Tutorial 6 (202)
Favour Ayoola-Ola
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.
rm(list=ls()); gc()## used (Mb) gc trigger (Mb) limit (Mb) max used (Mb)
## Ncells 527323 28.2 1171927 62.6 NA 669420 35.8
## Vcells 971366 7.5 8388608 64.0 16384 1851857 14.2# 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()
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## [1] "lubridate" "forcats" "stringr" "dplyr" "purrr" "readr"
## [7] "tidyr" "tibble" "ggplot2" "tidyverse" "stats" "graphics"
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## [1] "infer" "lubridate" "forcats" "stringr" "dplyr" "purrr"
## [7] "readr" "tidyr" "tibble" "ggplot2" "tidyverse" "stats"
## [13] "graphics" "grDevices" "utils" "datasets" "methods" "base"
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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")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
)## 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,] 0.1101248 0.3815066
## [2,] 0.1138739 0.3944946# Table of expected frequencies
ftab <- N * ptab
cat("Table of Expected Frequencies:\n")## Table of Expected Frequencies:print(ftab)## [,1] [,2]
## [1,] 2059.444 7134.556
## [2,] 2129.556 7377.444# 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: 166.1849# 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 = 166.18, df = 1, p-value < 2.2e-16answer: P-value < 2.2e-16- the Chi-squared test associated p-value. Strong evidence is present against the null hypothesis, as indicated by the extremely small p-value, which is almost zero. Stated differently, the variables within the dataset have a statistically significant correlation with one another.
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).
france_data <- ess %>%
# Filter for France
filter(cntry == "FR") %>%
#Recoding to the corresponding 5 categories
mutate(
sbbsntx = case_when(
sbbsntx == 1 ~ "Agree strongly",
sbbsntx == 2 ~ "Agree",
sbbsntx == 3 ~ "Neither agree nor disagree",
sbbsntx == 4 ~ "Disagree",
sbbsntx == 5 ~ "Disagree strongly",
sbbsntx %in% c(7, 8, 9) ~ NA_character_, # Handle other specific cases where you want to set it as NA
TRUE ~ as.character(sbbsntx) # Keep other values as-is but ensure they are characters
),
# Recoding to urban and rural
domicil = case_when(
domicil == 1 ~ "Urban",
domicil == 2 ~ "Urban",
domicil == 3 ~ "Rural",
domicil == 4 ~ "Rural",
domicil == 5 ~ "Rural",
domicil %in% c(7, 8, 9) ~ NA_character_, # Handle other specific cases where you want to set it as NA
TRUE ~ as.character(domicil) # Keep other values as-is but ensure they are characters
)
)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 = "Chisq") # replace in between quotation marks appropriate test statistic## Warning: Removed 14999 rows containing missing values.## X-squared
## 19.97187null_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 = "Chisq")## Warning: Removed 14999 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 = "greater") # would only replace test_stat if assigned another name in Step 1p_val## # A tibble: 1 × 1
## p_value
## <dbl>
## 1 0.001answer: Frequentist hypothesis testing assumes that a p-value of 0 indicates substantial evidence to reject the null hypothesis.
Task 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.
conf_int <- null_distribution %>%
get_confidence_interval(level = 0.95, type = "percentile")
null_distribution %>%
visualize() +
shade_p_value(obs_stat = test_stat, direction = "greater") +
shade_confidence_interval(endpoints = conf_int)
null_distribution## Response: sbbsntx (factor)
## Explanatory: domicil (factor)
## Null Hypothesis: independence
## # A tibble: 1,000 × 2
## replicate stat
## <int> <dbl>
## 1 1 1.07
## 2 2 6.76
## 3 3 4.24
## 4 4 1.24
## 5 5 1.17
## 6 6 8.01
## 7 7 6.46
## 8 8 7.10
## 9 9 2.02
## 10 10 2.87
## # ℹ 990 more rowsanswer: The image’s distribution is right-skewed, which means that its median, mean, and mode are all smaller than 10. Furthermore, the confidence intervals fall between 0 and about 10, suggesting that this range is most likely where the actual value is. Additionally, because the red line is at 20 and does not fall into the null distribution, it provides evidence against the null hypothesis.
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 'lrscale'
mutate(
lrscale = 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
),
ccrdprs = ifelse(ccrdprs %in% c(66, 77, 88, 99), NA, ccrdprs), # Set values 66, 77, 88, 99 in 'ccrdprs' to NA
)test_stat <- france_data %>%
specify(explanatory = lrscale, # change variable name for explanatory variable
response = ccrdprs) %>% # change variable name for outcome of interest
hypothesize(null = "independence") %>%
calculate(stat = "F") ## Warning: Removed 15413 rows containing missing values.print(test_stat$stat)## [1] 15.41871null_distribution <- france_data %>%
specify(explanatory = lrscale,
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 = "F")## Warning: Removed 15413 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 = "greater") # 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 0answer: According to frequentist hypothesis testing assumptions, there is strong evidence to reject the null hypothesis inside the null distribution with a p-value of 0.
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.
conf_int <- null_distribution %>%
get_confidence_interval(level = 0.95, type = "percentile")
null_distribution %>%
visualize() +
shade_p_value(obs_stat = test_stat, direction = "greater") +
shade_confidence_interval(endpoints = conf_int)## Warning in min(diff(unique_loc)): no non-missing arguments to min; returning
## Inf
null_distribution## Response: ccrdprs (numeric)
## Explanatory: lrscale (factor)
## Null Hypothesis: independence
## # A tibble: 1,000 × 2
## replicate stat
## <int> <dbl>
## 1 1 0.140
## 2 2 0.479
## 3 3 0.608
## 4 4 0.223
## 5 5 2.45
## 6 6 1.38
## 7 7 0.219
## 8 8 0.417
## 9 9 0.316
## 10 10 1.42
## # ℹ 990 more rowsanswer:The graph illustration displays a left-skewed distribution, or a negative skew.indicating that less than five is the value of its mean, mode, and median. ## End.