Question

To study how social media may influence the products consumers​ buy, researchers collected the opening weekend box office revenue​ (in millions of​ dollars) for 23 recent movies and the social media message rate​(average number of messages referring to the movie per​ hour). The data are available below. Conduct a complete simple linear regression analysis of the relationship between revenue​ (y) and message rate​ (x).

Message Rate		Revenue​ ($millions)
1363.2		146
1219.2		79
681.2		67
583.6		37
454.7		35
413.9		34
306.2		21
289.8		18
245.1		18
163.9		17
148.9		16
147.4		15
147.3		15
123.6		14
118.1		13
108.9		13
100.1		12
90.3		11
89.1		6
70.1		6
56.2		5
41.6		3
8.4		1

The least squares regression equation is y=−0.031+l0.086x. ​(Round to three decimal places as​ needed.)

Check the usefulness of the hypothesized model. What are the hypotheses to​ test?

A.H0​: β1≠0 against Ha​:β1=0

B.H0​β0:=0 againstHa​:β0≠0

C.H0​:β0≠0 against Ha​:β0=0

D.H0​:β1=0 againstHa​:β1≠0 Your answer is correct.

Determine the estimate of the standard deviation.

s=9.59 ​(Round to two decimal places as​ needed.)

What is the test statistic for the​ hypotheses?

t=15.14 ​(Round to two decimal places as​ needed.)

What is the​ p-value for the test​ statistic?

​p-value=0​(Round to three decimal places as​ needed.)State the conclusion at α=0.05.

Since the​ p-value is less than α​, there is sufficient evidence to reject H0.

Conclude there is

a linear relationship between revenue and message rate.What is the value for the coefficient of determination

r2​?

r2=0.92 ​(Round to two decimal places as​ needed.)Interpret the value of r2 in the context of this problem.

A.r2 is the proportion of the total sample variability around message rate that is explained by the linear relationship between the revenue and the message rate.

B.r2 is the proportion of the sample points that do not fit within the​ 95% confidence interval.

C. r2 is the proportion of the sample points that fit on the estimated linear regression line.

D.r2 is the proportion of the total sample variability around mean revenue that is explained by the linear relationship between the revenue and the message rate.

Answer 1

Regression Analysis
	r²	0.916	n	23
	r	0.957	k	1
	Std. Error	9.592	Dep. Var.	Revenue ($Millions)
ANOVA table
Source	SS	df	MS	F	p-value
Regression	21,097.1386	1	21,097.1386	229.30	8.96E-13
Residual	1,932.1658	21	92.0079
Total	23,029.3043	22
Regression output					confidence interval
variables	coefficients	std. error	t (df=21)	p-value	95% lower	95% upper	std. coeff.
Intercept	-0.0309						0.000
Message Rate	0.0865	0.0057	15.143	8.96E-13	0.0746	0.0983	0.957

(a) Regression equation is y = -0.031 + 0.087x

(b) D.H0​:β1=0 against Ha​:β1≠0

(c) 9.59

(d) 15.14

(e) p- value = 0.000

Since the​ p-value is less than α​, there is sufficient evidence to reject H0.

Conclude there is a linear relationship between revenue and message rate

(f) 0.92

(g) D. r^2 is the proportion of the total sample variability around mean revenue that is explained by the linear relationship between the revenue and the message rate.