Question

- ස ය ත Table 3: Advertising Pages(P), in Hundreds, and Advertising Revenue (R), in Millions of Dollars) for 41 Magazines in 1986 Magazine P R Magazine PR Cosmopolitan 25 50.0 Town and Country 1 7.0 Redbook 15 19.7 True Story 77 6.6 Glamour 20 34.0 | Brides 13 6.2 Southern Living 17 30.7 | Book Digest Magazine 5 5.8 Vogue 23 27.0 | W 7 5.1 Sunset 17 26.3 Yankee 13 4.1 House and Garden 14 24.6 Playgirl New York Magazine 22 16.9 Saturday Review 6 3.9 House Beautiful 12 16.7 New Woman 3 3.5 Mademoiselle 15 14.6 Ms. 6 3.3 Psychology Today 8 13.8 Cuisine 4 3.0 Life Magazine 7 13.2 Mother Earth News Smithsonian 9 13.1 1001 Decorating Ideas 3 2.3 Rolling Stone 12 10.6 Self 5 2.3 Modern Bride 1 8.8 Decorating and Craft Ideas 1.8 Parents Saturday Evening Post 4 1.5 Architectural Digest 128.5 McCall's Needlework and Craft 3 1.3 Harper's Bazaar 8.3 Weight Watchers 3 1.3 Apartment Life 7 8.2 High Times 4 1.0 Bon Appetit Soap Opera Digest 20.3 Gourmet 7 7.3 උය උය ය ත 2.5 8.2
Using R.

Answer 1

(a)

Loaded the data into magazines dataframe. Below command used to fit the linear regression on the data to predict revenue based on advertising pages.

model = lm(R~P, data = magazines)

> model

Call:
lm(formula = R ~ P, data = magazines)

Coefficients:
(Intercept) P
7.6041 0.3527

Getting the summary of the model R-Sq by below command, we see that R-Sq is 0.1263 which is very low. That is the model only explains 12.63% of variation of advertising revenue. So, the fit is poor.

> summary(model)

Call:
lm(formula = R ~ P, data = magazines)

Residuals:
Min 1Q Median 3Q Max
-28.162 -6.362 -2.773 2.322 36.805

Coefficients:
Estimate Std. Error t value Pr(>|t|)   
(Intercept) 7.6041 2.4061 3.160 0.00304 **
P 0.3527 0.1486 2.374 0.02262 *
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 11.6 on 39 degrees of freedom
Multiple R-squared: 0.1263,   Adjusted R-squared: 0.1039
F-statistic: 5.636 on 1 and 39 DF, p-value: 0.02262

By running the below command, we can further evaluate the fit. Residual vs Fitted shows a pattern (all data points are concentrated at one place), which shows that fit is bad and there could be a non-linear relationship between predictor variables and an outcome variable.

> op <- par(mfrow=c(2,2),mar=c(2,3,1.5,0.5))
> plot(model)

(b)

Plotting the advertising pages on x axis and advertising revenue on y-axis, we see that the relationship is not linear and advertising revenue rises sharply with advertising pages. Also, there are lot of outliers in the data.

plot(magazines$P,magazines$R, xlab = "Advertising Pages", ylab = "Advertising Revenue", pch = 16)

Transforming the variable P and R to log(P) and log(R), we get the r-squared of the model as 0.4203

> summary(lm(log(R)~log(P), data = magazines))$r.squared
[1] 0.420323

Although the model is somewhat improved from part (a) but the fit is still average.

(c)

We will delete the outliers to further improve the model.

First, we will calculate the outlier for the advertising revenue (R).

> summary(magazines$R)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.30 3.30 7.30 11.36 13.80 50.00
> IQR = 13.80 - 3.30
> 13.80 + 1.5 * IQR
[1] 29.55

So, any advertising revenue greater than 29.55 is considered as outlier.

We will delete the below entries from the dataframe.

Magazine P R
1 Cosmopolitan 25 50.0
2 Redbook 15 49.7
3 Glamour 20 34.0
4 SouthernLiving 17 30.7

Similarly, we will delete the entry for the outlier of advertising pages.

23 TrueStory 77 6.6

We will get the data in new dataframe magazines.new

> magazines.new = magazines[c(-1:-4,-23),]

Running the regression after deleting the outliers, we get R-sq as 63.44% which is a good fit.

> summary(lm(R~P, data = magazines.new))$r.squared
[1] 0.6344904

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