Question

1. Part of an Excel output relating 15 observations of X (independent variable) and Y (dependent variable) is shown below. Provide the values for a-e shown in the table below. (See section 15.5)

Summary Output

ANOVA
	df	SS	MS	F	Significance F
Regression	1	2.7500	-d-	-e-	0.632
Residual	-a-	-b-	11.45
Total	14	-c-

5. A Company has recorded data on daily demand for its product (y in thousands of units) and the unit price (x in hundreds of dollars). A sample of 10 days demand and associated prices resulted in the following summary data. Use the information to answer the questions below.

Ex= 90 E(x-x̄)(y- ȳ )=466

Ey= 170 E(x-x̄)^2=234

E(y- ȳ)^2=1434 SSE=505.98

                                      

                                     

                                                   

1. Compute b₁
   
   
2. Compute b₀
   
3. Use the least squares method to develop and write the estimated regression the equation.

4. What type of relationship exists between x and y? Explain.

5. Would the demand ever reach zero? If yes, at what price would the demand be zero?

6. Compute the coefficient of determination and the correlation coefficient?

Answer 1

ANOVA

n = 15, k = 1

a = df Residual = df Total-df Regression = 14-k = 14-1 = 13

b = SS Residual = df Residual*MS Residual = 11.45*13 = 148.85

c = SS Total = SS Regression + SS Residual = 2.75+148.85 = 151.6

d = MS Regression = SS Regresion/df Regression = 2.75/1 = 2.75

e = MS Regression/MS Residual = 2.75/11.45 = 0.24

Demand and Price

a. b₁ = (E(x-x̄)(y- ȳ))/(E(x-x̄)^2) = 466/234 = 1.991

b. b₀ = ((E(y)-b₁*E(x))/n = (170-1.991*90)/10 = -0.919

c. Least squares method: y = b₀ + b₁x

y = -0.919 + 1.99x

d. Type of relationship between x and y can be explained by correlation coefficient: r

r = E(x-x̄)(y- ȳ)/(E(x-x̄)²*E(y- ȳ)²)^0.5 = 466/(234*1434)^0.5 = 0.804

So, relationship between x and y is strong and positive.

e. y: Demand = 0

y = -0.919 + 1.991x

0 = -0.919 + 1.991x

1.991x = 0.919

x = 0.919/1.991 = 0.462

Price = 0.462

e. Coefficient of correlation = 0.804

Coefficient of determination = r*r = 0.804*0.804 = 0.646