Question

Ten communities roughly comparable in size, wealth, and purchasing powers were selected to investigate the effect of advertising expenditures on sales of electric vehicles. For each community, the advertising outlay (X, in thousands of dollars) and the sales (Y, in units sold) are shown below: dvertising Expenditure Sales 3.9 7.8 1.1 6.6 3.4 2.5 6.0 32 25 23 30 4.6 25 2.0 1. 2. 3. Find the regression equation for predicting sales from advertising expenditure. Construct the 99 percent confidence interval for the slope of the regression line. Test at the 5 percent significance level the hypothesis that no sales would be recorded if there were not advertising. (Clearly, this should be a one-sided test.) Perform the analysis of variance for this data and carry out the associated test at the 5 percent level of significance. Report the p-value for this test (from Excel, Minitab, or similar) 4. 5. If the prediction equation derived in problem 1 were to be applied to the population as a whole, what proportion of the variability in sales figures could be predicted from knowledge of advertising expenditure?

Answer 1

1)

x<- c(3.9,7.8,1.1,6.6,3.4,2.5,6,9.1,4.6,2)
y <- c(17,32,6,25,23,18,30,33,25,11)

model <- lm (y ~x)

summary(model)

Call:
lm(formula = y ~ x)

Residuals:
   Min     1Q Median     3Q    Max 
-4.681 -2.754 -1.116  3.215  5.088 

Coefficients:
            Estimate Std. Error t value Pr(>|t|)    
(Intercept)   7.2220     2.5324   2.852 0.021419 *  
x             3.1442     0.4764   6.600 0.000169 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 3.742 on 8 degrees of freedom
Multiple R-squared:  0.8449,    Adjusted R-squared:  0.8255 
F-statistic: 43.56 on 1 and 8 DF,  p-value: 0.0001693

y^= 7.222 + 3.1442 x

2)

Using Excel

SUMMARY OUTPUT
Regression Statistics
Multiple R	0.919158799
R Square	0.844852898
Adjusted R Square	0.825459511
Standard Error	3.741928104
Observations	10
ANOVA
	df	SS	MS	F	Significance F
Regression	1	609.9837925	609.9837925	43.56396678	0.000169334
Residual	8	112.0162075	14.00202593
Total	9	722
	Coefficients	Standard Error	t Stat	P-value	Lower 99.0%	Upper 99.0%
Intercept	7.222042139	2.532438664	2.851813251	0.021419261	-1.275270473	15.71935475
x	3.144246353	0.476379272	6.600300507	0.000169334	1.545809378	4.742683329

99% confidence interval for slope

1.545809

4.742683

3)

p-value for one-tailed test = 0.00016/2 = 0.00008

p-value < alpha

we reject the null hypothesis

4)

p-value = 0.00017 < alpha

hence we reject the null hypothesis

5)

R^2 = 0.8449

hence 84.49 % of variation is explained by this model