DATA303/473 Test 1 Fish markets -Solved

DATA 303/473 Test 1: Fish markets 
1 April 2021 
1. Data were collected on 158 fifish of some species commonly sold in fifish markets. The variables in the 
dataset are: 
• species: Name of fifish species 
• weight: Weight of fifish in grams 
• vert.len: Vertical length in cm 
• diag.len: Diagonal length in cm 
• cross.len: Cross length in cm 
• height: Height in cm 
• width: Diagonal width in cm 
A marine scientist is interested in developing a model that can be used to predict the weight of a fifish with 
species, vert.len, diag.len, cross.len, height and width as potential predictors. Part of the analysis 
carried out is shown below. Use the results to answer the questions that follow. 
fish<-read.csv("fishmarket.csv", header=T, stringsAsFactors = TRUE) 
summary(fish) 
## species weight vert.len diag.len 
## Bream :35 Min. : 5.9 Min. : 7.50 Min. : 8.40 
## Parkki :11 1st Qu.: 121.2 1st Qu.:19.15 1st Qu.:21.00 
## Perch :56 Median : 281.5 Median :25.30 Median :27.40 
## Pike :17 Mean : 400.8 Mean :26.29 Mean :28.47 
## Roach :19 3rd Qu.: 650.0 3rd Qu.:32.70 3rd Qu.:35.75 
## Smelt :14 Max. :1650.0 Max. :59.00 Max. :63.40 
## Whitefish: 6 
## cross.len height width 
## Min. : 8.80 Min. : 1.728 Min. :1.048 
## 1st Qu.:23.20 1st Qu.: 5.941 1st Qu.:3.399 
## Median :29.70 Median : 7.789 Median :4.277 
## Mean :31.28 Mean : 8.987 Mean :4.424 
## 3rd Qu.:39.67 3rd Qu.:12.372 3rd Qu.:5.587 
## Max. :68.00 Max. :18.957 Max. :8.142 
## 
fit1<-lm(weight~species+ vert.len + diag.len + cross.len + 
height + width, data=fish) 
library(pander) 
pander(summary(fit1), caption="") 
Estimate Std. Error t value Pr(>|t|) 
(Intercept) 
-912.7 127.5 -7.161 3.638e-11 
speciesParkki 
160.9 75.96 2.119 0.03582 
speciesPerch 
133.6 120.6 1.107 0.27 
speciesPike 
-209 135.5 -1.543 0.1251 
speciesRoach 
104.9 91.46 1.147 0.2532 
speciesSmelt 
442.2 119.7 3.695 0.0003108 
1Estimate Std. Error t value Pr(>|t|) 
speciesWhitefifish 
91.57 96.83 0.9456 0.3459 
vert.len 
-79.84 36.33 -2.198 0.02955 
diag.len 
81.71 45.84 1.783 0.07675 
cross.len 
30.27 29.48 1.027 0.3062 
height 
5.807 13.09 0.4435 0.6581 
width 
-0.7819 23.95 -0.03265 0.974 
Observations Residual Std. Error 
R2 
Adjusted R2 
158 93.95 0.9358 0.931 
BIC(fit1) 
## [1] 1937.243 
fit2<-lm(log(weight)~species+ vert.len + diag.len + cross.len + 
height + width, data=fish) 
pander(summary(fit2), caption="") 
Estimate Std. Error t value Pr(>|t|) 
(Intercept) 
2.427 0.2937 8.263 7.889e-14 
speciesParkki 
0.1541 0.175 0.8803 0.3801 
speciesPerch 
0.1668 0.2779 0.6002 0.5493 
speciesPike 
0.05541 0.3122 0.1775 0.8594 
speciesRoach 
0.1273 0.2108 0.6039 0.5468 
speciesSmelt 
-1.13 0.2758 -4.097 6.921e-05 
speciesWhitefifish 
0.3183 0.2231 1.427 0.1558 
vert.len 
0.09876 0.08372 1.18 0.2401 
diag.len 
-0.1081 0.1056 -1.024 0.3078 
cross.len 
0.06333 0.06794 0.9321 0.3528 
height 
0.06754 0.03017 2.239 0.0267 
width 
0.1973 0.05518 3.575 0.0004756 
Observations Residual Std. Error 
R2 
Adjusted R2 
158 0.2165 0.9752 0.9733 
BIC(fit2) 
## [1] 18.19266 
library(mgcv) 
gam1<-gam(log(weight)~species+ s(vert.len) + s(diag.len) + s(cross.len) + 
s(height) + s(width), data=fish, method="REML") 
summary(gam1) 
## 
## Family: gaussian 
## Link function: identity 
## 
2## Formula: 
## log(weight) ~ species + s(vert.len) + s(diag.len) + s(cross.len) + 
## s(height) + s(width) 
## 
## Parametric coefficients: 
## Estimate Std. Error t value Pr(>|t|) 
## (Intercept) 5.32479 0.09485 56.139 <2e-16 *** 
## speciesParkki 0.12221 0.08759 1.395 0.1652 
## speciesPerch 0.17259 0.13261 1.301 0.1953 
## speciesPike 0.11770 0.16907 0.696 0.4875 
## speciesRoach 0.10859 0.10506 1.034 0.3032 
## speciesSmelt -0.21700 0.15076 -1.439 0.1524 
## speciesWhitefish 0.23760 0.10467 2.270 0.0248 * 
## --- 
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
## 
## Approximate significance of smooth terms: 
## edf Ref.df F p-value 
## s(vert.len) 1.000 1.000 1.553 0.214862 
## s(diag.len) 1.000 1.000 0.050 0.823955 
## s(cross.len) 7.700 8.496 11.347 < 2e-16 *** 
## s(height) 3.128 3.999 5.486 0.000404 *** 
## s(width) 3.189 4.151 7.133 2.62e-05 *** 
## --- 
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 
## 
## R-sq.(adj) = 0.996 Deviance explained = 99.6% 
## -REML = -123.27 Scale est. = 0.0075377 n = 158 
par(mfrow=c(2,2)) 
gam.check(gam1) 
−0.2 
−0.1 
0.0 
0.1 
0.2 
theoretical quantiles 
2 
3 
4 
5 
6 
7 
Resids vs. linear pred. 
linear predictor 
Histogram of residuals 
Residuals 
−0.4 
−0.2 
0.0 
0.2 
2 
3 
4 
5 
6 
7 
Response vs. Fitted Values 
Fitted Values 
## 
## Method: REML Optimizer: outer newton 
## full convergence after 10 iterations. 
3 
−0.4 
0.0 
deviance residuals 
−0.4 
0.0 
residuals 
Frequency 
0 
30 60 
2 
4 
6 
Response## Gradient range [-4.641604e-05,9.053582e-05] 
## (score -123.2722 & scale 0.007537743). 
## Hessian positive definite, eigenvalue range [2.131058e-05,73.19175]. 
## Model rank = 52 / 52 
## 
## Basis dimension (k) checking results. Low p-value (k-index<1) may 
## indicate that k is too low, especially if edf is close to k'. 
## 
## k' edf k-index p-value 
## s(vert.len) 9.00 1.00 0.92 0.14 
## s(diag.len) 9.00 1.00 0.97 0.34 
## s(cross.len) 9.00 7.70 1.03 0.62 
## s(height) 9.00 3.13 1.03 0.60 
## s(width) 9.00 3.19 0.92 0.12 
gam2<-gam(log(weight)~species+ s(cross.len) + s(height) + s(width), 
data=fish,method="REML") 
gam3<-gam(log(weight)~species+ cross.len + height + width, 
data=fish, method="REML") 
modname<-c("GAM1", "GAM2", "GAM3") 
mod.compare<-data.frame(modname, 
c(AIC(gam1), AIC(gam2),AIC(gam3)), 
c(BIC(gam1), BIC(gam2), BIC(gam3))) 
names(mod.compare)<-c("Model", "AIC", "BIC") 
library(pander) 
pander(mod.compare,round=3, align='c') 
Model AIC BIC 
GAM1 -296.7 -216.9 
GAM2 -298 -223.9 
GAM3 -24.07 9.615 
a. [2 marks] Based on the summary output for the model in fit1 which species had the lowest expected 
weight? Explain your answer brieflfly. 
b. [2 marks] Does it make practical sense to interpret the intercept of the model in fit1? Explain your 
answer brieflfly. 
c. [2 marks] Using the fit1 model equation, predict weight for a fifish with the following characteristics: 
species=Bream, vert.len=7.5, diag.len= 28.4, cross.len=29.0, height=7.8, and width=4.2. 
d. [2 marks] BIC values for models fit1 and fit2 are calculated as shown above. If the analyst said 
that based on these results, the preferred model is the one with the lower BIC value, would you agree? 
Explain your answer brieflfly. 
e. [3 marks] Give a mathematical interpretation of the effffect of speciesPike on weight for the model 
in fit2. 
f. [3 marks] A GAM is fifitted as shown in gam1. Comment on the non-linearity and signifificance of smooth 
terms. 
g. [3 marks] Is there evidence that more basis functions are required for any of the smooth terms? 
Explain your answer brieflfly. 
h. [3 marks] AIC and BIC values are calculated for three models and presented in a table as shown 
above. State the preferred model according to each criterion. Given that the aim of this modeling 
exercise is to make accurate predictions of weight, which of the three models would you choose as your 
preferred model? Explain your answer brieflfly. 
42. Write TRUE or FALSE about the following statements. Where you select FALSE, explain why you 
think the statement is not true. 
a. [2 marks] Log transformations of the response variable are mainly used to deal with violation of the 
assumption of normality. 
b. [2 marks] Preservation of hierarchy is required for polynomial and log transformations. 
c. [2 marks] The following model equation is an example of a local basis function: 
Y = β0 + β1(1/X1) + β2X2 + β3X22 + . 
d. [2 marks] In a linear model, checking of model assumptions can be carried out using the response 
variable Y instead of the residuals. 
e. [2 marks] The following model is an example of a non-linear model: 
Y = β0 + β1(1/X1) + β2 log(X2) + β3X3 + . 
Test total: 30 marks 
5