Question

Listed below are the log body weights and log brain weights of the primates species in the data set ”mammals”. Find the equation of the least squares line with y = log brain weight and x = log body weight. Do it by hand, by constructing a table like the one in Example 9.1. Then do it with your calculator as efficiently as possible. Finally, use the lm function in R to do it by creating a linear model object ”primates.lm”. The model formula is ”log(brain)∼log(body)”. You can select the primates and put them in a new data frame by first listing the primate species names:

> primatenames=c(”Owl monkey”, ”Patas monkey”, ”Gorilla”, etc.)

and then

> primates=mammals[primatenames, ]

Your ”data” argument in calling lm would be ”data=primates”, as in

> primates.lm=lm(log(brain)∼log(body),data=primates)

Alternatively, you can just use ”mammals[primatenames, ]” as the data argument in lm, that is,

> primates.lm=lm(log(brain)∼log(body), data=mammals[primatenames,])

log body               log brain

Owl monkey                       -0.7339692         2.740840

Patas monkey                    2.3025851           4.744932

Gorilla                                   5.3327188           6.006353

Human                                 4.1271344           7.185387

Rhesus monkey                                1.9169226           5.187386

Chimpanzee                                       3.9543159           6.086775

Baboon                                                 2.3561259           5.190175

Verbet                                                  1.4327007           4.060443

Galago                                                  -1.6094379        1.609438

Slow loris                                             0.3364722           2.525729

Answer 1

The dataset is not given hence we use the mammalssleep data set ,

the r code is as below

data("mammalsleep")

primatenames <- c("Owl monkey", "Patas monkey", "Gorilla")

primates= mammalsleep %>% filter(species %in% primatenames)
#Your ”data” argument in calling lm would be ”data=primates”, as in
primates.lm=lm(log(brw)~log(bw),data=primates)

summary(primates.lm)

The results of the regression is

summary(primates.lm)

Call:

lm(formula = log(brw) ~ log(bw), data = primates)

Residuals:

1 2 3

-0.1233 -0.1231 0.2464

Coefficients:

Estimate Std. Error t value Pr(>|t|)  

(Intercept) 3.25902 0.23780 13.705 0.0464 *

log(bw) 0.53831 0.07035 7.652 0.0827 .

---

Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.3018 on 1 degrees of freedom

Multiple R-squared: 0.9832, Adjusted R-squared: 0.9664

F-statistic: 58.55 on 1 and 1 DF, p-value: 0.08273

The regression equation is formed using the

as

log(brw) = 3.25 +0.5383*log(bw)

the r2 value is 0.9832 , this means that the model is very good, higher the value better the model range is 0 to 1

also note that r2 means that about 98.32 % variation in data is explained by the model