Question

A tele-marketing company wants to know if sales go up as they call more people per day but spend less time per call. The following is the data from nine randomly selected days. The information is the number of calls the salesperson makes in a day and the total amount of sales (in thousands of dollars).

Calls             25      29      33      37      43      48      52      55      67

Sales            3.7     4.2     4.2     5.0     4.7     5.3     4.9     5.6     5.9

a. Calculate ∑X, ∑X², ∑Y, ∑Y², ∑XY

b. Calculate SS_XX, SS_YY, and SS_XY.

c. Use the information from parts (a) and (b) to generate the estimated OLS line.

d. Interpret the estimated slope coefficient from your line in part c.

e. Predict the amount of sales for the fourth observation in the data set.

f. Calculate the residual for that observation.

g. Construct the ANOVA table for this situation.

h. Calculate the coefficient of determination.

i. Interpret the coefficient of determination.

j. Using alpha = 0.05, use a model test to see if a linear relationship exists between the number of calls and sales.

k. A positive relationship is anticipated between these two variables. At alpha = 0.05, test to see if the evidence supports that anticipated sign.

l. Construct a 98% confidence interval for the population slope coefficient.

Answer 1

a)

X	Y	XY	X²	Y²
25	3.7	92.5	625	13.69
29	4.2	121.8	841	17.64
33	4.2	138.6	1089	17.64
37	5	185	1369	25
43	4.7	202.1	1849	22.09
48	5.3	254.4	2304	28.09
52	4.9	254.8	2704	24.01
55	5.6	308	3025	31.36
67	5.9	395.3	4489	34.81

	X	Y	XY	X²	Y²
total sum	389.000	43.500	1952.500	18295	214.33

b)

sample size ,   n =   9      
here, x̅ =Σx/n =   43.2222   ,   ȳ = Σy/n =   4.833333333
              
SSxx =    Σx² - (Σx)²/n =   1481.556      
SSxy=   Σxy - (Σx*Σy)/n =   72.333      
SSyy =    Σy²-(Σy)²/n =   4.080      

c)

estimated slope , ß1 = SSxy/SSxx =   72.333   /   1481.556   =   0.0488
                  
intercept,   ß0 = y̅-ß1* x̄ =   2.7231          
                  
so, regression line is   Ŷ =   2.7231   +   0.0488   *x

d)

for every unit increase in calls, the sales get increase by $48.8

e)

Predicted Y at X=   37   is                  
Ŷ =   2.723   +   0.049   *   37   =   4.530

predicted sale = $4530

f)

residual = 5000-4530=470

g)

Anova table
variation	SS	df	MS	F-stat	p-value
regression	3.531	1	3.531	45.069	0.0003
error,	0.549	7	0.078
total	4.080	8

H)

R² =    (Sxy)²/(Sx.Sy) =    0.8656

i)

about 86.56% of variation in observation of sales is explained by variable,calls

j)

F stat=45.069

p value=0.0003

p value <α=0.05,

a linear relationship exists between the number of calls and sales.

k)

Ho:   ß1=   0          
H1:   ß1 >   0          
n=   9              
alpha=   0.05              
estimated std error of slope =Se(ß1) = Se/√Sxx =    0.280   /√   1482   =   0.0073
                  
t stat = estimated slope/std error =ß1 /Se(ß1) =    0.0488   /   0.0073   =   6.713
                  
Degree of freedom ,df = n-2=   7              
p-value =    0.0001              
decision :    p-value<α , reject Ho              
Conclusion:   Reject Ho and conclude that slope is significantly positive

  

l)

confidence interval for slope                  
α=   0.02              
t critical value=   t α/2 =    2.998   [excel function: =t.inv.2t(α/2,df) ]      
estimated std error of slope = Se/√Sxx =    0.27992   /√   1481.56   =   0.007
                  
margin of error ,E= t*std error =    2.998   *   0.007   =   0.022
estimated slope , ß^ =    0.0488              
                  
                  
lower confidence limit = estimated slope - margin of error =   0.0488   -   0.022   =   0.027
upper confidence limit=estimated slope + margin of error =   0.0488   +   0.022   =   0.071