
| Adwords | Revenue |
| 50 | 427 |
| 75 | 472 |
| 100 | 467 |
| 125 | 529 |
| 150 | 518 |
| 175 | 543 |
I have attached the complete R code below as well as the output
> x=c(50,75,100,125,150,175)
> y=c(427,472,467,529,518,543)
> model=lm(y~x);model
Call:
lm(formula = y ~ x)
Coefficients:
(Intercept) x
392.3810 0.8914
> summary(model)
Call:
lm(formula = y ~ x)
Residuals:
1 2 3 4 5 6
-9.952 12.762 -14.524 25.190 -8.095 -5.381
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 392.3810 19.9443 19.674 3.94e-05 ***
x 0.8914 0.1657 5.378 0.00578 **
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 17.33 on 4 degrees of freedom
Multiple R-squared: 0.8785, Adjusted R-squared: 0.8481
F-statistic: 28.93 on 1 and 4 DF, p-value: 0.005775
> cor(x,y)
[1] 0.9372906
a) 392.38
As it can be observed from the output that the intercept is 392.38
b) Option D
The regression line is given by
y=392.38+0.8914*x
So when x that is Adwords is equal to 0 then the expected revenue is 392.38.
Hence option D is correct
c) 0.94
It can be seen from the output that the sample correlation is equal to 0.94 after rounding up which implies that both x and y are highly correlated with each other that is when one increases both also increases.
d) 0.89
The slope of the regression line after rounding up is 0.89 which implies that for a unit increase in Adwords the expected revenue increases by a 0.89.
e) Option C
As seen in the output the p-value for the final regression line is 0.005775 which is smaller than 0.05 which implies that the regression line is significant.
As the regression line is significant we can use the line for prediction purpose.
Do comment if you have any doubt.
Thank you !!
Adwords Revenue 50 427 75 472 100 467 125 529 150 518 175 543 A small...
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