Use the data given in the table to answer the following questions. The data represents the...

Use the data given in the table to answer the following questions. The data represents the average number of miles that a salesperson travels in a day verses the number of sales made each month.

Miles, x	24	32	75	41	76	107	32	46	112
Sales, y	76	58	190	112	141	235	24	147	188

(a) What is the value of the correlation coefficient for this set of data? Round to 3 decimal places.

(b) What is the equation of the Regression Line for this set of data? Round values to two decimal places.

(c) Predict the number of sales an associate could expect to make if he travelled an average of 108 miles each day. Round to two decimal places.

Expert Solution

a)

x	y	(x-x̅)²	(y-ȳ)²	(x-x̅)(y-ȳ)
24	76	1336.31	2928.01	1978.06
32	58	815.42	5200.01	2059.17
75	190	208.64	3586.68	865.06
41	112	382.42	328.01	354.17
76	141	238.53	118.57	168.17
107	235	2157.09	11001.68	4871.51
32	24	815.42	11259.57	3030.06
46	147	211.86	285.23	-245.83
112	188	2646.53	3351.12	2978.06

	ΣX	ΣY	Σ(x-x̅)²	Σ(y-ȳ)²	Σ(x-x̅)(y-ȳ)
total sum	545	1171	8812.222222	38058.9	16058.44
mean	60.56	130.11	SSxx	SSyy	SSxy

correlation coefficient , r = Sxy/√(Sx.Sy) = 0.877

b)

sample size ,   n =   9
here, x̅ = Σx / n=   60.56   ,     ȳ = Σy/n =   130.11

SSxx =    Σ(x-x̅)² =    8812.2222
SSxy=   Σ(x-x̅)(y-ȳ) =   16058.4

estimated slope , ß1 = SSxy/SSxx =   16058.4   /   8812.222   =   1.8223

intercept,   ß0 = y̅-ß1* x̄ =   19.7612

so, regression line is   Ŷ =   19.76 +   1.82 *x

c)

Predicted Y at X= 108 is
Ŷ = 19.76119 + 1.822292 * 108 = 216.57

milcah answered 7 months ago

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