Homework 3
2024-03-06
# List of packages
packages <- c("tidyverse", "infer", "fst") # add any you need here
# Install packages if they aren't installed already
new_packages <- packages[!(packages %in% installed.packages()[,"Package"])]
if(length(new_packages)) install.packages(new_packages)
# Load the packages
lapply(packages, library, character.only = TRUE)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.4
## ✔ forcats 1.0.0 ✔ stringr 1.5.1
## ✔ ggplot2 3.4.4 ✔ tibble 3.2.1
## ✔ lubridate 1.9.3 ✔ tidyr 1.3.0
## ✔ purrr 1.0.2
## ── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors## [[1]]
## [1] "lubridate" "forcats" "stringr" "dplyr" "purrr" "readr"
## [7] "tidyr" "tibble" "ggplot2" "tidyverse" "stats" "graphics"
## [13] "grDevices" "utils" "datasets" "methods" "base"
##
## [[2]]
## [1] "infer" "lubridate" "forcats" "stringr" "dplyr" "purrr"
## [7] "readr" "tidyr" "tibble" "ggplot2" "tidyverse" "stats"
## [13] "graphics" "grDevices" "utils" "datasets" "methods" "base"
##
## [[3]]
## [1] "fst" "infer" "lubridate" "forcats" "stringr" "dplyr"
## [7] "purrr" "readr" "tidyr" "tibble" "ggplot2" "tidyverse"
## [13] "stats" "graphics" "grDevices" "utils" "datasets" "methods"
## [19] "base"ess <- read_fst("All-ESS-Data.fst")Homework 3 (5%): due by next lecture on March 12th
Instructions: Start a new R markdown for the homework and call it “Yourlastname_Firstname_Homework_2”.
Copy everything below from Task 1 to Task 5. Keep the task prompt and questions, and provide your code and answer underneath.
Remember: you need all the steps for your code to work, including loading your data – otherwise it will not knit.
To generate a new code box, click on the +C sign above. Underneath your code, provide your answer to the task question.
When you are done, click on “Knit” above, then “Knit to Html”. Wait for everything to compile. If you get an error like “Execution halted”, it means there are issues with your code you must fix. When all issues are fixed, it will prompt a new window. Then click on “Publish” in the top right, and then Rpubs (the first option) and follow the instructions to create your Rpubs account and get your Rpubs link for your document (i.e., html link as I provide for the tutorial).
Note: Make sure to provide both your markdown file and R pubs link. If you do not submit both, you will be penalized 2 pts. out of the 5 pts. total.
Task 1
Based on the recoding of lrscale for “left” and “right” and omitting “moderates” (see Tutorial 4), and educ.ba from this tutorial, do the long form coding for Chi-Square. Using the steps outlined in the tutorial, first generate the tables of expected proportions and frequencies. Determine and interpret the critical value for independence. Finally, determine and interpret both the Pearson’s chi-squared statistic and the p-value. Discuss main takeaways.
france_data <- ess %>%
# Filter for France
filter(cntry == "FR") %>%
# Recoding clsprty and cleaning trstplt (will use later)
mutate(
clsprty = case_when(
clsprty == 1 ~ "Yes", # Recode 1 to "Yes"
clsprty == 2 ~ "No", # Recode 2 to "No"
clsprty %in% c(7, 8, 9) ~ NA_character_, # Handle other specific cases where you want to set it as NA
TRUE ~ as.character(clsprty) # Keep other values as-is but ensure they are characters
),
trstplt = ifelse(trstplt %in% c(77, 88, 99), NA, trstplt),
# Recoding gender
gndr = case_when(
gndr == 1 ~ "Male",
gndr == 2 ~ "Female",
gndr == 9 ~ NA_character_,
TRUE ~ as.character(gndr)
),
# Recoding education with case_when for consistency
educ.ba = case_when(
essround < 5 & edulvla == 5 ~ "BA or more",
essround >= 5 & edulvlb > 600 ~ "BA or more",
TRUE ~ "No BA"
)
) %>%
# Ensure all NAs are handled uniformly
mutate(
edulvla = ifelse(edulvla %in% c(77, 88), NA_integer_, edulvla),
edulvlb = ifelse(edulvlb %in% c(5555, 7777, 8888), NA_integer_, edulvlb),
educ.ba = as.character(educ.ba) # Force educ.ba to be treated as character
)# using unique to identify
unique(france_data$gndr)## [1] "Female" "Male"# does not return NA, but the following does
unique(france_data$clsprty)## [1] "Yes" "No" NA# here's how to make sure it's really cleaned
france_data <- france_data %>% filter(!is.na(clsprty))
france_data <- france_data %>% filter(!is.na(trstplt))
# double-checking + quick distribution
unique(france_data$clsprty)## [1] "Yes" "No"table(france_data$clsprty)##
## No Yes
## 9098 9480## Not necessary, but always good to rename when doing operations so you can backtrack to clean dataset
tab_dat <- france_data
# Create a crosstable and save it into the object mytab
mytab <- table(tab_dat$clsprty, tab_dat$educ.ba)
# Calculate row-wise sums of counts in table
rsums <- rowSums(mytab)
# Calculate column-wise sums of counts in table
csums <- colSums(mytab)
# Get the total count of the table (N)
N <- sum(mytab)
# Generate table of expected proportions
ptab <- tcrossprod(rsums/N, csums/N)
cat("Table of Expected Proportions:\n")## Table of Expected Proportions:print(ptab)## [,1] [,2]
## [1,] 0.1099746 0.3797444
## [2,] 0.1145921 0.3956889# Table of expected frequencies
ftab <- N * ptab
cat("Table of Expected Frequencies:\n")## Table of Expected Frequencies:print(ftab)## [,1] [,2]
## [1,] 2043.108 7054.892
## [2,] 2128.892 7351.108# Critical Value for Independence:
alpha <- 0.05 # significance level
c_val <- qchisq(alpha, df = 1, lower.tail = FALSE)
cat("Critical Value:", round(c_val, 3), "\n")## Critical Value: 3.841# Pearson's chi-squared statistic:
test_stat <- sum((mytab - ftab)^2 / ftab)
cat("Pearson's X^2:", round(test_stat, 4), "\n")## Pearson's X^2: 160.4057# p-value:
p_val <- pchisq(test_stat, df = 1, lower.tail = F)
cat("p-value:", round(p_val, 4), "\n")## p-value: 0# Using chisq.test for validation:
cat("Chi-squared test result:\n")## Chi-squared test result:print(chisq.test(mytab, correct = F))##
## Pearson's Chi-squared test
##
## data: mytab
## X-squared = 160.41, df = 1, p-value < 2.2e-16The small p value tells us that we reject the null hypothesis and that the two variables are independent.
Task 2
Do Steps 1 to 3 using the infer package (as we did in the tutorial), for the variables sbbsntx (recoded to the corresponding 5 categories) as the response and domicil (recoded to urban and rural, setting 2 to urban instead of peri-urban) as the explanatory variable for the country of France.
See variable info here: https://ess.sikt.no/en/datafile/ffc43f48-e15a-4a1c-8813-47eda377c355/92?tab=1&elements=[%22874a86e2-a0f6-40b4-aef7-51c8eed98a7d/1%22])
Provide interpretations of the output (consider the variable info).
test_stat <- france_data %>%
specify(explanatory = educ.ba, # change variable name for explanatory variable
response = sbbsntx) %>% # change variable name for outcome of interest
hypothesize(null = "independence") %>%
calculate(stat = "t") # replace in between quotation marks appropriate test statistic## Warning: Removed 14528 rows containing missing values.## Warning: The statistic is based on a difference or ratio; by default, for
## difference-based statistics, the explanatory variable is subtracted in the
## order "BA or more" - "No BA", or divided in the order "BA or more" / "No BA"
## for ratio-based statistics. To specify this order yourself, supply `order =
## c("BA or more", "No BA")` to the calculate() function.print(test_stat$stat)## t
## 4.450111null_distribution <- france_data %>%
specify(explanatory = educ.ba,
response = sbbsntx) %>%
hypothesize(null = "independence") %>%
generate(reps = 1000, type = "permute") %>% # only line we are adding, use reps = 1000 as standard but change if too computationally demanding
calculate(stat = "t")## Warning: Removed 14528 rows containing missing values.## Warning: The statistic is based on a difference or ratio; by default, for
## difference-based statistics, the explanatory variable is subtracted in the
## order "BA or more" - "No BA", or divided in the order "BA or more" / "No BA"
## for ratio-based statistics. To specify this order yourself, supply `order =
## c("BA or more", "No BA")` to the calculate() function.p_val <- null_distribution %>% # replace name here if you assigned something other than null_distribution above
get_pvalue(obs_stat = test_stat, direction = "two-sided") # would only replace test_stat if assigned another name in Step 1## Warning: Please be cautious in reporting a p-value of 0. This result is an approximation
## based on the number of `reps` chosen in the `generate()` step.
## ℹ See `get_p_value()` (`?infer::get_p_value()`) for more information.p_val## # A tibble: 1 × 1
## p_value
## <dbl>
## 1 0Task 3
Now do Step 4 (i.e., the null distribution visualization with confidence intervals) for the same variables and country as Task 2, and interpret the output.
null_distribution %>%
visualize() +
shade_p_value(obs_stat = test_stat, direction = "two-sided")
null_distribution## Response: sbbsntx (numeric)
## Explanatory: educ.ba (factor)
## Null Hypothesis: independence
## # A tibble: 1,000 × 2
## replicate stat
## <int> <dbl>
## 1 1 0.541
## 2 2 0.0886
## 3 3 0.861
## 4 4 1.12
## 5 5 -1.38
## 6 6 0.187
## 7 7 -0.331
## 8 8 0.0345
## 9 9 0.727
## 10 10 -0.393
## # ℹ 990 more rowsThe hypothesis is not disagreed with because
Task 4
Conduct Steps 1 to 3 using the infer package (as we did in the tutorial), for the variables ccrdprs (leaving it on the 0-10 numeric scale) as the response (or outcome) variable and lrscale (recoded as left and right, omitting “moderates” as we did in Task 1) as the explanatory variable.
Variable info here: https://ess.sikt.no/en/datafile/ffc43f48-e15a-4a1c-8813-47eda377c355/92?tab=1&elements=[%2283b08cf2-508e-49a0-9fc8-c3cc281290ec/4%22]
Provide interpretations of the output (consider the variable info).
test_stat <- france_data %>%
specify(explanatory = educ.ba, # change variable name for explanatory variable
response = ccrdprs) %>% # change variable name for outcome of interest
hypothesize(null = "independence") %>%
calculate(stat = "t") # replace in between quotation marks appropriate test statistic## Warning: Removed 14645 rows containing missing values.## Warning: The statistic is based on a difference or ratio; by default, for
## difference-based statistics, the explanatory variable is subtracted in the
## order "BA or more" - "No BA", or divided in the order "BA or more" / "No BA"
## for ratio-based statistics. To specify this order yourself, supply `order =
## c("BA or more", "No BA")` to the calculate() function.print(test_stat$stat)## t
## -1.545298null_distribution <- france_data %>%
specify(explanatory = educ.ba,
response = ccrdprs) %>%
hypothesize(null = "independence") %>%
generate(reps = 1000, type = "permute") %>% # only line we are adding, use reps = 1000 as standard but change if too computationally demanding
calculate(stat = "t")## Warning: Removed 14645 rows containing missing values.## Warning: The statistic is based on a difference or ratio; by default, for
## difference-based statistics, the explanatory variable is subtracted in the
## order "BA or more" - "No BA", or divided in the order "BA or more" / "No BA"
## for ratio-based statistics. To specify this order yourself, supply `order =
## c("BA or more", "No BA")` to the calculate() function.p_val <- null_distribution %>% # replace name here if you assigned something other than null_distribution above
get_pvalue(obs_stat = test_stat, direction = "two-sided") # would only replace test_stat if assigned another name in Step 1
p_val## # A tibble: 1 × 1
## p_value
## <dbl>
## 1 0.164A p value of o.1 gives strong evidence to reject the null hypothesis.
Task 5
Now do Step 4 (i.e., the null distribution visualization with confidence intervals) for the same variables and country as Task 4, and interpret the output.
null_distribution %>%
visualize() +
shade_p_value(obs_stat = test_stat, direction = "two-sided")
null_distribution## Response: ccrdprs (numeric)
## Explanatory: educ.ba (factor)
## Null Hypothesis: independence
## # A tibble: 1,000 × 2
## replicate stat
## <int> <dbl>
## 1 1 -2.20
## 2 2 0.557
## 3 3 -1.00
## 4 4 -1.02
## 5 5 -1.04
## 6 6 1.21
## 7 7 -2.46
## 8 8 -1.47
## 9 9 -0.749
## 10 10 -2.35
## # ℹ 990 more rowsThe null hypothesis is rejected because it is outside the peramiters as shown through the shading. It spills over into the deviation. ## End.
I don’t know what’s going on in all honesty. This is the hardest thing I’ve done to this date bruh. Only God can help us :(