HW3
natalia zagoriti
2024-03-05
getwd()## [1] "/Users/nataliazagoriti/Desktop/HW3"Setting up environment
# 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 2.1.5
## ✔ 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"
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## [1] "infer" "lubridate" "forcats" "stringr" "dplyr" "purrr"
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## [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")rm(list=ls()); gc() ## used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
## Ncells 1048464 56.0 2135508 114.1 NA 1460992 78.1
## Vcells 1807379 13.8 1046271028 7982.5 16384 1304805798 9954.9ess <- read_fst("All-ESS-Data.fst")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") %>%
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$gndr)## [1] "Female" "Male"unique(france_data$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 9480tab_dat <- france_data
mytab <- table(tab_dat$clsprty, tab_dat$educ.ba)
rsums <- rowSums(mytab)
csums <- colSums(mytab)
N <- sum(mytab)
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 0.3956889ftab <- 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 critical value depends on the significance level chosen for the test and the degrees of freedom.The Pearson’s chi-squared statistic, which is 160.41 in this case, measures the extent of discrepancy between the observed and expected frequencies in the contingency table. A higher chi-squared statistic implies a larger discrepancy, indicating a stronger association between the variables being analyzed. The p-value, represented as “p-value < 2.2e-16” (or 2.2 × 10^(-16)), is extremely small and close to zero.This indicates strong evidence against the null hypothesis of independence between the variables.Therefore, we reject the null hypothesis and infer that there is a significant association between the two categorical variables analyzed in the dataset.
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 = domicil, # change variable name for explanatory variable
response = sbbsntx) %>% # change variable name for outcome of interest
hypothesize(null = "independence") %>%
calculate(stat = "slope") # replace in between quotation marks appropriate test statistic## Warning: Removed 14528 rows containing missing values.## Message: The independence null hypothesis does not inform calculation of the
## observed statistic (a slope) and will be ignored.print(test_stat$stat)## domicil
## -0.0768133null_distribution <- france_data %>%
specify(explanatory = domicil,
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 = "slope")## Warning: Removed 14528 rows containing missing values.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: domicil (numeric)
## Null Hypothesis: independence
## # A tibble: 1,000 × 2
## replicate stat
## <int> <dbl>
## 1 1 0.0158
## 2 2 0.0689
## 3 3 0.0318
## 4 4 0.0132
## 5 5 0.00949
## 6 6 -0.000345
## 7 7 0.00411
## 8 8 0.0435
## 9 9 -0.0261
## 10 10 0.0119
## # ℹ 990 more rowsconf_int <- null_distribution %>%
get_confidence_interval(level = 0.95, type = "percentile")
null_distribution %>%
visualize() +
shade_p_value(obs_stat = test_stat, direction = "two-sided") +
shade_confidence_interval(endpoints = conf_int)
The observed test statistics (the red line) falls within the null distribution,meaning that there is no significant association between the two datasets.
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).
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
)unique(france_data$gndr)## [1] "Female" "Male"unique(france_data$clsprty)## [1] "Yes" "No" NAfrance_data <- france_data %>% filter(!is.na(clsprty))
france_data <- france_data %>% filter(!is.na(trstplt))unique(france_data$clsprty)## [1] "Yes" "No"table(france_data$clsprty)##
## No Yes
## 9098 9480france_clean <- france_data %>%
mutate(
clsprty = ifelse(clsprty == 2, 0, ifelse(clsprty %in% c(7, 8, 9), NA, clsprty)), # If 'clsprty' is 2, set it to 0. If it's 7, 8, or 9, set it to NA.
stfdem = ifelse(stfdem %in% c(77, 88, 99), NA, stfdem), # For 'stfdem', set values 77, 88, and 99 to NA.
trstplt = ifelse(trstplt %in% c(77, 88, 99), NA, trstplt), # For 'trstplt', set values 77, 88, and 99 to NA.
)
france_clean <- france_clean %>% filter(!is.na(clsprty), !is.na(stfdem), !is.na(trstplt))france_clean <- france_clean %>%
mutate(
# Recoding 'lrscale' into a new variable 'ID'
ID = case_when(
lrscale %in% 0:3 ~ "Left", # Values 0 to 3 in 'lrscale' are categorized as "Left"
lrscale %in% 7:10 ~ "Right", # Values 7 to 10 in 'lrscale' are categorized as "Right"
lrscale %in% 4:6 ~ "Moderate", # Values 4 to 6 in 'lrscale' are categorized as "Moderate"
lrscale %in% c(77, 88, 99) ~ NA_character_ # Values 77, 88, 99 in 'lrscale' are set as NA
),
# Recoding 'yrbrn' (year born) into a new variable 'Gen'
Gen = case_when(
yrbrn %in% 1900:1945 ~ "1", # Birth years from 1900 to 1945 categorized as "1"
yrbrn %in% 1946:1964 ~ "2", # Birth years from 1946 to 1964 categorized as "2"
yrbrn %in% 1965:1979 ~ "3", # Birth years from 1965 to 1979 categorized as "3"
yrbrn %in% 1980:1996 ~ "4", # Birth years from 1980 to 1996 categorized as "4"
TRUE ~ as.character(yrbrn) # Other years remain unchanged
),
# Converting 'Gen' to a factor with meaningful labels
Gen = factor(Gen, levels = c("1", "2", "3", "4"),
labels = c("Interwar", "Baby Boomers", "Gen X", "Millennials")) # Assigning labels to the levels in 'Gen'
)test_stat <- france_data %>%
specify(explanatory = lrscale,
response = ccrdprs) %>%
hypothesize(null = "independence") %>%
calculate(stat = "slope") ## Warning: Removed 14645 rows containing missing values.## Message: The independence null hypothesis does not inform calculation of the
## observed statistic (a slope) and will be ignored.print(test_stat$stat)## lrscale
## 0.007179428null_distribution <- france_data %>%
specify(explanatory = lrscale,
response = ccrdprs) %>%
hypothesize(null = "independence") %>%
generate(reps = 1000, type = "permute") %>%
calculate(stat = "slope")## Warning: Removed 14645 rows containing missing values.p_val <- null_distribution %>%
get_pvalue(obs_stat = test_stat, direction = "two-sided")
p_val## # A tibble: 1 × 1
## p_value
## <dbl>
## 1 0.19A p-value of 0.254 shows that there is not enough evidence to reject the null hypothesis, thus the statistic that was observed was not different than what was expected under 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: lrscale (numeric)
## Null Hypothesis: independence
## # A tibble: 1,000 × 2
## replicate stat
## <int> <dbl>
## 1 1 -0.0109
## 2 2 -0.00337
## 3 3 -0.00244
## 4 4 0.00418
## 5 5 -0.0118
## 6 6 0.0000535
## 7 7 -0.00460
## 8 8 0.0000576
## 9 9 -0.00211
## 10 10 0.00337
## # ℹ 990 more rowsconf_int <- null_distribution %>%
get_confidence_interval(level = 0.95, type = "percentile")
null_distribution %>%
visualize() +
shade_p_value(obs_stat = test_stat, direction = "two-sided") +
shade_confidence_interval(endpoints = conf_int)