Chi-squared and t-test
2BK team: Bakhareva, Borisenko, Kireeva, Kuzmicheva
24/02/2019
Indentifying topic and describing individual contribution
Hello. We are 2BK. Our topic is “Politics”. The country we have chosen for studying is Ireland (round 8). Team members are Bakhareva Anastasia, Borisenko Iana, Kireeva Irina, Kuzmicheva Daria. We have focused on the results of the surveys connected both with politics and personal information on Ireland.
- During our work, our hypotheses and conclusions will be marked in this way.
As for individual contribution, there it is done as follows:
Anastasia Bakhareva: Constucting stacked barplot в„–2, The conclusion about the groups representativeness, construction of histogram and conclusion about distribution by histogram , creating report
Iana Borisenko: Constructing contigency table, constucting stacked barplot в„–1, assocplot building and conclusion about residuals, overall conclusion, creating report
Irina Kireeva: Manipulations with data set, creating new variable “politics13$lr”, boxplot conduction and conclusion, descriptive statistics for T-test, Wilcoxon test conduction and conclusions
Daria Kuzmicheva: Describing variables with different measurement scales, Chi-test conduction and conclusion, conclusion about distribution by skew Рё kurtosis, Q-Q plot construction and conclusions
Preparing data for analysis.
First of all, we run all the libraries necessary for the analysis.
library(readr)
library(ggplot2)
library(dplyr)
library(sjlabelled)
library(sjmisc)
library(sjstats)
library(ggeffects)
library(sjPlot)
library(knitr)
library(psych)
library(foreign)
Next, we load out dataset. It is a combination of data connected with both politics and personal information.
politics_gender <- read_spss("politics_gender.sav")
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Here we filter our data in order to delete all the observation useless for the analysis.
politics <- politics_gender %>%
select(agea, lrscale, sgnptit, vote)
politics <- politics[!is.na(politics$agea),]
politics <- politics[!is.na(politics$lrscale),]
politics <- politics[!is.na(politics$sgnptit),]
politics <- politics[!is.na(politics$vote),]
politics13 <- politics %>%
filter(lrscale != 77) %>%
filter(lrscale != 88) %>%
filter(lrscale != 99 )
politics13 <- politics13 %>%
filter(sgnptit != 7) %>%
filter(sgnptit != 8) %>%
filter(sgnptit != 9)
politics13 <- politics13 %>%
filter(agea != 999)
Describing variables
First, we modify one of the variables to make it comfortable for manipulations. Then, we update our dataset.
politics13$lr <- ifelse(politics13$lrscale <= 3, "Left",
ifelse(politics13$lrscale >= 7, "Right", "Middle"))
politics13 <- politics13 %>%
select(- lrscale)
Now, let`s look at the number of observations and the number of variables.
## [1] 2231 4
- Well, that`s all we need for conducting tests: 4 variables and enough number of observations.
Then, there is a description of chosen variables presented.
Label <- c("`sgnptit`", "`lr`", "`agea`", "`vote`" )
Meaning <- c("Signed petition last 12 months", "Placement on left right scale", "Age", "Voted last national election")
Level_Of_Measurement <- c("Nominal", "Nominal", "Ratio", "Nominal")
Test <- c("Chi-squared test", "Chi-squared test", "T-test for independet variables", "T-test for independet variables")
df <- data.frame(Label, Meaning, Level_Of_Measurement,Test, stringsAsFactors = FALSE)
kable(df)
sgnptit |
Signed petition last 12 months |
Nominal |
Chi-squared test |
lr |
Placement on left right scale |
Nominal |
Chi-squared test |
agea |
Age |
Ratio |
T-test for independet variables |
vote |
Voted last national election |
Nominal |
T-test for independet variables |
Exploratory analysis of categories
Firstly, we select variables necessary for chi-square test. Next, there is a contigency table presented.
politics_chi <- politics13 %>%
select(lr, sgnptit)
politics_chi$sgnptit <- factor(politics_chi$sgnptit, labels = c("Yes", "No"), ordered= F,exclude = NA)
ContigencyTable <- table(politics_chi$lr, politics_chi$sgnptit)
kable(ContigencyTable)
| Left |
141 |
188 |
| Middle |
281 |
1036 |
| Right |
95 |
490 |
In order to check whether our categories are successful to run chi-square test, we are going to create stacked barplots and analyze them.
Stacked barplot №1
ggplot() +
geom_bar(data = politics_chi, aes(x = lr, fill = sgnptit), position = "fill")+
coord_flip()+
xlab("Party affiliation") +
ylab("Percentage of people") +
ggtitle("Participation in signing petitions due to party affiliation")
Stacked barplot №2
sjp.xtab(politics_chi$lr, politics_chi$sgnptit, type = "bar", margin ="row",
bar.pos = "stack", title = "Participation in signing petitions due to party affiliation", title.wtd.suffix = NULL,
axis.titles = NULL, axis.labels = NULL, legend.title = NULL,
legend.labels = NULL, weight.by = NULL, rev.order = FALSE,
show.values = TRUE, show.n = TRUE, show.prc = TRUE, show.total = TRUE,
show.legend = TRUE, show.summary = TRUE, summary.pos = "r",
string.total = "Total", wrap.title = 50, wrap.labels = 15,
wrap.legend.title = 20, wrap.legend.labels = 20, geom.size = 0.7,
geom.spacing = 0.1, geom.colors = "Paired", dot.size = 3,
smooth.lines = FALSE, grid.breaks = 0.2, expand.grid = FALSE,
ylim = NULL, vjust = "bottom", hjust = "left", y.offset = NULL,
coord.flip = TRUE)
- After building two plots, we were convinced that, one way or another, the adherents of each category of political preferences signed the petitions. However, it can be noted that the tendency to take part in signing petitions is not very common in Ireland; The largest group supporting this trend is the Liberals, with 42.5% of those who signed any petitions. The “middle” and “right”, respectively, have in their ranks 21.1% and 15.3% respectively.
- Each observation is independent of all the others (i.e., one observation per subject)and no more than 20% of the expected counts are less than 5. (none of them, actually). Therefore, the data is appropriate to conduct a reliable chi-square test.
Chi-square test
With the introduction of the Internet, signing petitions has become available online. That is, the study of the largerst online petitions` platform (Change.org) has shown that this source is strongly biased toward liberal causes. 
In Ireland, the half of the political parties presented are right(conservatism) or centre(social democracy, liberal conservatism, populism) and another half is left(socialism, respublicanism), so citizens have a wide spectrum of different views to share. Accordingly, in order to find out whether this distribution of the petition signatories due to their political preferences is random, we decided to build a Chi- square test.
The following hypotheses were approved for this:
- H0 - there is no relation between the political preferences of respondents and their signing petition or not signing behavior
- H1 - there is a relation between the political preferences of respondents and their signing petition or not signing behavior
So, let`s run chi-square test.
colnames(ContigencyTable) <- c("Petition +", "Petition -")
rownames(ContigencyTable) <- c("L", "R", "C")
chi.test <- chisq.test(ContigencyTable)
chi.test
##
## Pearson's Chi-squared test
##
## data: ContigencyTable
## X-squared = 89.895, df = 2, p-value < 2.2e-16
- After carrying out the Chi-square test, we found out that its p-value is extremely small, which means that we do not have strong enough evidence to assert that there is no relation between these two variables. In this way, our H0 should be rejected and political preferences of respondents and their signing petition or not signing behavior are likely to be related.
Next, we have to look at residuals.
kable(chi.test$stdres) #for residuals
| L |
9.164265 |
-9.164265 |
| R |
-2.468595 |
2.468595 |
| C |
-4.627590 |
4.627590 |
assocplot(t(ContigencyTable), main="Residuals and number of observations" )
On the plot of residuals, we can see the confirmation of our conclusion on the Chi-test: the difference in the number of petitioners who belong to different political parties is too big to say that the variables are independent of each other. Especially distinguished are the liberals, in whose ranks the number of signatories of the petition for indicator 10 is greater than expected, if these variables were independent; as well as the “right” ones, where the indicator 6 is less than the expected number of people who signed the petitions, if these variables were independent.
Thus, we were convinced that, apparently, since the Chi-square test and the difference in the residuals indicate a lack of evidence in favor of the independence of these data, we can assert that the political preferences of the respondents and their desire to sign or not to sign petitions of any kinds are related.
Exploring data for T-test
Here we start with filtering data to delete values useless for our test.
politics_ttest <- politics13 %>%
select(agea, vote) %>%
filter(vote != 3) %>%
filter(vote != 7) %>%
filter(vote != 8)
Next, Let’s compare mean values with the help of boxplot.
politics_ttest$vote <- factor(politics_ttest$vote, labels = c("Yes", "No"), ordered= F,exclude = NA)
ggplot() +
geom_boxplot(data = politics_ttest, aes(x = vote, y = agea), fill="#A44200", col="#A44200", alpha = 0.5) +
scale_y_continuous(limits = c(0,100)) +
xlab("Voted last national election") +
ylab("Age") +
ggtitle("Participation in the election due to age")
- The median age of voters is higher than the median age of those who did not vote. The first box plot is taller than the second, so we can say that there is a greater variety of ages in the group of voters than in the group of those who refused to vote. The whiskers are pretty the same on both of the graphs. However, the graph of non-voters shows some outliers.
Cheking normality of distribution
There is the first way to check normality presented.
describeBy(politics_ttest, politics_ttest$vote)
##
## Descriptive statistics by group
## group: Yes
## vars n mean sd median trimmed mad min max range skew
## agea 1 1689 54.93 16.41 55 54.98 19.27 19 92 73 -0.03
## vote* 2 1689 1.00 0.00 1 1.00 0.00 1 1 0 NaN
## kurtosis se
## agea -0.77 0.4
## vote* NaN 0.0
## --------------------------------------------------------
## group: No
## vars n mean sd median trimmed mad min max range skew
## agea 1 408 40.69 15.87 38 39.19 14.83 16 96 80 0.87
## vote* 2 408 2.00 0.00 2 2.00 0.00 2 2 0 NaN
## kurtosis se
## agea 0.44 0.79
## vote* NaN 0.00
Skewness is a measure of the symmetry in a distribution. The normal distribution is symmetrical, so skew should be equal to 0 in normal distribution. In the group of voters skew equals to 7.49, and in the group of non-voters the skew is higher, 8.71. The distribution of age is more symmetrical in the group of voters, but still it is far away from normal. However, both of skews are greater than 1, so both of the groups have a high positive skewness (right).
Kurtosis tells us, whether the distribution is peaked or plain. The kurtosis of the age in voters group equals to 55.29, and in the non-voters group kurtosis equals to 75.76. That means that the distribution of the first group (voters) is less sharp than the distribution of the second group.
Next, we check normality with the help of histogram.
library(ggplot2)
ggplot(politics_ttest, aes(x = agea, fill = vote)) +
geom_histogram(aes(y=..density..), position = "identity", alpha = 0.7, binwidth = 3) +
geom_density(col = "yellow", fill = "white", alpha = 0.1) +
geom_vline(aes(xintercept = mean(politics_ttest$agea), color = 'mean'), linetype="dashed", size=1) +
geom_vline(aes(xintercept = median(politics_ttest$agea), color = 'median'), linetype="longdash", size=1) +
scale_color_manual(name = "Measurement", values = c(median = "#cb3f68", mean = "#824acd")) +
xlab("Age") +
ylab("Density") +
ggtitle("Age distribution of voters and non-voters")
- Now we can see the distribution and assume that both of the groups are not very close to normal distribution, but the group of voters is slightly closer to it.
Finally, we check normality with the help of Q-Q Plot.
#creating subgroups based on voting / non-voting
voteplus <- subset(politics_ttest[politics_ttest$vote == "Yes",])
voteminus <- subset(politics_ttest[politics_ttest$vote == "No",])
par(mfrow = c(1,2))
# y is limited from 18 because it is age at which the Irish are allowed to vote
qqnorm(voteplus$agea, ylim = c(18, 100), main = "Normal Q-Q Plot for vote+"); qqline(voteplus$agea,ylim = c(18 ,100), col= 2)
qqnorm(voteminus$agea, ylim = c(18 ,100), main = "Normal Q-Q Plot for vote-"); qqline(voteminus$agea, col= 2, ylim = c(18 ,100))
- Both Q-Q plots show the distributions that are skewed to the right (some data higher than the line) and sharp peaks (the shape of the line is not the same as the normality line). However, the first plot shows a slightly plainer peak.
- Thus, we can conclude that the first group (voters) is more normally distributed than the second group (non-voters). However, during a series of tests on the normality of the sample, it was proved that the resulting sample has an abnormal distribution.
Conducting T-test
Age is sometimes mentioned as one of the factor which can influence voting behaviour, but it seems that every country should be studied as a unique case.
As in the case with political preferences and signed petitions, we would like to find out whether voting behavior in Ireland is related to the respondent’s age, so, we should conduct a T-test. The following hypotheses were approved for this:
- H0: the mean age of people who voted and did not vote does not differ.
- H1: the mean age does differ and, thus, there is a relation between age and voting behavior
Now we are going to run T-test.
t.test(politics_ttest$agea ~ politics_ttest$vote)
##
## Welch Two Sample t-test
##
## data: politics_ttest$agea by politics_ttest$vote
## t = 16.147, df = 634.29, p-value < 2.2e-16
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
## 12.50096 15.96259
## sample estimates:
## mean in group Yes mean in group No
## 54.92540 40.69363
- Statistical conclusion: at the 5% significance level on the available data the null hypothesis should be rejected in favor of the alternative one (p-value <О±).
- Substantive conclusion : the average age of people is significantly different among those who voted, and those who refused to vote.
Double-checking results with non-parametric test
- H0: the two populations (voters and non-voters) have the same distribution with the same median age.
- H1: the two populations (voters and non-voters) have the different distribution with the different median age.
wilcox.test(agea ~ vote, data = politics_ttest)
##
## Wilcoxon rank sum test with continuity correction
##
## data: agea by vote
## W = 509930, p-value < 2.2e-16
## alternative hypothesis: true location shift is not equal to 0
- Statistical conclusion: according to the obtained p-value, which is really, really small, there are no strong enough evidence to assert that H0 is true. Thus, it should be rejected.
- Substantive conclusion: The Wilcoxon test also proves that the age of people from the considered groups is significantly different among those who voted, and those who refused to vote.
Overall conclusion
Thus, by operating on the data and having conducted several statistical tests, we can confidently assert the following:
- The activity of people in the field of signing petitions is extremely related to the political preferences of Irish citizens. The greatest number of petitions are signed by people of the left political orientation, the least - by the right one.
- Age is considered to be one of the important factor which influences voting behavior in Ireland. Young citizens in Ireland tend not to attend elections, although they have already reached the age of majority, while people after forty for the most part regularly attend the elections.