首页 > > 详细

代做STA303留学生作业、R程序设计作业调试、data课程作业代写、R语言作业代做 代做SPSS|代写R语言程序

STA303 - Assignment 1
Winter 2020
Due 2019-01-31
This assignment is worth 5% of your final grade. It is also intended as preparation for Test 1 (worth 20%)
and your final exam, so making a good effort here can help you get up to 33% of your final grade. You will
get your feedback on Assignment 1 before Test 1.
You should be able to do Question 1 by the end of week 1, Question 2 by the end of week 2, and Question
3 by the end of week 3.
• Question 1 uses data about the TV ratings for crime shows, (crime_show_ratings.RDS). You will
need to download this from the Assignment 1 Quercus page.
• Question 2 uses smoking data, (smoking.RData) and instructions for obtaining the data are at the
beginning of the question.
• Question 3 uses Fiji birth data, (fiji.RData) and instructions for obtaining the data are at the beginning
of the question.
Note: You can use whatever packages are useful to you, i.e., tidyverse is not required if you prefer base R
or something else. Just make sure you show which packages you are loading in the libraries chunk. Some
example code in this assignment is shown with tidyverse functions.
Libraries used:
library(tidyverse)
1
Question 1: ANOVA as a linear model
A random sample of 55 crime shows was taken from each decade (1990s, 2000s, 2010s). The following
variables are provided in crime_show_ratings.RDS:
Variable Description
season_number Season of show
title Name of show
season_rating Average rating of episodes in the given season
decade Decade this season is from (1990s, 2000s, 2010s)
genres Genres this shows is part of
Question of interest: We want to know if the average season rating for crime shows is the same
decade to decade.
Question 1a
Write the equation for a linear model that would help us answer our question of interest AND state the
assumptions for the ANOVA.
Question 1b
Write the hypotheses for an ANOVA for the question of interest in words. Make it specific to this context
and question.
Question 1c
Make two plots, side-by-side boxplots and facetted historgrams, of the season ratings for each decade. Briefly
comment on which you prefer in this case and one way you might improve this plot (you don’t have to make
that improvement, just briefly decribe it). Based on these plots, do you think there will be a significant
difference between any of the means?
# load crimeshow data
# (have the .RDS downloaded to the same location your assignment .Rmd is saved)
crime_show_data <- readRDS("crime_show_ratings.RDS")
# Side by side box plots
crime_show_data %>%
ggplot(aes(x = decade, y = season_rating)) +
geom_boxplot() +
ggtitle("Boxplots of average rating by decade for crime TV shows")
# Facetted histograms
crime_show_data %>%
ggplot(aes(x = season_rating)) +
geom_histogram(bins=20) +
facet_wrap(~decade) +
ggtitle("Histograms of average rating by decade for crime TV shows")
2
(a) box (b) hist
Figure 1: question 1c
Question 1d
Conduct a one-way ANOVA to answer the question of interest above. Show the results of summary() on
your ANOVA and briefly interpret the results in context (i.e., with respect to our question of interest).
Question 1e
Update the code below to create two plots and the standard deviation of season rating by decade. Briefly
comment on what each plot/output tells you about the assumptions for conducting an ANOVA with this
data.
Note: there are specific tests for equality of variances, but for the purposes of this course we will just
consider a rule of thumb from Dean and Voss (Design and Analysis of Experiments, 1999, page 112): if the
ratio of the largest within-in group variance estimate to the smallest within-group variance estimate does
not exceed 3, 𝑠
2
𝑚𝑎𝑥/𝑠2
𝑚𝑖𝑛 < 3 , the assumption is probably satisfied.
# add your ANOVA object's name below (from Q1d)
plot(, 1)
plot(, 2)
# Note: this is the tidyverse way you can use a different method if you wish,
# but you're not required to write any code here
crime_show_data %>%
group_by(decade) %>%
summarise(var_rating = sd(season_rating)^2)
Question 1f
Conduct a linear model based on the question of interest. Show the result of running summary() on your
linear model. Interpret the coefficients from this linear model in terms of the mean season ratings for each
decade. From these coefficients, calculate the observed group means for each decade, i.e., 1990𝑠 ̂𝜇 , 2000𝑠 ̂𝜇 , and
2010𝑠 ̂𝜇
3
Question 2: Generalised linear models - Binary
Data from the 2014 American National Youth Tobacco Survey is available on http://pbrown.ca/teaching/
303/data, where there is an R version of the 2014 dataset smoke.RData, a pdf documentation file
2014-Codebook.pdf, and the code used to create the R version of the data smokingData.R.
You can obtain the data with:
smokeFile = 'smokeDownload.RData'
if(!file.exists(smokeFile)){
download.file(
'http://pbrown.ca/teaching/303/data/smoke.RData',
smokeFile)
}
(load(smokeFile))
## [1] "smoke" "smokeFormats"
The smoke object is a data.frame containing the data, the smokeFormats gives some explanation of the variables.
The colName and label columns of smokeFormats contain variable names in smoke and descriptions
respectively.
smokeFormats[
smokeFormats[,'colName'] == 'chewing_tobacco_snuff_or',
c('colName','label')]
## colName
## 151 chewing_tobacco_snuff_or
## label
## 151 RECODE: Used chewing tobacco, snuff, or dip on 1 or more days in the past 30 days
Consider the following model and set of results
# get rid of 9, 10 year olds and missing age and race
smokeSub = smoke[which(smoke$Age > 10 & !is.na(smoke$Race)), ]
smokeSub$ageC = smokeSub$Age - 16
smokeModel = glm(chewing_tobacco_snuff_or ~ ageC + RuralUrban + Race + Sex,
data=smokeSub, family=binomial(link='logit'))
knitr::kable(summary(smokeModel)$coef, digits=3)
Estimate Std. Error z value Pr(>|z|)
(Intercept) -2.700 0.082 -32.843 0.000
ageC 0.341 0.021 16.357 0.000
RuralUrbanRural 0.959 0.088 10.934 0.000
Raceblack -1.557 0.172 -9.068 0.000
Racehispanic -0.728 0.104 -6.981 0.000
Raceasian -1.545 0.342 -4.515 0.000
Racenative 0.112 0.278 0.404 0.687
Racepacific 1.016 0.361 2.814 0.005
SexF -1.797 0.109 -16.485 0.000
4
logOddsMat = cbind(est=smokeModel$coef, confint(smokeModel, level=0.99))
oddsMat = exp(logOddsMat)
oddsMat[1,] = oddsMat[1,] / (1+oddsMat[1,])
rownames(oddsMat)[1] = 'Baseline prob'
knitr::kable(oddsMat, digits=3)
est 0.5 % 99.5 %
Baseline prob 0.063 0.051 0.076
ageC 1.407 1.334 1.485
RuralUrbanRural 2.610 2.088 3.283
Raceblack 0.211 0.132 0.320
Racehispanic 0.483 0.367 0.628
Raceasian 0.213 0.077 0.466
Racenative 1.119 0.509 2.163
Racepacific 2.761 0.985 6.525
SexF 0.166 0.124 0.218
Question 2a
Write down and explain the statistical model which smokeModel corresponds to, defining all your variables.
It is sufficient to write 𝑋𝑖𝛽 and explain in words what the variables in 𝑋𝑖 are, you need not write 𝛽1𝑋𝑖1 +
𝛽2𝑋𝑖2 + ….
Question 2b
Write a sentence or two interpreting the row “baseline prob” in the table above. Be specific about which
subset of individuals this row is referring to.
Question 2c
If American TV is to believed, chewing tobacco is popular among cowboys, and cowboys are white, male and
live in rural areas. In the early 1980s, when Dr. Brown was a child, the only Asian woman ever on North
American TV was Yoko Ono, and Yoko Ono lived in a city and was never seen chewing tobacco. Consider the
following code, and recall that a 99% confidence interval is roughly plus or minus three standard deviations.
newData = data.frame(Sex = rep(c('M','F'), c(3,2)),
Race = c('white','white','hispanic','black','asian'),
ageC = 0, RuralUrban = rep(c('Rural','Urban'), c(1,4)))
smokePred = as.data.frame(predict(smokeModel, newData, se.fit=TRUE, type='link'))[,1:2]
smokePred$lower = smokePred$fit - 3*smokePred$se.fit
smokePred$upper = smokePred$fit + 3*smokePred$se.fit
smokePred
## fit se.fit lower upper
## 1 -1.740164 0.05471340 -1.904304 -1.576024
## 2 -2.699657 0.08219855 -2.946253 -2.453062
## 3 -3.427371 0.10692198 -3.748137 -3.106605
## 4 -6.053341 0.19800963 -6.647370 -5.459312
## 5 -6.041103 0.35209311 -7.097383 -4.984824
5
expSmokePred = exp(smokePred[,c('fit','lower','upper')])
knitr::kable(cbind(newData[,-3],1000*expSmokePred/(1+expSmokePred)), digits=1)
Sex Race RuralUrban fit lower upper
M white Rural 149.3 129.6 171.4
M white Urban 63.0 49.9 79.2
M hispanic Urban 31.5 23.0 42.8
F black Urban 2.3 1.3 4.2
F asian Urban 2.4 0.8 6.8
Write a short paragraph addressing the hypothesis that rural white males are the group most likely to use
chewing tobacco, and there is reasonable certainty that less than half of one percent of ethnic-minority urban
women and girls chew tobacco.
6
Question 3: Generalised linear models - Poisson
Data from the Fiji Fertility Survey of 1974 can be obtained as follows.
fijiFile = 'fijiDownload.RData'
if(!file.exists(fijiFile)){
download.file(
'http://pbrown.ca/teaching/303/data/fiji.RData',
fijiFile)
}
(load(fijiFile))
## [1] "fiji" "fijiFull"
The monthsSinceM variable is the number of months since a woman was first married. We’ll make the overly
simplistic assumption that a woman’s fertility rate is zero before marriage and constant thereafter until
menopause. Only pre-menopausal women were included in the survey sample. The residence variable has
three levels, with ‘suva’ being women living in the capital city of Suva. Consider the following code.
# get rid of newly married women and those with missing literacy status
fijiSub = fiji[fiji$monthsSinceM > 0 & !is.na(fiji$literacy),]
fijiSub$logYears = log(fijiSub$monthsSinceM/12)
fijiSub$ageMarried = relevel(fijiSub$ageMarried, '15to18')
fijiSub$urban = relevel(fijiSub$residence, 'rural')
fijiRes = glm(
children ~ offset(logYears) + ageMarried + ethnicity + literacy + urban,
family=poisson(link=log), data=fijiSub)
logRateMat = cbind(est=fijiRes$coef, confint(fijiRes, level=0.99))
knitr::kable(cbind(
summary(fijiRes)$coef,
exp(logRateMat)),
digits=3)
Estimate Std. Error z value Pr(>|z|) est 0.5 % 99.5 %
(Intercept) -1.181 0.017 -69.196 0.000 0.307 0.294 0.321
ageMarried0to15 -0.119 0.021 -5.740 0.000 0.888 0.841 0.936
ageMarried18to20 0.036 0.021 1.754 0.079 1.037 0.983 1.093
ageMarried20to22 0.018 0.024 0.747 0.455 1.018 0.956 1.084
ageMarried22to25 0.006 0.030 0.193 0.847 1.006 0.930 1.086
ageMarried25to30 0.056 0.048 1.159 0.246 1.057 0.932 1.195
ageMarried30toInf 0.138 0.098 1.405 0.160 1.147 0.882 1.462
ethnicityindian 0.012 0.019 0.624 0.533 1.012 0.964 1.061
ethnicityeuropean -0.193 0.170 -1.133 0.257 0.824 0.514 1.242
ethnicitypartEuropean -0.014 0.069 -0.206 0.837 0.986 0.822 1.171
ethnicitypacificIslander 0.104 0.055 1.884 0.060 1.110 0.959 1.276
ethnicityroutman -0.033 0.132 -0.248 0.804 0.968 0.675 1.336
ethnicitychinese -0.380 0.121 -3.138 0.002 0.684 0.492 0.920
ethnicityother 0.668 0.268 2.494 0.013 1.950 0.895 3.622
literacyno -0.017 0.019 -0.857 0.391 0.984 0.936 1.034
urbansuva -0.159 0.022 -7.234 0.000 0.853 0.806 0.902
urbanotherUrban -0.068 0.019 -3.513 0.000 0.934 0.888 0.982
7
fijiSub$marriedEarly = fijiSub$ageMarried == '0to15'
fijiRes2 = glm(
children ~ offset(logYears) + marriedEarly + ethnicity + urban,
family=poisson(link=log), data=fijiSub)
logRateMat2 = cbind(est=fijiRes2$coef, confint(fijiRes2, level=0.99))
knitr::kable(cbind(
summary(fijiRes2)$coef,
exp(logRateMat2)),
digits=3)
Estimate Std. Error z value Pr(>|z|) est 0.5 % 99.5 %
(Intercept) -1.163 0.012 -93.674 0.000 0.313 0.303 0.323
marriedEarlyTRUE -0.136 0.019 -7.189 0.000 0.873 0.832 0.916
ethnicityindian -0.002 0.016 -0.154 0.877 0.998 0.958 1.039
ethnicityeuropean -0.175 0.170 -1.034 0.301 0.839 0.524 1.262
ethnicitypartEuropean -0.014 0.068 -0.202 0.840 0.986 0.823 1.171
ethnicitypacificIslander 0.102 0.055 1.842 0.065 1.107 0.957 1.273
ethnicityroutman -0.038 0.132 -0.285 0.775 0.963 0.672 1.330
ethnicitychinese -0.379 0.121 -3.130 0.002 0.684 0.493 0.921
ethnicityother 0.681 0.268 2.545 0.011 1.976 0.907 3.667
urbansuva -0.157 0.022 -7.162 0.000 0.855 0.808 0.904
urbanotherUrban -0.066 0.019 -3.414 0.001 0.936 0.891 0.984
lmtest::lrtest(fijiRes2, fijiRes)
## Likelihood ratio test
##
## Model 1: children ~ offset(logYears) + marriedEarly + ethnicity + urban
## Model 2: children ~ offset(logYears) + ageMarried + ethnicity + literacy +
## urban
## #Df LogLik Df Chisq Pr(>Chisq)
## 1 11 -9604.3
## 2 17 -9601.1 6 6.3669 0.3834
Question 3a
Write down and explain the statistical model which fijiRes corresponds to, defining all your variables. It is
sufficient to write 𝑋𝑖𝛽 and explain in words what the variables in 𝑋𝑖 are, you need not write 𝛽1𝑋𝑖1+𝛽2𝑋𝑖2+….
Question 3b
Is the likelihood ratio test performed above comparing nested models? If so what constraints are on the
vector of regression coefficients 𝛽 in the restricted model?
Question 3c
It is hypothesized that improving girls’ education and delaying marriage will result in women choosing
to have fewer children and increase the age gaps between their children. An alternate hypothesis is that
contraception was not widely available in Fiji in 1974 and as a result there was no way for married women
8
to influence their birth intervals. Supporters of each hypothesis are in agreement that fertility appears to be
lower for women married before age 15, likely because these women would not have been fertile in the early
years of their marriage.
Write a paragraph discussing the results above in the context of these two hypotheses.
9

联系我们
  • QQ:99515681
  • 邮箱:99515681@qq.com
  • 工作时间:8:00-23:00
  • 微信:codinghelp
热点文章
程序代写更多图片

联系我们 - QQ: 99515681 微信:codinghelp
© 2014 www.7daixie.com
程序代写网!