Drzymala_Gabriela_Homework_3
2024-03-05
# 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
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## ✖ dplyr::filter() masks stats::filter()
## ✖ dplyr::lag() masks stats::lag()
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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"
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## [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 <- read.fst("france_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
)# 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 9480Homework 3 (5%): due by next lecture on March 12th
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_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% c(77, 88, 99) ~ NA_character_ # Values 77, 88, 99 in 'lrscale' are set as NA
),
# 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
)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: 3.841Since the critical value is 3.841, if the Chi-squared test statistic exceeds 3.841, then we reject the null hypothesis at the 0.05 significance level. This would mean that there is a statistically significant association between feeling close to a party and education level in your dataset.
france_data <- france_data %>% filter(!is.na(lrscale))test_stat <- sum((mytab - ftab)^2 / ftab)
cat("Pearson's X^2:", round(test_stat, 4), "\n")## Pearson's X^2: 160.4057p_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 observed X-squared value of 160.4057 is much larger than the critical value of 3.841. Therefore, there is a statistically significant association between ideological self-placement and one’s educational level. This means that the probability of ideological self-placement is not the same across different education levels. A p-value of 0 means that we reject the null hypothesis because the two variables are independent of each other.
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.
france_data <- ess %>%
filter(cntry == "FR") %>%
mutate(
domicil = case_when(
domicil == 1 ~ "Rural",
domicil == 2 ~ "Urban",
domicil %in% c(77, 88, 99) ~ NA_character_,
TRUE ~ as.character(domicil)
),
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_,
TRUE ~ as.character(sbbsntx)
),
)test_stat <- france_data %>%
specify(explanatory = domicil,
response = sbbsntx) %>%
hypothesize(null = "independence") %>%
calculate(stat = "Chisq")## Warning: Removed 14998 rows containing missing values.print(test_stat$stat)## X-squared
## 40.3495null_distribution <- france_data %>%
specify(explanatory = domicil,
response = sbbsntx) %>%
hypothesize(null = "independence") %>%
generate(reps = 1000, type = "permute") %>%
calculate(stat = "Chisq")## Warning: Removed 14998 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 0.006Provide interpretations of the output (consider the variable info).
A t-statistic of 40.3495 is quite large, indicating a significant difference in social benefits costing business too much depending on a person’s domicile. It suggests that the mean spending level in for businesses is different for the two groups. Under the assumptions of frequentist hypothesis testing, a p value this close to 0 means there is strong 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.
null_distribution %>%
visualize() +
shade_p_value(obs_stat = test_stat, direction = "two-sided")## Warning: Chi-Square usually corresponds to right-tailed tests. Proceed with
## caution.
null_distribution## Response: sbbsntx (factor)
## Explanatory: domicil (factor)
## Null Hypothesis: independence
## # A tibble: 1,000 × 2
## replicate stat
## <int> <dbl>
## 1 1 12.0
## 2 2 32.9
## 3 3 16.9
## 4 4 16.5
## 5 5 21.0
## 6 6 24.9
## 7 7 21.6
## 8 8 31.4
## 9 9 16.8
## 10 10 27.5
## # 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)## Warning: Chi-Square usually corresponds to right-tailed tests. Proceed with
## caution.
null_distribution## Response: sbbsntx (factor)
## Explanatory: domicil (factor)
## Null Hypothesis: independence
## # A tibble: 1,000 × 2
## replicate stat
## <int> <dbl>
## 1 1 12.0
## 2 2 32.9
## 3 3 16.9
## 4 4 16.5
## 5 5 21.0
## 6 6 24.9
## 7 7 21.6
## 8 8 31.4
## 9 9 16.8
## 10 10 27.5
## # ℹ 990 more rowsSince the observed statistic does not fall within the null distribution, this visualization indicates that we should reject the null.
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.
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% c(77, 88, 99) ~ NA_character_ # Values 77, 88, 99 in 'lrscale' are set as NA
),
)infer_test <- france_data %>%
specify(explanatory = lrscale,
response = ccrdprs) %>%
hypothesize(null = "independence") %>%
calculate(stat = "slope") ## Warning: Removed 14991 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)## X-squared
## 40.3495null_distribution <- france_data %>%
specify(explanatory = lrscale,
response = ccrdprs) %>%
hypothesize(null = "independence") %>%
generate(reps = 1000, type = "permute") %>%
calculate(stat = "slope") ## Warning: Removed 14991 rows containing missing values.p_val <- null_distribution %>%
get_pvalue(obs_stat = test_stat, direction = "two-sided") ## 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 0Provide interpretations of the output (consider the variable info). A t-statistic of 13.99345 is large, indicating a significant difference in the extent to which people feel responsibility to reduce climate change depending on if a person is left or right winged. It suggests that the mean spending level of responsibility is different for the two groups. Under the assumptions of frequentist hypothesis testing, a p value of 0 means there is 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.
conf_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)
null_distribution## Response: ccrdprs (numeric)
## Explanatory: lrscale (numeric)
## Null Hypothesis: independence
## # A tibble: 1,000 × 2
## replicate stat
## <int> <dbl>
## 1 1 -0.0128
## 2 2 -0.00590
## 3 3 -0.00425
## 4 4 0.000956
## 5 5 0.000656
## 6 6 -0.00238
## 7 7 -0.00572
## 8 8 0.00220
## 9 9 0.00469
## 10 10 -0.00379
## # ℹ 990 more rowsThis indicates that we should reject the null.