66.19\ \%

The proportion of Y variance explained by the linear relationship between X and Y is  66.19 %66.19\ \% .


iv. The regression equation is:


Y=72.2857144.142857XY=72.285714-4.142857X

v. X=17X=17


Y=72.2857144.142857(17)2Y=72.285714-4.142857(17)\approx2

vi. If there is a significant linear relationship between the independent variable X and the dependent variable Y, the slope will not equal zero.

H0:m=0H_0: m=0

H1:m0H_1:m\not=0


s(m)=i(yiyˉ)2(n2)i(xixˉ)2=s(m)=\sqrt{\dfrac{\sum_i(y_i-\bar{y})^2}{(n-2)\sum_i(x_i-\bar{x})^2}}=

=1452(72)(56)2.2772=\sqrt{\dfrac{1452}{(7-2)(56)}}\approx2.2772

This corresponds to a two-tailed test, for which a t-test for one mean, with unknown population standard deviation will be used.

Based on the information provided, the significance level is α=0.05,\alpha=0.05, and the critical value for a two-tailed test df=n2=5df=n-2=5 is tc=2.570543t_c=2.570543

The t-statistic is computed as follows:


t=m0s=4.1428572.27721.8193t=\dfrac{m-0}{s}=\dfrac{-4.142857}{2.2772}\approx-1.8193

Using the P-value approach: The p-value is p=0.128575,p=0.128575, and since p=0.128575>0.05,p=0.128575>0.05, it is concluded that the null hypothesis is not rejected. Therefore, there is not enough evidence to claim that the slope mm  is different than 0, at the 0.05 significance level.



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Comments

Assignment Expert
15.06.20, 01:19

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desmond
12.06.20, 04:24

thanks

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