Perez-Vasquez_Nicole_Homework3
Nicole
2024-03-11
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.
packages <- c("tidyverse", "infer", "fst")france_data <- ("france_data.fst") lapply(packages, library, character.only = TRUE)## ── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
## ✔ dplyr 1.1.4 ✔ readr 2.1.5
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## ✔ 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## Warning: package 'infer' was built under R version 4.3.3## [[1]]
## [1] "lubridate" "forcats" "stringr" "dplyr" "purrr" "readr"
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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"france_data <- read_fst("france_data.fst")france_data <- france_data %>%
filter(cntry == "FR") %>%
mutate(
educ.ba = case_when(
essround < 5 & edulvla == 5 ~ "BA or more",
essround >= 5 & edulvlb > 600 ~ "BA or more",
TRUE ~ "No BA"
)
) %>%
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)
)france_data <- france_data %>%
filter(cntry == "FR") %>%
mutate
ID = case_when
lrscale %in% 0:3 ~ "Left" ## lrscale %in% 0:3 ~ "Left" lrscale %in% 7:10 ~ "Right" ## lrscale %in% 7:10 ~ "Right" lrscale %in% 4:6 ~ "Moderate" ## lrscale %in% 4:6 ~ "Moderate" lrscale %in% c (77, 88, 99) ~ NA_character_ ## lrscale %in% c(77, 88, 99) ~ NA_character_ Gen = case_when
yrbrn %in% 1900:1945 ~ "0" ## yrbrn %in% 1900:1945 ~ "0" yrbrn %in% 1946:1964 ~ "1" ## yrbrn %in% 1946:1964 ~ "1" yrbrn %in% 1965:1979 ~ "2" ## yrbrn %in% 1965:1979 ~ "2" yrbrn %in% 1980:1996 ~ "3" ## yrbrn %in% 1980:1996 ~ "3" TRUE ~ as.character(yrbrn) ## TRUE ~ as.character(yrbrn)table(france_data$lrscale)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 945 499 1269 1996 1838 5274 1594 1749 1458 409 686 381 940table(france_data$educ.ba)##
## BA or more No BA
## 4252 14786cat("Table of Expected Frequencies:\n")## Table of Expected Frequencies: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#to be honest this was a very confusing coding set to work with, however we can observe that the critical value is set at 3.841 for which appoints to where the null hypothesis is rejected. there is very little information as to the association between educ.ba and lrscale.
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 <- read.fst("france_data.fst")
table(france_data$domicil)##
## 1 2 3 4 5 7 8
## 3584 2361 6257 5668 1162 3 3table(france_data$sbbsntx)##
## 1 2 3 4 5 7 8
## 757 1538 804 640 301 3 100france_clean <- france_data %>%
mutate(
domicil = ifelse(domicil == 2, 0, ifelse(domicil %in% c(7, 8, 9), NA, domicil)),
sbbsntx = ifelse(sbbsntx %in% c(77, 88, 99), NA, sbbsntx),) france_clean <- france_clean %>% filter(!is.na(domicil), !is.na(sbbsntx),)table(france_clean$domicil)##
## 0 1 3 4 5
## 487 699 1462 1249 245table(france_clean$sbbsntx)##
## 1 2 3 4 5 7 8
## 756 1538 804 640 301 3 100france_data <- france_data %>%
mutate(
sbbsntx = recode(as.character(sbbsntx),
'1' = "agree strongly",
'2' = "agree",
'3' = "neither agree or disagree",
'4' = "dissagree",
'5' = "strongly disagree",
'7' = NA_character_,
'8' = NA_character_,
'9' = NA_character_)
)%>%
filter(!is.na(sbbsntx))france_data <- france_data %>%
mutate(
domicil = recode(as.character(domicil),
'1' = "urban",
'2' = "urban",
'3' = "rural",
'4' = "rural",
'5' = "rural",
'7' = NA_character_,
'8' = NA_character_,
'9' = NA_character_)
)%>%
filter(!is.na(domicil))france_data <- france_data %>%
filter(!is.na(domicil) & !is.na(sbbsntx))
france_data <- france_data %>%
mutate(
domicil = factor(domicil),
sbbsntx = factor(sbbsntx)
) test_stat <- france_data %>%
specify(explanatory = domicil,
response = sbbsntx) %>%
hypothesize(null = "independence") %>%
calculate(stat = "chisq")
print(test_stat$stat)## X-squared
## 19.97187null_distribution <- france_data %>%
specify(explanatory = domicil,
response = sbbsntx) %>%
hypothesize(null = "independence") %>%
generate(reps = 1000, type = "permute") %>%
calculate(stat = "chisq")p_val1 <- 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_val1## # A tibble: 1 × 1
## p_value
## <dbl>
## 1 0a squared value of 19.87 after cleaning up the data. This data appoints that if relationships between the data being tested were not related there would be some chnace or percentage that observing the data for extreme results.
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 0.369
## 2 2 4.18
## 3 3 8.35
## 4 4 4.43
## 5 5 3.96
## 6 6 2.03
## 7 7 3.47
## 8 8 7.81
## 9 9 3.66
## 10 10 8.87
## # ℹ 990 more rowsthere are higher counts in the lower stat areas that range between 1-4 number stats. The stats have a higher encountering between the 2 stat number, the numbers do not range to any higher stats instead they increase more the more you decrease the number.
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 <- read.fst("france_data.fst")
table(france_data$lrscale)##
## 0 1 2 3 4 5 6 7 8 9 10 77 88
## 945 499 1269 1996 1838 5274 1594 1749 1458 409 686 381 940table(france_data$ccrdprs)##
## 0 1 2 3 4 5 6 7 8 9 10 66 77 88 99
## 72 26 64 78 114 478 403 736 952 358 708 27 1 29 1france_clean <- france_data %>%
mutate(
lrscale = ifelse(lrscale == 2, 0, ifelse(lrscale %in% c(7, 8, 9), NA, lrscale)),
ccrdprs = ifelse(ccrdprs %in% c(77, 88, 99), NA, ccrdprs),)france_clean <- france_clean %>% filter(!is.na(lrscale), !is.na(ccrdprs),)table(france_clean$lrscale)##
## 0 1 3 4 5 6 10 77 88
## 395 87 412 340 1098 327 156 110 257table(france_clean$ccrdprs)##
## 0 1 2 3 4 5 6 7 8 9 10 66
## 55 21 50 58 85 378 318 554 743 295 604 21france_data <- france_data %>%
mutate(
ccrdprs = recode(as.character(ccrdprs),
'1' = "not at all",
'2' = "agree",
'3' = "neither agree or disagree",
'4' = "dissagree",
'5' = "strongly disagree",
'7' = NA_character_,
'8' = NA_character_,
'9' = NA_character_)
)%>%
filter(!is.na(ccrdprs))france_data <- france_data %>%
filter(cntry == "FR") %>%
mutate(
# Recoding 'lrscale' into a new variable 'LR'
LR = case_when(
lrscale %in% 0:3 ~ "Left", # Values 0 to 3 in 'lrscale' are categorized as "Left"
lrscale %in% 7:10 ~ "Right", # Values 7 to 10 in 'lrscale' are categorized as "Right"
lrscale %in% 4:6 ~ "Moderate", # Values 4 to 6 in 'lrscale' are categorized as "Moderate"
lrscale %in% c(77, 88, 99) ~ NA_character_ # Values 77, 88, 99 in 'lrscale' are set as NA
),
# Recoding 'yrbrn' (year born) into a new variable 'Gen'
Gen = case_when(
yrbrn %in% 1900:1945 ~ "1", # Birth years from 1900 to 1945 categorized as "1"
yrbrn %in% 1946:1964 ~ "2", # Birth years from 1946 to 1964 categorized as "2"
yrbrn %in% 1965:1979 ~ "3", # Birth years from 1965 to 1979 categorized as "3"
yrbrn %in% 1980:1996 ~ "4", # Birth years from 1980 to 1996 categorized as "4"
TRUE ~ as.character(yrbrn) # Other years remain unchanged
),
# Converting 'Gen' to a factor with meaningful labels
Gen = factor(Gen, levels = c("1", "2", "3", "4"),
labels = c("Interwar", "Baby Boomers", "Gen X", "Millennials")) # Assigning labels to the levels in 'Gen'
)%>%
filter(!is.na(lrscale))france_data <- france_data %>%
filter(!is.na(lrscale) & !is.na(ccrdprs))
france_data <- france_data %>%
mutate(
lrscale = factor(lrscale),
ccrdprs = factor(ccrdprs)
) test_stat <- france_data %>%
specify(explanatory = lrscale,
response = ccrdprs) %>%
hypothesize(null = "independence") %>%
calculate(stat = "chisq")
print(test_stat$stat)## X-squared
## 215.049null_distribution <- france_data %>%
specify(explanatory = lrscale,
response = ccrdprs) %>%
hypothesize(null = "independence") %>%
generate(reps = 1000, type = "permute") %>%
calculate(stat = "chisq")p_val1 <- null_distribution %>% # replace name here if you assigned something other than null_distribution above
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_val1## # A tibble: 1 × 1
## p_value
## <dbl>
## 1 0Task 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") ## Warning: Chi-Square usually corresponds to right-tailed tests. Proceed with
## caution.
null_distribution## Response: ccrdprs (factor)
## Explanatory: lrscale (factor)
## Null Hypothesis: independence
## # A tibble: 1,000 × 2
## replicate stat
## <int> <dbl>
## 1 1 113.
## 2 2 108.
## 3 3 136.
## 4 4 109.
## 5 5 142.
## 6 6 109.
## 7 7 98.6
## 8 8 121.
## 9 9 159.
## 10 10 133.
## # ℹ 990 more rowsThis data seems to be left skewed from obersvation making the outcome negative. The stats are all in between the numbers of 100s range but are left skewed. The null distribution therefore is negative.
End.
Provide interpretations of the output (consider the variable info).