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日期:2020-02-01 11:06

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, ??

2

??????/??2

?????? < 3 , the assumption is probably satisfied.

# add your ANOVA object's name below (from Q1d)

plot(<name of anova object here>, 1)

plot(<name of anova object here>, 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., 1990?? ??? , 2000?? ??? , and

2010?? ???

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 ?????? and explain in words what the variables in ???? are, you need not write ??1????1 +

??2????2 + ….

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 ?????? and explain in words what the variables in ???? are, you need not write ??1????1+??2????2+….

Question 3b

Is the likelihood ratio test performed above comparing nested models? If so what constraints are on the

vector of regression coefficients ?? 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


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