University of Western Australia SCIE4402 Agricultural Quiz Solution

This document presents the solutions to a quiz from the University of Western Australia's SCIE4402 Agricultural and Resource Economics course. The quiz covers various topics, including linear regression models for pig growth, random intercept and slope models, and tests for functional form. It also explores power calculations using rat weight gain data and the application of partial adjustment and error correction models to time series data, such as UK food expenditure. The solutions include specific answers to the quiz questions, such as model selection, elasticity estimates, and statistical test results. This assignment allows students to test their understanding of the material covered in the computer lab, focusing on applying econometric techniques and interpreting the results.

University of Western Australia
School of Agriculture and Environment
Agricultural and Resource Economics
SCIE4402
DAY FIVE STREAM C
DATA SETS
The data sets can be found in the LMS. All the questions in the quiz relate to concepts covered in
the computer lab.
PURPOSE OF THE QUIZ
This is an opportunity for you to test your understanding of the material covered. The questions
essentially ask you to go back to the script you worked through in the lab and consider similar
models or small changes to the examples given in the script

QUESTION 1
In the computer lab you considered ways to model the growth of 30 pigs through time. The
following questions relate to a different set of 18 pigs. The issues considered are the same as those
considered in the computer lab. The models to consider are of the same form i.e. linear-linear
models where we use time (measured in weeks) to explain pig weight. Reflecting my limited
ability to come up with file names the data is in the pigwt2 data file.
i. For the data set estimate a pooled simple linear regression model that ignores the fact that
we have repeat measures on individual pigs. For this model conduct a reset test to check for
functional form. Does the model fail this functional form test? Answer question in yes or
no. (This is question 1). (Marks 0.5)
Answer =
ii. Using appropriate methods compare the random intercept model, the random slope model,
and the random slope and random intercept model. (These are the same models considered
in the computer lab). Decide which of these three models is the most appropriate model (At
this stage do not consider the issue of heteroskedasticity and autocorrelation (This is
question 2). (Marks 0.5)
Answer = put answer in this box i.e. name of
appropriatemodel
iii. For the model you decide is most appropriate as part of Question 1 part (ii), first test to see
whether you need to consider an AR(1) process for the error terms; and second, consider
whether you need to allow for heteroskedasticty in the model, where, as is the computer lab,
the heteroskedasticty is related to the fitted values. For the model that you decide is most
appropriate, what is the point estimate for the slope. Report the value to THREE decimal
places. (This is question 3). Note you may need to use ctrl <- lmeControl(opt='optim') to reset
the nlme settings so that the model estimates. Check the relevant section of the example
script files for how to implement this (Marks 0.5)
Answer = put the point of estimate for slope in the box(report value to three decimal places)
No
Random slope model
6.489

QUESTION 2 POWER OF THE TEST
The RatsQuiz2.csv file contains measurements on the weight gain (in grams) of rats on two
different diets over a fixed period of time. From a biological point of view, a difference of 3 grams
is significant. Use the sample data and the formula in the script to generate an estimate of the
pooled standard deviation. Use the estimate of the pooled standard deviation and 3 grams as the
difference in the mean to generate an effect size estimate. Using this effect size estimate determine
the power of the test to detect this difference. (This is question 4).
(Report your estimate of test power to four decimal points in box below) (Marks 0.5)
Answer =
QUESTION 3 PARTIAL ADJUSTMENT MODEL (PILOT STUDY)
The UKFood.csv file contains the following variables:
FOOD = expenditure on food in billions
DPI = disposable private income in billions
PFOOD = index of cost of a basket of food
PTPE = index of general private expenditure costs
PRFOOD = index of the relative price of food, found as PFOOD/PTPE x 100, and this is
approximately a real cost index for food
The data are annual and start in 1956.
i. Estimate a partial adjustment model in log format. Decide, for this model specification,
whether: (i) no time trend is required; (ii) a linear time trend is required; or (iii) a quadratic
time trend is required (This is question 5). (Marks 0.5)
Put the right option in box below
Answer =
ii. For the partial adjustment model you select as most appropriate in part (i) (ie no trend,
linear trend, or quadratic trend) what is the long-run price elasticity. (This is question 6).
No time trend
0.8528

(Marks 0.5) put the answer in the box value
Answer =
iii. For the partial adjustment model, you select as most appropriate in part (i) (i.e. no trend,
linear trend, or quadratic trend) what is the SE for the long-run income elasticity. (This is
question 7). (Marks 0.5) put the value in the box below
Answer =
iv. Estimate an ADL(1,1) model in the Error Correction Model (ECM) format in log format.
Decide, for this model specification, whether: (i) no time trend is required; (ii) a linear time
trend is required; or (iii) a quadratic time trend is required (This is question 8). (Marks 0.5)
Put the right option in the box below
Answer =
v. For the ECM model you select as most appropriate in part (iv) (ie no trend, linear trend, or
quadratic trend) what is the long-run income elasticity. (This is question 9). (Marks 0.5)
Put the answer in the box below
Answer =
vi. For the ECM model you select as most appropriate in part (iv) (ie no trend, linear trend, or
quadratic trend) what is the SE for the long-run price elasticity. (This is question 10).
(Marks 0.5)
Put the value in the box below
Answer =
0.009829203
0.006585202
No time trend required
-0.0284629
0.02030319