Question

Mr. James McWhinney, president of Daniel James Financial Services, believes there is a relationship between the number of client contacts and the dollar amount of sales. To document this assertion, Mr. McWhinney gathered the following sample information. The X column indicates the number of client contacts last month and the Y column shows the value of sales ($ thousands) last month for each client sampled. Number of Contacts, x 14 12 20 16 46 Sales ($ thousands) 24 14 28 30 80 Number of Contacts, 23 48 50 55 50 Sales ($ thousands), y 30 90 85 120 110 a. Determine the standard error of estimate. (Round your answer to 2 decimal places.) Standard error of estimate b. Determine the coefficient of determination. (Round your answer to 2 decimal places.) Coefficient of determination c. Interpret the coefficient of determination. (Round your answer to the nearest whole number.) percent of the variation in sales can be explained by the variation in the number of contacts

Answer 1

Answer:-

Given That:-

Mr. James McWhinney, president of Daniel-James Financial Services, believes there is a relationship between the number of client contacts and the dollar amount of sales. To document this assertion, Mr. McWhinney gathered the following sample information.

(a) Determine the standard error of estimate.

Number of contacts (X)	Sales (Y)
14	24
12	14
20	28
16	30
46	80
23	30
48	90
50	85
55	120
50	110

$\mathrm{Total = 7 24 + 6 14 + 10 28 + 8 30 + 23 80 + 11.5 30 + 24 90 + 25 85 + 27.5*120 + 25*110 = 13292}$

Mean (u) 13292 611 = 21.7545

Mean() 21.7545

and

(24 (7)2 + 14 *(6)2 + 28 * (10)2 + 30 * (8)2 + 80 (23) + 30 + (11.5)2 + 90 * (24)2 + 85 * (25)2 2 + 120 (27.5)2 + 110 (25)) – 611 * (21.7545)? Standard deviation (0) = 1 610

(317152.5-611 * (21.754512) 610

$=\sqrt{45.888 }$

=6.774

$\Rightarrow \sigma =6.774$

$\mathrm{Standard\: error\: of\:estimate\:=}\:\frac{\sigma }{\sqrt{n}}=\frac{6.774}{\sqrt{10}}=2.14$

(b) Determine the coefficient of determination.

$\begin{array}{c|cccccccccc}X&14&12&20&16&46&23&48&50&55&50\\Y&24&14&28&30&80&30&90&85&120&110\end{array}$

X	Y	X*Y	X*X	Y*Y
14	24	336	196	576
12	14	168	144	196
20	28	560	400	784
16	30	480	256	900
46	80	3680	2116	6400
23	30	690	529	900
48	90	4320	2304	8100
50	85	4250	2500	7225
55	120	6600	3025	14400
50	110	5500	2500	12100

$\sum{X}=334 ,$

$\sum{Y}=611 ,$

$\sum{X * Y}=26584 ,$

$\sum{X^2}=13970 ,$

and

$\sum{Y^2}=51581$

(c) Interpret the coefficient of determination.

$\mathrm{Correlation\: coefficient\:(r)\::}$

$\begin{aligned}r~&=~\frac{n*\sum{XY} - \sum{X}*\sum{Y}} {\sqrt{\left[n \sum{X^2}-\left(\sum{X}\right)^2\right] \cdot \left[n \sum{Y^2}-\left(\sum{Y}\right)^2\right]}} \\r~&=~\frac{ 10 26584 - 334 611 } {\sqrt{\left[ 10 13970 - 334^2 \right] \left[ 10 * 51581 - 611^2 \right] }} \end{aligned}$

$=\frac{61766}{\sqrt{\left(139700-111556\right)\left(515810-373321\right)}}$

$=\frac{61766}{\sqrt{28144\left(515810-373321\right)}}$

$=\frac{61766}{\sqrt{142489*28144}}$

$=\frac{61766}{\sqrt{4010210416}}$

$\mathrm{Rationalize\:}\frac{61766}{\sqrt{4010210416}}:$

$=\frac{61766\sqrt{4010210416}}{\sqrt{4010210416}\sqrt{4010210416}}$

$=\frac{61766\sqrt{4010210416}}{4010210416}$

$=\frac{347\sqrt{4010210416}}{22529272}$

=0.9754

$\Rightarrow \mathrm{Correlation\: coefficient\:(r)\:}=0.9754$

Now,

$\mathrm{Correlation\: of\:determination\:(r^{2})\:}=r*r=0.9754*0.9754$

$=0.95140516\simeq0.95$

=0.95

$\mathrm{Hence,\:\:Correlation\: of\:determination\:(r^{2})\:}=0.95$

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