| Characteristic of Scale | Nominal | Ordinal | Interval | Ratio |
|---|---|---|---|---|
| Applies names or numbers to categories? | Yes | Yes | Yes | Yes |
| Orders categories according to quantity? | Yes | Yes | Yes | |
| Displays equal intervals between consecutive numbers? | Yes | Yes | ||
| Diplays a “true zero point?” | Yes |
## 'data.frame': 395 obs. of 33 variables:
## $ school : chr "GP" "GP" "GP" "GP" ...
## $ sex : chr "F" "F" "F" "F" ...
## $ age : int 18 17 15 15 16 16 16 17 15 15 ...
## $ address : chr "U" "U" "U" "U" ...
## $ famsize : chr "GT3" "GT3" "LE3" "GT3" ...
## $ Pstatus : chr "A" "T" "T" "T" ...
## $ Medu : int 4 1 1 4 3 4 2 4 3 3 ...
## $ Fedu : int 4 1 1 2 3 3 2 4 2 4 ...
## $ Mjob : chr "at_home" "at_home" "at_home" "health" ...
## $ Fjob : chr "teacher" "other" "other" "services" ...
## $ reason : chr "course" "course" "other" "home" ...
## $ guardian : chr "mother" "father" "mother" "mother" ...
## $ traveltime: int 2 1 1 1 1 1 1 2 1 1 ...
## $ studytime : int 2 2 2 3 2 2 2 2 2 2 ...
## $ failures : int 0 0 3 0 0 0 0 0 0 0 ...
## $ schoolsup : chr "yes" "no" "yes" "no" ...
## $ famsup : chr "no" "yes" "no" "yes" ...
## $ paid : chr "no" "no" "yes" "yes" ...
## $ activities: chr "no" "no" "no" "yes" ...
## $ nursery : chr "yes" "no" "yes" "yes" ...
## $ higher : chr "yes" "yes" "yes" "yes" ...
## $ internet : chr "no" "yes" "yes" "yes" ...
## $ romantic : chr "no" "no" "no" "yes" ...
## $ famrel : int 4 5 4 3 4 5 4 4 4 5 ...
## $ freetime : int 3 3 3 2 3 4 4 1 2 5 ...
## $ goout : int 4 3 2 2 2 2 4 4 2 1 ...
## $ Dalc : int 1 1 2 1 1 1 1 1 1 1 ...
## $ Walc : int 1 1 3 1 2 2 1 1 1 1 ...
## $ health : int 3 3 3 5 5 5 3 1 1 5 ...
## $ absences : int 6 4 10 2 4 10 0 6 0 0 ...
## $ G1 : int 5 5 7 15 6 15 12 6 16 14 ...
## $ G2 : int 6 5 8 14 10 15 12 5 18 15 ...
## $ G3 : int 6 6 10 15 10 15 11 6 19 15 ...## [1] "There are 395 observations in our student data set."\[ x^2\;=\;\sum\frac{(O_{i}\;-\;E_{i})^2}{E_{i}} \\ \\ \] \[ x^2 = chi-squared\\ O_{i}=\;observed\;value \\ E_{i}=\;expected\;value \]
Here we want to investigate if there is an association between the type of job a mother has and access to the internet.
##
## no yes
## at_home 22 37
## health 2 32
## other 27 114
## services 12 91
## teacher 3 55##
## Pearson's Chi-squared test
##
## data: student$Mjob and student$internet
## X-squared = 28.861, df = 4, p-value = 8.341e-06mjob_internet<- xtabs(formula = ~Mjob+internet,data = student)
chisq_result<- chisq.test(mjob_internet ,correct = FALSE)
chisq_result##
## Pearson's Chi-squared test
##
## data: mjob_internet
## X-squared = 28.861, df = 4, p-value = 8.341e-06## internet
## Mjob no yes
## at_home 22 37
## health 2 32
## other 27 114
## services 12 91
## teacher 3 55## internet
## Mjob no yes
## at_home 9.858228 49.14177
## health 5.681013 28.31899
## other 23.559494 117.44051
## services 17.210127 85.78987
## teacher 9.691139 48.30886total_observed<- chisq_result$observed[,1]+chisq_result$observed[,2]
observed_expected<- data.frame(chisq_result$observed[,1],chisq_result$expected[,1],chisq_result$observed[,2],chisq_result$expected[,2],total_observed)
colnames(observed_expected) <- c("Observed","Expected","Observed","Expected","")total_observed<- chisq_result$observed[,1]+chisq_result$observed[,2]
###Create data frame from chi-square results
observed_expected<- data.frame(chisq_result$observed[,1],round(chisq_result$expected[,1],digits = 3),chisq_result$observed[,2],round(chisq_result$expected[,2],digits = 3),total_observed)
overall_total<- rbind(observed_expected, c(colSums(round(observed_expected[,1:5]))))
colnames(overall_total) <- c("Observed","Expected","Observed","Expected","")
rownames(overall_total) <- c("at home","health","other","services","teacher","Total")
###Create chi-square table
kable(x = overall_total) %>%
kable_styling(row_label_position = "l", full_width = F) %>%
footnote(general = paste0("",sprintf(r'($X^2$)')," = (",chisq_result$parameter,", n = 395) = ",round(chisq_result$statistic,3)," p = ",round(chisq_result$p.value,digits = 12))) %>%
add_header_above(c("Mjob", "Internet No" = 2, "Internet Yes" = 2,"Total"=1))| Observed | Expected | Observed | Expected | ||
|---|---|---|---|---|---|
| at home | 22 | 9.858 | 37 | 49.142 | 59 |
| health | 2 | 5.681 | 32 | 28.319 | 34 |
| other | 27 | 23.559 | 114 | 117.441 | 141 |
| services | 12 | 17.210 | 91 | 85.790 | 103 |
| teacher | 3 | 9.691 | 55 | 48.309 | 58 |
| Total | 66 | 67.000 | 329 | 328.000 | 395 |
| Note: | |||||
| \(X^2\) = (4, n = 395) = 28.861 p = 8.341138e-06 |
alpha <- 0.05
degrees_of_freedom <- 4
###Calculate critical value
###qchisq is from the base stats package in R
critical_value <- qchisq(p = 1-alpha,df=degrees_of_freedom)
print(paste("Chi-square critical value is ",round(critical_value,3)))## [1] "Chi-square critical value is 9.488"We can see that our \(X^2\) value is 28.861 and the p-value is \(p>.05\). We check our \(X^2\) value against the critical table value of 9.488. Since our \(X^2\) value 28.861 is above our critical table value of 9.488 we cannot reject our null hypothesis. Therefore there is no significant association between Mjob type and internet.
Next we can investigate if there is an association between the type of guardian a student has and access to the internet.
###Chi-square calculation
guardian_internet<- xtabs(formula = ~guardian+internet,data = student)
guard_int_chisq_result<- chisq.test(guardian_internet ,correct = FALSE)
guard_int_chisq_result##
## Pearson's Chi-squared test
##
## data: guardian_internet
## X-squared = 1.401, df = 2, p-value = 0.4963| no | yes | |
|---|---|---|
| father | 12 | 78 |
| mother | 47 | 226 |
| other | 7 | 25 |
| no | yes | |
|---|---|---|
| father | 15.037975 | 74.96203 |
| mother | 45.615190 | 227.38481 |
| other | 5.346835 | 26.65316 |
total_observed<- guard_int_chisq_result$observed[,1]+guard_int_chisq_result$observed[,2]
###Create data frame from chi-square results
observed_expected<- data.frame(guard_int_chisq_result$observed[,1],round(guard_int_chisq_result$expected[,1],digits = 3),guard_int_chisq_result$observed[,2],round(guard_int_chisq_result$expected[,2],digits = 3),total_observed)
overall_total<- rbind(observed_expected, c(colSums(round(observed_expected[,1:5]))))
colnames(overall_total) <- c("Observed","Expected","Observed","Expected","")
rownames(overall_total) <- c("father","other","mother","Total")
###Create chi-square table
kable(x = overall_total) %>%
kable_styling(row_label_position = "l", full_width = F) %>%
footnote(general = paste0("",sprintf(r'($X^2$)')," = (",guard_int_chisq_result$parameter,", n = 395) = ",round(guard_int_chisq_result$statistic,3),"p =",round(guard_int_chisq_result$p.value,4))) %>%
add_header_above(c("Guardian", "Internet No" = 2, "Internet Yes" = 2,"Total"=1))| Observed | Expected | Observed | Expected | ||
|---|---|---|---|---|---|
| father | 12 | 15.038 | 78 | 74.962 | 90 |
| other | 47 | 45.615 | 226 | 227.385 | 273 |
| mother | 7 | 5.347 | 25 | 26.653 | 32 |
| Total | 66 | 66.000 | 329 | 329.000 | 395 |
| Note: | |||||
| \(X^2\) = (2, n = 395) = 1.401p =0.4963 |
alpha <- 0.05
degrees_of_freedom <- 2
###Calculate critical value
###qchisq is from the base stats package in R
critical_value <- qchisq(p = 1-alpha,df=degrees_of_freedom)
print(paste("Chi-square critical value is ",round(critical_value,3)))## [1] "Chi-square critical value is 5.991"We can see that our \(X^2\) value is 1.401 and the p-value is \(p>.05\). We check our \(X^2\) value against the critical table value of 5.991. Since our \(X^2\) value 1.401 is below our critical table value of 5.991, therefore we cannot reject our null hypothesis. Therefore there is no significant relationship between guardian type and internet.
\[\Large r_{xy} = \frac{\sum(x_{i}-\bar{x})(y_{i}-\bar{y})}{\sqrt{\sum(x_{i}-\bar{x})^2\sum(y_{i}-\bar{y})^2}}\]
The following assumptions are in place when utilizing the Pearson’s
\(r\) when calculating a correlation
coefficient.
| Correlation Coefficient | Effect Size |
|---|---|
| Small Effect | r +/- .10 |
| Medium Effect | r +/- .30 |
| Large Effect | r +/- .50 |
##
## Pearson's product-moment correlation
##
## data: student$absences and student$G1
## t = -0.6149, df = 393, p-value = 0.539
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## -0.12927845 0.06787576
## sample estimates:
## cor
## -0.0310029Here we can see that our correlation \(r\) = -0.031 and our \(p>.05\). Since our \(p-value\) is greater than \(0.05\) we can say there is no relationship between \(absences\) and \(G1\).
##
## Pearson's product-moment correlation
##
## data: student$absences and student$G3
## t = 0.67933, df = 393, p-value = 0.4973
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## -0.06464215 0.13247070
## sample estimates:
## cor
## 0.03424732Here we can see that our correlation \(r\) = 0.034 and our \(p>.05\). Since our \(p-value\) is greater than \(0.05\) we can say there is no relationship between \(absences\) and \(G3\).
Here we are looking at a true dichotomy sex (male vs female) and the association with final \(G3\) grade. We need to recode the variable into a binary variable of \(1\) and \(0\) with \(1\) being \(Male\) and \(0\) being \(Female\).
##
## Pearson's product-moment correlation
##
## data: student$G3 and student$sexrecoded
## t = 2.062, df = 393, p-value = 0.03987
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## 0.004833966 0.200084200
## sample estimates:
## cor
## 0.1034556Here we can look at whether there is an association between a students family support and their final grade. Again we need to re-code the variable so it is binary e.g. \(0\) and \(1\)
###Recode the variable into a binary variable
student$familysuprecoded <- ifelse(student$famsup=="yes",1,0)
cor.test(student$G3,student$familysuprecoded)##
## Pearson's product-moment correlation
##
## data: student$G3 and student$familysuprecoded
## t = -0.77686, df = 393, p-value = 0.4377
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## -0.13729770 0.05974472
## sample estimates:
## cor
## -0.03915715Here we can look at if there is an association between \(extra\;educational\; support\) \(schoolsup\) and final grade \(G3\) + Again we need to re-code the variable so it is binary e.g. \(0\) and \(1\)
student$schoolsuprecoded <- ifelse(student$schoolsup=="yes",1,0)
cor.test(student$G3,student$schoolsuprecoded)##
## Pearson's product-moment correlation
##
## data: student$G3 and student$schoolsuprecoded
## t = -1.6469, df = 393, p-value = 0.1004
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## -0.17998895 0.01601362
## sample estimates:
## cor
## -0.08278821Here we can look at if there is an association between those who attended nursery school \(nursery\) and final grade \(G3\) + We need to re-code the variable so it is binary e.g. \(0\) and \(1\)
student$nurseryrecoded <- ifelse(student$nursery=="yes",1,0)
cor.test(student$G3,student$nurseryrecoded)##
## Pearson's product-moment correlation
##
## data: student$G3 and student$nurseryrecoded
## t = 1.0237, df = 393, p-value = 0.3066
## alternative hypothesis: true correlation is not equal to 0
## 95 percent confidence interval:
## -0.04734402 0.14947834
## sample estimates:
## cor
## 0.0515679Spearman correlation coefficient for ranked data (Spearman’s Rho or \(r_s\)) – displays the correlation between two variables whose values have been ranked.
In the student data set we see that studytime and traveltime have been ranked. Therefore we can utilize the Spearman’s Rho \(r_s\) to see the association between these variables and final grade \(G3\).
## Warning in cor.test.default(student$studytime, student$traveltime, method =
## "spearman"): Cannot compute exact p-value with ties##
## Spearman's rank correlation rho
##
## data: student$studytime and student$traveltime
## S = 11360053, p-value = 0.03526
## alternative hypothesis: true rho is not equal to 0
## sample estimates:
## rho
## -0.1059694## Warning in cor.test.default(student$traveltime, student$goout, method =
## "spearman"): Cannot compute exact p-value with ties##
## Spearman's rank correlation rho
##
## data: student$traveltime and student$goout
## S = 10286267, p-value = 0.9774
## alternative hypothesis: true rho is not equal to 0
## sample estimates:
## rho
## -0.001429844Assumptions for phi \(\phi\) coefficient:
Phi coefficient (\(\phi\))– appropriate when both variables are true dichotomous variables
E.g subject sex (males versus female) with sport-team status (team-member vs not a team-member)
| \(y=1\) | \(y=0\) | \(total\) | |
|---|---|---|---|
| \(x=1\) | \(n_{11}\) | \(n_{10}\) | \(n_{1\cdot}\) |
| \(x=0\) | \(n_{01}\) | \(n_{00}\) | \(n_{0\cdot}\) |
| \(total\) | \(n_{\cdot1}\) | \(n_{\cdot0}\) | \(n\) |
\[\Large Phi\;equation: \phi = \frac{n_{11} \cdot n_{00}\;-\;n_{10} \cdot n_{01}}{\sqrt{n_{1\cdot}n_{0\cdot} n_{\cdot0} n_{\cdot1}} }\]
| \(y=1\) | \(y=0\) | |
|---|---|---|
| \(x=1\) | \(A\) | \(B\) |
| \(x=0\) | \(C\) | \(D\) |
\[Phi\;equation: \phi = \frac{A \cdot D\;-\;B \cdot C}{\sqrt{(A+B)(C+D)(A+C)(B+D)} }\]
First we will want to create a contingency of the table. We can do this in R simply by utilizing the \(table\) function and specifying two columns of \(dichotomous\) or \(binary\) variables.
##
## no yes
## F 38 170
## M 28 159##
## R U
## F 44 164
## M 44 143How to calculate a \(phi\) coefficient manually for \(gender\) and \(internet\)
# Φ = (AD-BC) / √(A+B)(C+D)(A+C)(B+D)
phi_coefficient = ((38*159)-(170*28))/sqrt((38+170)*(28+159)*(38+28)*(170+159))
print(phi_coefficient)## [1] 0.0441129We can measure the \(Phi\) coefficient similarly as to how we measure a Pearson’s \(r\) correlation. The closer to \(0\) there is no relationship. The closer to \(-1\) to \(1\) the stronger the relationship between the two dichotomous variables.
Here we can utilize the \(psych\) package to perform a \(Phi\) \(\phi\) coefficient analysis.
## [1] 0.0441129Hatcher, L. (2013). Advanced statistics in research: Reading, understanding, and writing up data analysis results. Shadow Finch Media.
Student Peformance (2014). https://archive.ics.uci.edu/dataset/320/student+performance
https://www.r-bloggers.com/2021/07/how-to-calculate-phi-coefficient-in-r/
William Revelle (2023). psych: Procedures for Psychological, Psychometric, and Personality Research. Northwestern University, Evanston, Illinois. R package version 2.3.9, https://CRAN.R-project.org/package=psych