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STA238 Embedded R chunks
Creation date：2024-04-30 17:01:24
Belonging category：统计学statistics
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Embedded R chunks

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STA238 Embedded R chunks

In this R Lab, you will be guided by your TA on how to adapt bootstrapping and simulation to construct
confidence intervals and perform hypothesis tests on two-sample data when normality assumptions fail. At
this point, it is assumed that you will have decent familiarity in creating histograms using ggplot to assess
distribution shape. You are expected to have access to R + R Studio + LaTeX in some capacity by this
tutorial, whether it be locally installed or accessed through JupyterHub.
Note: You are expected to join your tutorial meeting with your U of T licensed Office 365
account. Your TAs will not be letting guests join the meeting!
R Lab Learning Goals
• Review: Using filter in dplyr (a package already contained in tidyverse) to filter data in dataframe
according to some criteria.
• Review: Using histograms to gauge normality, and a visual check on equal-variance assumption.
• Use empirical bootstrap to generate confidence intervals for the difference in population means (for
simplicity, you will use the percentile method instead of the studentized method).
• Use permutation test to test for equal population means under assumptions of equal variance.
Use the tasks in this R lab to complete the exercises at the end of this document.
Some additional resources you may find helpful as you learn is the R Markdown Cheatsheet and the LaTeX
Cheatsheet. As you work more in LaTeX, the notation will get easier and become second nature!
Submission Instructions
Your submission should be completed in a NEW R Markdown file that is set to ‘knit to pdf’ that contains
only the exercise portion found at the end of this document. Your final submission should include:
• The original questions (copy and paste is fine)
• Embedded R chunks
• Corresponding outputs for the exercises. Note that you must not include a printout of large data sets!
• Required graphs should include informative and clearly communicated axis labels and titles. Adjust
your R chunk options to produce reasonably sized graphs in your output
• Discussion questions should be answered using proper punctuation and grammar, to the best of your
abilities. Treat these responses as a formal write-up, even if it is short!
• You must submit both your .rmd and knitted .pdf file on Quercus for your current tutorial assignment
as it appears on Quercus.
• The grading rubric for the R Lab is available on your R Lab # 4 submission page.
1
-1. Note
The presentation of this tutorial is adapted from section 10.6 of Modern Mathematical Statistics with
Applications. Refer to this if you are interested in the material presented here as well as non-parametric
approaches to inference on paired data and variances!
0. Motivation
The data we will be working with today is from a study reported in A Longitudinal Study of the Development
of Elementary School Children’s Private Speech. In this study, researchers recorded instances where children
were engaged in private speech during a randomly chosen 10 second interval. Private speech refers to speech
spoken aloud that is either addressed to oneself or to no particular listener. Many such intervals were observed
for each child, and the percentage of intervals where private speech occurred for each child was recorded.
One of the outcomes of interest for this study and the focus of our tutorial is the following research question:
Does the amount of private speech differ between boys and girls?
To put this question into a mathematical context, let (X1, ...Xm) and (Y1, ..., Yn) denote the percentage of
private speech observed for boys and girls, respectively. Furthermore, assume that each observation Xi comes
from a population with mean µ1 and Yj from a population with mean µ2. This problem can then be viewed
as that of making inferences about differences in means µ1 − µ2, or determining whether boys and girls have
different engagements in private speech.
1. Exploratory data analysis
To begin exploring this question, let us first load in the study data and the relevant libraries for our analysis.
library(tidyverse)
library(latex2exp)
library(gridExtra)
# Load in the speech data
# Remember to set your working directory and ensure data files are stored in the same location
speech <- read.csv("speech.csv")
Next we will plot the data for boys and girls separately to get an idea of their distribution. For the purposes
of this analysis, the labels boy and girl refer to a child being labeled as male or female in the data set.
###########################################
### Remove the eval=F from the R chunk! ###
###########################################
# Plot the samples for boys and girls using gridExtra
# Density histograms to compare the density histogram of the two samples
plot.boys <- speech %>%
## INPUT ## %>%
ggplot(aes(x = percentage, y =..density..)) + #density histogram
geom_histogram(binwidth = 0.5, center = 0.25, #bin aesthetics
color = "black",
fill = "thistle2") +
labs(title = "Private speech: Boys",
x = "Percentage",
y = "Density") +
xlim(c(0,30)) #common range
2
plot.girls <- speech %>%
## INPUT ## %>%
ggplot(aes(x = percentage, y =..density..)) +
geom_histogram(binwidth = 0.5, center = 0.25,
color = "black",
fill ="thistle2") +
labs(title = "Private speech: Girls",
x = "Percentage",
y = "Density") +
xlim(c(0,30))
grid.arrange(plot.boys, plot.girls)
From these plots it is clear that the data is not normal and with the limited data size, any confidence intervals
or hypothesis tests based on the assumption of normality will be misleading. In such situations, we typically
relax the assumption of normal data and turn to non-parametric approaches to form inferences about the
quantities of interest.
In the following sections, we will demonstrate how confidence intervals can be obtained using a bootstrap
approach and how the so-called permutation test can be used to perform hypothesis testing with minimal
distributional assumptions.
3
2. Bootstrap confidence interval for the difference in means
In this first section we will demonstrate how to construct a confidence interval for the difference in means
µ1 − µ2 using a bootstrap approach. The general approach is very similar to that of the one-sample problem:
• Draw a bootstrap sample (X∗1 , ..., X∗m) with replacement from the first group (X1, ..., Xm)
• Draw a bootstrap sample (Y ∗1 , ..., Y ∗m) with replacement from the second group (Y1, ..., Ym)
• Compute the difference in means X¯∗ − Y¯ ∗ for the bootstrap samples
• Repeat this process many times to obtain the bootstrapped sampling distribution
• Construct a bootstrap confidence interval (C∗lower, C∗upper) from the sampling distribution.
In this presentation, we will be using the percentile method to construct the (1 − α)-level confidence
intervals in the final step, where C∗lower and C∗upper are taken to be the α/2 and 1− α/2 sample quantiles of
the bootstrap sampling distribution, respectively. This method is simple, and makes few assumptions about
the shape of the bootstrap sampling distribution, though the resulting interval is typically biased. To see
another approach similar to the studentized approach from the previous tutorial, see the discussion in section
10.6 of the textbook.
2.1 Implementation in R
Let’s use this recipe to construct a bootstrapped confidence interval in R:
###########################################
### Remove the eval=F from the R chunk! ###
###########################################
# Create vectors containing the observed data for each group
# Define the bootstrap experiment
The 95% bootstrap confidence interval for the difference in means is then
###########################################
### Remove the eval=F from the R chunk! ###
###########################################
# Compute the 95% percentile confidence interval
ci.mean.diff <- quantile(boot.mean.diff, probs = c(0.025, 0.975))
The corresponding confidence interval based on the assumption of normal data can be calculated to be
(0.55, 9.63), which has a smaller lower limit compared to our bootstrapped interval. Given that our data is
definitely not normal, we have more confidence in the validity of the bootstrap-based interval.
4
3. Permutation test for hypothesis testing
Another common approach to studying the difference in means between two groups is the hypothesis testing
approach. Here, we consider the following null and alternative hypotheses:
• H0 : µ1 − µ2 = 0
• H1 : µ1 − µ2 6= 0
Since our data is not normally distributed, we can no longer use the two-sample t−test approach, where the
null distribution of the sample statistic is a known t−distribution.
3.1 The permutation test
If we are comfortable assuming that the only possible difference in the distribution of the two samples is in
their means µ1, µ2, then we can use what is called the permutation test to perform this hypothesis test.
Under the null hypothesis µ1 = µ2, where both sampling distributions are the same, every observation
could equally well belong to either group and so the group labels have no effect on the outcome. Since the
group labels are exchangeable, we can randomly assign them to the observed values and we will obtain what
looks like a new observation from the null distribution with the same probability as the original sample. By
comparing our observed difference X¯ − Y¯ to the difference measured on these new permuted data, we can get
a sense of how unlikely it is to observe such a difference under the null hypothesis.
Following this reasoning, the recipe for the permutation test is as follows:
• Compute the test statistic Tobs = X¯ − Y¯ on the original data (this test statistic is labeled T, but does
not imply that it has a T-distribution).
• Create a permuted data set by randomly assigning the group labels to the observed values and compute
the test statistic T ∗ = X¯∗ − Y¯ ∗. Note: this is not the same as sampling with replacement.
Rather we are relabeling each of the data points as a ’male’ or ’female’.
• Repeat this process for every permutation of the data (or a very large number of permutations)
• Compute p−value of Tobs by computing the proportion of permuted test statistics T ∗ that are at least
as “extreme” as Tobs.
3.2 Implementation in R
We implement the permutation test for our data as follows:
# Compute the observed difference between genders
# Permutation test
The p−value for the our particular value of (FILL THIS IN) is then
# Compute p-value
where we have defined “extreme” to mean |T | ≥ |Tobs|, which is a two-tailed test. Therefore we reject the
null hypothesis that µ1 = µ2 at the 5% level.
Note
The assumption that both distributions are the same in every possible way except perhaps the mean is a very
strong assumption that should typically be checked. In practice we usually require this be approximately
true by checking that the sample variances are roughly equal. If we check this for our the speech data set, we
find that this is unlikely satisfied and that we may want to use another approach for testing this hypothesis.
5
4. Exercises
In the exercises you will be working with a data set from a study of survival times for patients with stomach
and breast cancer, both treated with ascorbate (Vitamin C). The data is a modified version of the found
in the article Supplemental ascorbate in the supportive treatment of cancer: Re-evaluation of prolongation
of survival times in terminal human cancer by E. Cameron and L. Pauling (1978). The outcome we are
interested in studying is the difference in average survival time in days between the two groups.
1.) Load the cancer data from cancer.csv using read.csv
# Load in cancer data
2.) Create a new column survival.time.log in the data.frame equal to the logarithm of the survival time.
(We make this transformation to make the sample variances roughly equal for the permutation test)
# Define logarithm of survival time
3.) a.) Plot and label a histogram of the log survival time for each type of cancer and combine them using
grid.arrange. Calculate the sample standard deviation of log survival time for each cancer type.
# Plot survival times for each cancer type
b.) Comment on the distribution of the samples.
4.) a.) Perform a permutation test and obtain the p-value of the observed difference in survival times based
on the null hypothesis H0 : µ1 = µ2, where µ1 and µ2 are the mean survival times in log-days for stomach and
breast cancer patients, respectively. Include in-line comments that explain the purpose of your lines of code.
b.) Can a conclusion be drawn about any differences in survival times of the two cancer types? Why or why
not?
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MAS316/414 ACCT2522 MSIN0181 ATMS 201 COMP3425 MATH7861 SOLA2060 MA 325 7CCEMCTH ECON 432 INFT2051 SOLA 2060 ECOS2002 Stats141XP BIOA02H3 S2022 MA4601 K5 COMP5329 EG 284 IB3A70 WECO1020 COMP3027 Energy ACS61011 MAST20026 STAT 533 MATA31 ENVS257 BEM2031 EC 486 COMP0172 CS3214 EECE5644 IB2B20 COSC2536 COSC1147 COMP6214 5CCM242A COMP0130 AMME2000 MTMG50: EG238 ACF6006 QBUS2810 COSC2391 ECON7560 MATH377 STA305 CS 486 COMP1117B EL648 ELEC3909 ECON 241 COMP 4901W BUSI4428 ENEN20002 ENGN4524 ELEC3115 CSC 413 PSYC 100B ECON1002 MATH256 ANSC20005 MTHS2008 MATH40082 COSC 2406 FIT3174 STA 238 CMT202 Differential Equations MTHM017 GARCH 101 MATH2069 CSC336S ACSC12 ETF2121 CEME 2001 DB B474F ARHT1001 EECS 3311 3220 STATS 786 CS 484 COMP20008 MATH4091 ECE 455 CSCI 204 BUS B200F BA 875 BU5562 COSC 1147 X11 GEOG 101 IMM250F CSE 584A MATH38172 LAB 4 ELEC5882M FINS5523 EE 430 COMP 465 6CCS3PRE 7CCSMNSE FINANCE ACMB02H3 ENGG2112 ELEC5616 SCM B371 CHEM3120 CS22 ECMT2130 MAST30011 ADSPBF706 CENG0013 ACCOUNTING BUSS1040 CHEM 181 MAT136H1 GY 6183 COMP3170 AGRI90089 PHYSICS 1002 ECON10003 IEOR E4004 ACS61012 MATH 40550 MKTG2113 TELE4653 ENSC1004 ECE 232E MK726 MTRX2700 SHEET 3 CSCA08 COMP30027 Math 2XX3 statistic MMAN3200 BISM7213 ECMM109 ECON 2022 FNCE 20005 GR5067 CSCB09 ME 3017 CS 5100 PHYS 2903 COMS 4771 COMP222 COMP 1046 CSC148H1 ISYS2038 FINS2615 OT5206 COMP2048 A6 MECH5770M MA826/20 CMPT 726 STA457 INFS6071 INFS588 integral ECON 216 IRDR 0017 MA323 PA 15213 PSYC2001 ENV222 Statistics ACCT5930 ACCT12 LAND2121 ME 5405 COMM1180 TELE3113 A7 ECE 568 ACCT2019 INFS1200 ENGG1340 CS 1400 ASST3 FINM7409 SWEN30006 INFO2222 FIN20150 SOSS1000 CSSE7231 FINS5513 COMS3200 CECN 702 ECON4060H MN3515 HW2 PSYA02H3 ECON 2112 COMP90051 CS 1652 MA30059 AF1605 MKTG7501 ECO-6003B COM4521 CTM ECON3350 COMP 3804 ELEC9762 BISM 7255 ST22 MATH20014 MNE3122 COMP343 MATH5165 ELE7029 ENGR30002 SEEM3500 ECON7350 RISK5002 FINM7405 COMP90059 MOS 3320B STAT 431 BISM7221 MATH3014 MTH783P EBA 35302 ACS133 BUS2206 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COMSM0093 COMP281 AFM 333 IBUS1001 CP2501 Math 105A VE320 TEC101 SOCY2166 COM240 EFIM20033 ENGI4467 EFIM20034 EC 979 BE432 7EJ502 CSSE6400 COMP90025 Econ 312 ENG2031 ECO0006 EXERSCI 207 CAN209 ECO 137W FT312 ECO 2199 MATH32052 BDW4213 APS 360 SAS 6 CHME0017 PO92Q Economics 01AV MATH39512 MATH 5486 COMP8620 ISE102 ECON235 ECON 206 FI 345 PHAY0020 MATH0099 MATH0085 EFB106 PHY 134 MAE115 circuits Chemistry CSE12 MATH-UA 233 PHAS0038 COMP2140 CONMGNT 7049 COMP151101 ECON 485 ACX5903 ENGN6536 ARCH1162 Law PHAS0042 CCF1C03 MATH 351 Phys 139 AG942 MGT001371 APH106 ECON1051 MA4AM MA3Z7 NUMBER THEORY REST0006 SESS0026 STAT0023 TABL 2741 CMP4008Y EFIM10015 Acct 560 PSYC735 GEOG 5 ACCT3104 CSEC5616 PHIL1012 ITLS6111 ECON8022 5CCM226A compX123 MA4XJ R001 EWB LLP120 COMP643 ENG 103 ECON 23950 ACCT 207 FINC 616 Bio99 Math 137A Psychology BCPM0054 MATH20212 EC3450 CENV6141 PSY 205 CE335 CS365 Calculus ACSD INFT6800 MCIT INFO6030 PHYS 111LV H63EEJ ECON2151 ENGI 2181 MTRX1701 PLIT08009 CS240 Math 270 CMP7000A CMP-7038B Math 263 LAW3016 CS371 STAT 334 FIN0008 ACC3004 STAT 371 ​COMP1921 MTH201 ACCT2000 UN3412 CIV6000 MATHS 3021 MATE7014 ECON1 CIBC6035 LAW121G COMM 321 CSC8204R TTM2033 CSCI 2600 MAE 115 LNG302 CENV6175W1 ECON30019 CPS2015 BST969 DDES9015 AMATH242 CS 1151 N1592 ACCT600 EG503G SMM519 EC318 Ecstatistics EEE 471 MARK2051 Topology COM00037H ACCT3000 MTHM501 EG501V COM00055H BUSN9085 GGR273 MATH 2B ITD102 MA214 ECON0060 ECE3114 ECMM447 Mechanics CCT361H5S BSAN3210 WELF2001 Fin3001 RESP5001 STA 137 Linguistics 101 CSCI368 CMPSC497 LUBS3655 OMBA 5355 BUSM2562 ​BFF5902 CYBR7001 CULT2005 ECON 7720 MTRX3760 FSE598 ​LUE1001 MGSC 7100 CS3240 COSC7500 ACCT3091 INFOGR TRAN 2080 DESIGN CIVE215001 CIVE2360 CPSC 8430 FNCE2003 Stoichiometry INFS6023 Quantitative Method Math 21D ECE5179 Design ​ECON8022 Chemical DMS 213 ADM1370 BFF5902 BFF2701 AGRI10051 AMN400 ECON7322 ​FINC5090 ​MATH1052 ​SciComp Elec4622 ​FINS5516 ENG 101 MMM143 CS431CA2 BSAN2201 EBSC7070 ES9v9-10 PLIN0009 DATA 8 MATH 101 BCPM0097 ​MAT 237 BISM2201 EEE8099 Physics 11 ECE2238B Biostat 234 ​ECON30009 INF10025 CHEM252 CCMA4008 CIVE 3007 STAT0031 ELEC ENG 3108 MGMT90140 CHEM 252 STAT3016 MMM095 ​MMGT6016 Legal STAT 311 ELEC9764 COMP 370 TCS 3053 EECS 1021 FIT1051 C/C++ EE3C COMP2003J COM398 COMP51915 Simulator FIN42330 8PRO102 PGD ELEC5160 XJCO1921 COMP5123M INFS 2042 CHE1148H ECA5313 IN3329 Microsoft Modeling B15 2TT COMP3687 COMP2432 DMS2030 KAN4TSF CFKG CSE 5322 CSE 445 STATS 3DA3 CHC5223 COMP1048 EL3105 EECS 3221 CS231 ME 4434 ITP4510 ITP4501 CSCI 5708 Controller EECE 5699 CPEN 231L CPN programs CSC8016 MATU9D2 ENGN4213 2. SVM. (10 points) Consider an SVM classifier using RBF kernel, and finish the following python code to search for the optimum kernel parameter γ. You are required to calculate the RBF function and search yourself, instead of using functions in Scikit-Learn (sklearn). def parameter search(gamma list ,Xtr,ytr,Xts,yts ,...): """ Parameters gamma list: A list of RBF kernel parameter gamma to search for Xtr: Training data, shape (ntr,nrow,ncol) ytr: Training labels, shape (ntr,) Xts: Test data, shape (nts,nrow,ncol) yts: Test labels, shape (nts,) ... Add other parameters as needed Returns: gamma optimum: The gamma with minimum error on Xts, yts """ ... return gamma optimum
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