Question

BBA 403 TEST 3: Name: Score:_ SOLVE AND THOROUGHLY INTERPRET YOUR ANSWER 1. A marketing manager conducted a study to determine whether there is a linear relationship between many spent on advertising and company sales. The data are shown on the table below. Display the data in a scatter plot, calculate the correlation coefficient, state a conclusion, and interpret the result Adverstising Company Expense, (1000s of s) (1000s of ) 2.4 16 2.0 26 14 1.6 2.0 2.2 184 220 240 180 184 186 215 ble above, you computed r. Test the significance of the correlation coefficient using a 0.05. (Be sure to identify the null and alternative hypotheses, degrees of freedom, critical value(s) and the rejection region(s), find the standardized test statistic, make a decision to reject or fail to reject the null hypothesis, and interpret the decision in the context of the original claim.) 3. Find the equation for the regression line for the advertising expenditures and company sales. Predict the company sales for 1.8 thousand dollars. 4. Calculate the coefficient of determination r2. What does it tell you about the explained variation of the data about the regression line and about the unexplained variation? 5. Find the standard error of estimate se and interpret the result 6, Construct a 95% prediction interval for the company sales when advertising expenses are $2100. What can you conclude?

Answer 1

1)

300 250 200 150 100 50 1.5 2.5 0.5 2 3

R = 0.9129

there is strong and positive correlation between two variables

2)

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.9129
R Square	0.8334
Adjusted R Square	0.8056
Standard Error	10.2903
Observations	8
ANOVA
	df	SS	MS	F	Significance F
Regression	1	3178.1559	3178.1559	30.0136	0.0015
Residual	6	635.3441	105.8907
Total	7	3813.5000
	Coefficients	Standard Error	t Stat	P-value	Lower 95%
Intercept	104.0607	18.6462	5.5808	0.0014	58.4351
x	50.7287	9.2597	5.4785	0.0015	28.0712

Ho:ρ=0

$Ha : \rho \neq 0$

df = 6

critical value = =t.inv.2t(0.05,6) = 2.4469

rejection region
|TS| > 2.4469

TS = 5.4785

p-value = 0.0015

since p-value < alpha

we reject the null hypothesis

we conclude that correlation is significant

3)

x = 1.8

y^ = 104.0607 + 50.7287*x

= 104.0607 + 50.7287*1.8

= 195.3724

4)

r^2 = 0.8334

which means 83.34 % of variation in y is explained by this model