Question

Marketing date on sales is presented for youtube. data are the advertising budget in thousands of dollars along with the sales. The experiment has been repeated 200 times with different budgets and the observed sales have been recorded. The simple linear regression model was fitted:

## ## Call: ## lm(formula = sales ~ youtube, data = marketing) ## ## Residuals: ## Min 1Q Median 3Q Max ## -10.06 -2.35 -0.23 2.48 8.65 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 8.43911 0.54941 15.4 <2e-16 *** ## youtube 0.04754 0.00269 17.7 <2e-16 *** ## --- ## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 ## ## Residual standard error: 3.91 on 198 degrees of freedom ## Multiple R-squared: 0.612, Adjusted R-squared: 0.61 ## F-statistic: 312 on 1 and 198 DF, p-value: <2e-16

a) Calculate the Correlation coefficient r of the sales and youtube (respectively, y and x), SST, SSR, and SSE

b) Write the CI Formula and the number form for the slope coefficient.

c) Interpret the slope parameter

d) Write up the individual hypothesis test and its results for the slope parameter.

Answer #1

**Solution**:

**Part
A**:

The value of Coefficient of Determination (R^{2}) =
0.612

So, Correlation Coefficient (r) = 0.612^{1/2}

Or, **r = 0.782**

In order to calculate the SST, SSR, and SSE, we need the data set which is missing in the question...

**Part
B**:

The formula for Confidence Interval is:

Here, the slope coefficient is 0.04754.

**Part C and
D**:

Hypothesis:

H_{0}: β_{1} = 0 (Null Hypothesis)

H_{a}: β_{1} ≠ 0 (Alternate Hypothesis)

As the P value fro the slope of the variable "YouTube" is very
small (2 * 10^{-16}), which is smaller than 0.05, here, the
null hypothesis is rejected. The slope can't take the value of 0 as
it is smaller than 0.05.

End of the Solution..

> muncy = lm(hit_distance~launch_speed, data=muncy)
> summary(muncy)
Call:
lm(formula = hit_distance ~ launch_speed, data = muncy)
Residuals:
Min 1Q
Median 3Q
Max
-258.24 -105.23 23.29 116.06 174.73
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) -240.8429 36.6769 -6.567 1.46e-10
***
launch_speed 4.8800
0.4022 12.134 < 2e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' '
1
Residual standard error: 122.4 on 438 degrees of freedom
Multiple R-squared: 0.2516, Adjusted R-squared: 0.2499
F-statistic:...

The data set (Canvas: body.csv) contains records of CHEST_DIAM,
, CHEST_DEPTH, ANKLE_DIAM,WAIST_GIRTH, WRIST_GIRTH, WRIST_DIAM (all
in cm.), AGE (years), WEIGHT (kg.), HEIGHT (cm.), andGENDER
(1=male) for 108 individuals. We will be looking for the best set
of variables to (parsimoniously?) modelWEIGHT. Even though 6
explanatory variables only gives 29=512 possibilities
for “all possible” regressions, we’lltry to be more methodical
about it.
##question2
library(MASS)
##
## Attaching package: 'MASS'
## The following object is masked from
'package:olsrr':
##
## cement
body =...

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