Question

A motion picture industry analyst is studying movies based on epic novels. The following data were obtained for 10 Hollywood movies made in the past five years. Each movie was based on an epic novel. For these data, x₁ = first-year box office receipts of the movie, x₂ = total production costs of the movie, x₃ = total promotional costs of the movie, and x₄ = total book sales prior to movie release. All units are in millions of dollars.

x1	x2	x3	x4
85.1	8.5	5.1	4.7
106.3	12.9	5.8	8.8
50.2	5.2	2.1	15.1
130.6	10.7	8.4	12.2
54.8	3.1	2.9	10.6
30.3	3.5	1.2	3.5
79.4	9.2	3.7	9.7
91.0	9.0	7.6	5.9
135.4	15.1	7.7	20.8
89.3	10.2	4.5	7.9

(a) Generate summary statistics, including the mean and standard deviation of each variable. Compute the coefficient of variation for each variable. (Use 2 decimal places.)

	x	s	CV
x1			%
x2			%
x3			%
x4			%

Relative to its mean, which variable has the largest spread of data values?

x₄

x₃    

x₂

x₁

Why would a variable with a large coefficient of variation be expected to change a lot relative to its average value? Although x₁ has the largest standard deviation, it has the smallest coefficient of variation. How does the mean of x₁ help explain this?

A variable with a large CV has large s relative to x. Here, x₁ has a small CV because we divide by a small mean.

A variable with a large CV has large s relative to x. Here, x₁ has a small CV because we divide by a large mean.    

A variable with a large CV has small s relative to x. Here, x₁ has a small CV because we divide by a large mean

.A variable with a large CV has small s relative to x. Here, x₁ has a small CV because we divide by a small mean.

(b) For each pair of variables, generate the correlation coefficient r. Compute the corresponding coefficient of determination r². (Use 3 decimal places.)

	r	r2
x1, x2
x1, x3
x1, x4
x2, x3
x2, x4
x3, x4

Which of the three variables x₂, x₃, and x₄ has the least influence on box office receipts?

x₄

x₃   

x₂

What percent of the variation in box office receipts can be attributed to the corresponding variation in production costs? (Use 1 decimal place.)
_________________%

(c) Perform a regression analysis with x₁ as the response variable. Use x₂, x₃, and x₄ as explanatory variables. Look at the coefficient of multiple determination. What percentage of the variation in x₁ can be explained by the corresponding variations in x₂, x₃, and x₄ taken together? (Use 1 decimal place.)
_____________ %

(d) Write out the regression equation. (Use 2 decimal places.)

x1 =

+   x2

+ x3

+   x4

Explain how each coefficient can be thought of as a slope.

If we hold all explanatory variables as fixed constants, the intercept can be thought of as a "slope."

If we look at all coefficients together, each one can be thought of as a "slope."    

If we look at all coefficients together, the sum of them can be thought of as the overall "slope" of the regression line.

If we hold all other explanatory variables as fixed constants, then we can look at one coefficient as a "slope."

If x₂ (production costs) and x₄ (book sales) were held fixed but x₃ (promotional costs) were increased by 0.6 million dollars, what would you expect for the corresponding change in x₁ (box office receipts)? (Use 2 decimal places.)________________


(e) Test each coefficient in the regression equation to determine if it is zero or not zero. Use level of significance 5%. (Use 2 decimal places for t and 3 decimal places for the P-value.)

	t	P-value
β2
β3
β4

Conclusion

Reject the null for β₃ and β₄. Fail to reject the null for β₂.

Reject the null for β₂ and β₃. Fail to reject the null for β₄.   

Reject the null for all tests.

Reject the null for β₂ and β₄. Fail to reject the null for β₃.

Explain why book sales x₄ probably are not contributing much information in the regression model to forecast box office receipts x₁.

From the previous tests, we can conclude that the coefficient for x₄ is different than 0. Thus it does not belong in the model.

From the previous tests, we can conclude that the coefficient for x₄ is not different than 0. Thus it does not belong in the model.    

From the previous tests, we can conclude that the coefficient for x₄ is different than 0. Thus it belongs in the model.

From the previous tests, we can conclude that the coefficient for x₄ is not different than 0. Thus it belongs in the model.

(f) Find a 90% confidence interval for each coefficient. (Use 2 decimal places.)

	lower limit	upper limit
β2
β3
β4

(g) Suppose a new movie (based on an epic novel) has just been released. Production costs were x₂ = 11.4 million; promotion costs were x₃ = 4.7 million; book sales were x₄ = 8.1 million. Make a prediction for x₁ = first-year box office receipts and find an 85% confidence interval for your prediction (if your software supports prediction intervals). (Use 1 decimal place.)

prediction
lower limit
upper limit

(h) Construct a new regression model with x₃ as the response variable and x₁, x₂, and x₄ as explanatory variables. (Use 2 decimal places.)

x3 =

+  x1

+  x2

+  x4

Suppose Hollywood is planning a new epic movie with projected box office sales x₁ = 100 million and production costs x₂ = 12 million. The book on which the movie is based had sales of x₄ = 9.2 million. Forecast the dollar amount (in millions) that should be budgeted for promotion costs x₃ and find an 80% confidence interval for your prediction.

prediction
lower limit
upper limit

Answer 1

[Used R-Software]

(a)

(See R-commands for computation)

	xbar	s	cv
x1	85.24	33.79	39.64%
x2	8.74	3.89	44.45%
x3	4.9	2.48	50.62%
x4	9.92	5.17	52.15%

Relative to its mean, which variable has the largest spread of data values? The variable that has highest value of coefficient of variation i.e. x4 has the largest spread of data values.

Why would a variable with a large coefficient of variation be expected to change a lot relative to its average value? Although x1 has the largest standard deviation, it has the smallest coefficient of variation. How does the mean of x1 help explain this?
A variable with a large CV has small s relative to x. Here, x1 has a small CV because we divide by a large mean.

(b)

For each pair of variables, generate the correlation coefficient r. Compute the corresponding coefficient of determination r². (See R-commands for computation)

	r	r2
x1, x2	0.917	0.842
x1, x3	0.93	0.865
x1, x4	0.475	0.225
x2, x3	0.79	0.624
x2, x4	0.429	0.184
x3, x4	0.299	0.089

Which of the three variables x₂, x₃, and x₄ has the least influence on box office receipts? Correlation between x₁ and x₄ is the least. Therefore, x₄ has the least influence on box office receipts.

What percent of the variation in box office receipts can be attributed to the corresponding variation in production costs? (Use 1 decimal place.) 91.7% (See R-commands for computation)

(c)

Perform a regression analysis with x₁ as the response variable. Use x₂, x₃, and x₄ as explanatory variables. Look at the coefficient of multiple determination. What percentage of the variation in x₁ can be explained by the corresponding variations in x₂, x₃, and x₄ taken together? (Use 1 decimal place.)

Coefficient of multiple determination will give percentage of the variation in x₁ that can be explained by the corresponding variations in x₂, x₃, and x₄ taken together. Thus, required percentage is 96.7%. (See R-commands for computation)

(d)

# Thus, regression equation is: x₁=7.68 + 3.66*x₂ + 7.62*x₃ + 0.83*x₄  (See R-commands for computation)

Explain how each coefficient can be thought of as a slope.
If we hold all other explanatory variables as fixed constants, then we can look at one coefficient as a "slope." As slope is change (rate of change) in one variable when there is unit change in another variable.

If x2 (production costs) and x4 (book sales) were held fixed but x3 (promotional costs) were increased by 0.6 million dollars, what would you expect for the corresponding change in x1 (box office receipts)? (Use 2 decimal places.)

# Thus, the new regression equation is: x1=3.10 + 3.66*x2 + 7.62*x3 + 0.83*x4

Only intercept changes when above situation appears. Betas remains the same.

R-commands and outputs:

x1=c(85.1,106.3,50.2,130.6,54.8,30.3,79.4,91.0,135.4,89.3)
x2=c(8.5,12.9,5.2,10.7,3.1,3.5,9.2,9.0,15.1,10.2)
x3=c(5.1,5.8,2.1,8.4,2.9,1.2,3.7,7.6,7.7,4.5)
x4=c(4.7,8.8,15.1,12.2,10.6,3.5,9.7,5.9,20.8,7.9)

#(a)
summary(x1)
Min. 1st Qu. Median Mean 3rd Qu. Max.
30.30 60.95 87.20 85.24 102.47 135.40
mean(x1)
sd(x1)
cvx1=sd(x1)/mean(x1)
cvx1

summary(x2)
Min. 1st Qu. Median Mean 3rd Qu. Max.
3.100 6.025 9.100 8.740 10.575 15.100
mean(x2)
sd(x2)
cvx2=sd(x2)/mean(x2)
cvx2

summary(x3)
Min. 1st Qu. Median Mean 3rd Qu. Max.
1.20 3.10 4.80 4.90 7.15 8.40
mean(x3)
sd(x3)
cvx3=sd(x3)/mean(x3)
cvx3

summary(x4)
Min. 1st Qu. Median Mean 3rd Qu. Max.
3.50 6.40 9.25 9.92 11.80 20.80
mean(x4)
sd(x4)
cvx4=sd(x4)/mean(x4)
cvx4

xibar=c(mean(x1),mean(x2),mean(x3),mean(x4))
xibar
[1] 85.24 8.74 4.90 9.92
si=c(sd(x1),sd(x2),sd(x3),sd(x4))
si
[1] 33.786361 3.885357 2.480143 5.173393
round(si,2)
[1] 33.79 3.89 2.48 5.17
cvxi=si/xibar
cvxi
[1] 0.3963675 0.4445489 0.5061517 0.5215114
cvxip=cvxi*100
cvxip
[1] 39.63675 44.45489 50.61517 52.15114
round(cvxip,2)
[1] 39.64 44.45 50.62 52.15

#(b)
cor(x1,x2)
[1] 0.9174448
cor(x1,x3)
[1] 0.9299678
cor(x1,x4)
[1] 0.4746911
cor(x2,x3)
[1] 0.7899575
cor(x2,x4)
[1] 0.4291329
cor(x3,x4)
[1] 0.2987613

## Crosscheck:
sum((x3-mean(x3))*(x4-mean(x4))/(10-1))/(sd(x3)*sd(x4))
[1] 0.2987613

round(cor(x1,x2),3)
[1] 0.917
round(cor(x1,x3),3)
[1] 0.93
round(cor(x1,x4),3)
[1] 0.475
round(cor(x2,x3),3)
[1] 0.79
round(cor(x2,x4),3)
[1] 0.429
round(cor(x3,x4),3)
[1] 0.299

round(cor(x1,x2)^2,3)
[1] 0.842
round(cor(x1,x3)^2,3)
[1] 0.865
round(cor(x1,x4)^2,3)
[1] 0.225
round(cor(x2,x3)^2,3)
[1] 0.624
round(cor(x2,x4)^2,3)
[1] 0.184
round(cor(x3,x4)^2,3)
[1] 0.089

round(100*cor(x1,x2),1)
[1] 91.7

#(c)
# x1=response
fit=lm(x1~x2+x3+x4)
fit
Call:
lm(formula = x1 ~ x2 + x3 + x4)

Coefficients:
(Intercept) x2 x3 x4
7.6760 3.6616 7.6211 0.8285

s=summary(fit)
s
Call:
lm(formula = x1 ~ x2 + x3 + x4)

Residuals:
Min 1Q Median 3Q Max
-12.4384 -3.1695 0.8499 3.5134 9.6207

Coefficients:
Estimate Std. Error t value Pr(>|t|)   
(Intercept) 7.6760 6.7602 1.135 0.2995   
x2 3.6616 1.1178 3.276 0.0169 *
x3 7.6211 1.6573 4.598 0.0037 **
x4 0.8285 0.5394 1.536 0.1754   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 7.541 on 6 degrees of freedom
Multiple R-squared: 0.9668, Adjusted R-squared: 0.9502
F-statistic: 58.22 on 3 and 6 DF, p-value: 7.913e-05

# Extracting r^2 from summary:
Rsq=s$r.squared
Rsq
[1] 0.9667888
round(100*Rsq,1)
[1] 96.7

#(d)
# Regression equation:
beta=coef(fit)
(Intercept) x2 x3 x4
7.6760280 3.6616044 7.6210501 0.8284682
beta=round(beta,2)
beta
(Intercept) x2 x3 x4
7.68 3.66 7.62 0.83
# Thus, regression equation is: x1=7.68 + 3.66*x2 + 7.62*x3 + 0.83*x4

newx3=x3+0.6
newx3
[1] 5.7 6.4 2.7 9.0 3.5 1.8 4.3 8.2 8.3 5.1
newfit=lm(x1~x2+newx3+x4)
newfit
Call:
lm(formula = x1 ~ x2 + newx3 + x4)

Coefficients:
(Intercept) x2 newx3 x4
3.1034 3.6616 7.6211 0.8285

news=summary(newfit)
news
Call:
lm(formula = x1 ~ x2 + newx3 + x4)
Residuals:
Min 1Q Median 3Q Max
-12.4384 -3.1695 0.8499 3.5134 9.6207
Coefficients:
Estimate Std. Error t value Pr(>|t|)   
(Intercept) 3.1034 6.9784 0.445 0.6721   
x2 3.6616 1.1178 3.276 0.0169 *
newx3 7.6211 1.6573 4.598 0.0037 **
x4 0.8285 0.5394 1.536 0.1754   
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 7.541 on 6 degrees of freedom
Multiple R-squared: 0.9668, Adjusted R-squared: 0.9502
F-statistic: 58.22 on 3 and 6 DF, p-value: 7.913e-05

newbeta=coef(newfit)
newbeta
(Intercept) x2 newx3 x4
3.1033980 3.6616044 7.6210501 0.8284682
nbeta=round(newbeta,2)
nbeta
(Intercept) x2 newx3 x4
3.10 3.66 7.62 0.83
# Thus, the new regression equation is: x1=3.10 + 3.66*x2 + 7.62*x3 + 0.83*x4