ROVELYN SUPERABLE GROUP STATISTICAL ANALYSIS
Kyle Kenneth Ruaya
2024-02-17
Data
New names:
• `Indicator 1` -> `Indicator 1...4`
• `Indicator 2` -> `Indicator 2...5`
• `Indicator 3` -> `Indicator 3...6`
• `Indicator 4` -> `Indicator 4...7`
• `Indicator 5` -> `Indicator 5...8`
• `Indicator 1` -> `Indicator 1...9`
• `Indicator 2` -> `Indicator 2...10`
• `Indicator 3` -> `Indicator 3...11`
• `Indicator 4` -> `Indicator 4...12`
• `Indicator 5` -> `Indicator 5...13`
• `Indicator 1` -> `Indicator 1...14`
• `Indicator 2` -> `Indicator 2...15`
• `Indicator 3` -> `Indicator 3...16`
• `Indicator 4` -> `Indicator 4...17`
• `Indicator 5` -> `Indicator 5...18`1. What is the demographic profile of the respondents in terms of:
Attaching package: 'dplyr'The following objects are masked from 'package:stats':
filter, lagThe following objects are masked from 'package:base':
intersect, setdiff, setequal, unionSex
Strand
The tables above provides the distributions of respondents in terms of sex, year level, and strand. It can be seen that there are 79 females and 41 males; 29 of which are from ABM, 31 from GAS, 30 from HUMSS, and 30 from STEM.
2. Is there a significant difference on the Academic Track Specialization, Facilities and Resources, and Community and Governance when grouped according to:
2.1 Sex
Call:
lm(formula = `Academic Track Specialization` ~ `Facilities and Resources` +
`Community and Governance`, data = Data)
Coefficients:
(Intercept) `Facilities and Resources`
2.19232 0.03515
`Community and Governance`
0.32750 
2.1.1 Sex and Academic Track Specialization
`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.
Attaching package: 'rstatix'The following object is masked from 'package:stats':
filterThe mean for male and female is 3.263 and 3.324, respectively.
The above graph shows the plotting of data by sex, which contains two sexes – male and female.
── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
✔ forcats 1.0.0 ✔ stringr 1.5.1
✔ lubridate 1.9.3 ✔ tibble 3.2.1
✔ purrr 1.0.2 ✔ tidyr 1.3.1
✔ readr 2.1.5
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ rstatix::filter() masks dplyr::filter(), stats::filter()
✖ dplyr::lag() masks stats::lag()
ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorsWarning: The following aesthetics were dropped during statistical transformation: fill
ℹ This can happen when ggplot fails to infer the correct grouping structure in
the data.
ℹ Did you forget to specify a `group` aesthetic or to convert a numerical
variable into a factor?
It clearly shows that there is no significant difference on the variable academic track specialization when grouped according to their sex.
Loading required package: carData
Attaching package: 'car'The following object is masked from 'package:purrr':
someThe following object is masked from 'package:dplyr':
recode
The histogram does not resemble a bell curve as seen above, means that the residuals do not have a normal distribution. Moreover, the points in the QQ-plots roughly follow the straight line, with the majority of them falling within the confidence bands. However, this does not guarantee that residuals follow a normal distribution since when based on the diagram on the left, it is the exact opposite of it. Thus, it is more convenient to observe the two.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.90788, p-value = 5.078e-07The Shapiro-Wilk p-value = 5.078e-07 on the residuals is less than the usual significance level of 0.05. Thus, we reject the hypothesis that residuals have a normal distribution.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.0461 0.8304
118 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Wilcoxon Rank Sum Test
Wilcoxon rank sum test
data: a and b
W = 1461, p-value = 0.366
alternative hypothesis: true location shift is not equal to 0Since the p-value is larger than 0.05, we fail to reject the null hypothesis, that is, there is no significant difference on the variable academic track specialization when grouped according to sex.
2.1.2 Sex and Facilities and Resources
`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.The mean for male and female is 2.956 and 2.886, respectively.
The above graph shows the plotting of data by sex, which contains two sexes – male and female.
Warning: The following aesthetics were dropped during statistical transformation: fill
ℹ This can happen when ggplot fails to infer the correct grouping structure in
the data.
ℹ Did you forget to specify a `group` aesthetic or to convert a numerical
variable into a factor?
It clearly shows that there is difference on the variable Facilities and Resources when grouped according to their sex. However, we still need to check the significance of this difference.
The histogram does not resemble a bell curve as seen above, means that the residuals do not have a normal distribution. Moreover, the points in the QQ-plots roughly follow the straight line, with the majority of them falling within the confidence bands. However, this does not guarantee that residuals follow a normal distribution since when based on the diagram on the left, it is the exact opposite of it. Thus, it is more convenient to observe the two.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.96814, p-value = 0.006036The Shapiro-Wilk p-value = 0.006036 on the residuals is less than the usual significance level of 0.05. Thus, we reject the hypothesis that residuals have a normal distribution.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.1105 0.7401
118 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Wilcoxon Rank Sum Test
Wilcoxon rank sum test
data: c and d
W = 1758, p-value = 0.435
alternative hypothesis: true location shift is not equal to 0Since the p-value is larger than 0.05, we fail to reject the null hypothesis, that is, there is no significant difference on the variable facilities and resources when grouped according to sex.
2.1.3 Sex and Community and Governance
`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.The mean for male and female is 3.107 and 3.066, respectively.
The above graph shows the plotting of data by sex, which contains two sexes – male and female.
Warning: The following aesthetics were dropped during statistical transformation: fill
ℹ This can happen when ggplot fails to infer the correct grouping structure in
the data.
ℹ Did you forget to specify a `group` aesthetic or to convert a numerical
variable into a factor?
It clearly shows that there is no significant difference on the variable Community and Governance when grouped according to their sex.
The histogram does not resemble a bell curve as seen above, means that the residuals do not have a normal distribution. Moreover, the points in the QQ-plots roughly follow the straight line, with the majority of them falling outside the confidence bands.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.92752, p-value = 6.763e-06The Shapiro-Wilk p-value = 6.763e-06 on the residuals is less than the usual significance level of 0.05. Thus, we reject the hypothesis that residuals have a normal distribution.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 2.092 0.1507
118 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Wilcoxon Rank Sum Test
Wilcoxon rank sum test
data: e and f
W = 1809.5, p-value = 0.273
alternative hypothesis: true location shift is not equal to 0Since the p-value is larger than 0.05, we fail to reject the null hypothesis, that is, there is no significant difference on the variable Community and Governance when grouped according to sex.
2.2 Strand
2.2.1 Strand and Academic Track Specialization
Normality Test
Shapiro-Wilk normality test
data: Data$`Academic Track Specialization`
W = 0.89176, p-value = 7.608e-08Since p-value = 7.608e-08 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 3 4.4458 0.005386 **
116
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The p-value is less than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is not met.
Attaching package: 'gplots'The following object is masked from 'package:stats':
lowessWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# A tibble: 4 × 11
Strand variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 STEM Academic Track … 30 2.8 4 3.6 0.95 3.47 0.447 0.082 0.167
2 ABM Academic Track … 29 2.4 4 3.4 0.4 3.38 0.425 0.079 0.162
3 HUMSS Academic Track … 30 2.8 4 3 0.55 3.25 0.379 0.069 0.141
4 GAS Academic Track … 31 2.8 3.8 3 0.2 3.12 0.23 0.041 0.084The mean of STEM, ABM, HUMSS, and GAS is 3.473, 3.379, 3.253, and 3.116, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Academic Track Specialization 120 13.7 3 0.0033 Kruskal-WallisBased on the p-value, there is significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 6 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Academic Tra… STEM ABM 30 29 -0.263 0.792 1 ns
2 Academic Tra… STEM HUMSS 30 30 -1.96 0.0498 0.299 ns
3 Academic Tra… STEM GAS 30 31 -3.23 0.00126 0.00754 **
4 Academic Tra… ABM HUMSS 29 30 -1.68 0.0926 0.556 ns
5 Academic Tra… ABM GAS 29 31 -2.93 0.00336 0.0202 *
6 Academic Tra… HUMSS GAS 30 31 -1.25 0.212 1 ns There is significant difference between, STEM and HUMSS so with STEM and GAS, ABM and GAS.
2.2.2 Strand and Facilities and Resources
Normality Test
Shapiro-Wilk normality test
data: Data$`Facilities and Resources`
W = 0.95209, p-value = 0.0003085Since p-value = 0.0003085 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 3 4.0718 0.00863 **
116
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The p-value is less than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is not met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# A tibble: 4 × 11
Strand variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 STEM Facilities and … 30 1.4 3.8 2.8 0.6 2.85 0.535 0.098 0.2
2 ABM Facilities and … 29 2.4 3.6 2.8 0.2 2.91 0.3 0.056 0.114
3 HUMSS Facilities and … 30 2.4 3.6 3 0.35 2.95 0.309 0.056 0.116
4 GAS Facilities and … 31 2.2 3.6 3 0.2 2.93 0.299 0.054 0.11 The mean of STEM, ABM, HUMSS, and GAS is 2.847, 2.910, 2.953, and 2.929, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Facilities and Resources 120 1.10 3 0.778 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 6 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Facilities and R… STEM ABM 30 29 0.274 0.784 1 ns
2 Facilities and R… STEM HUMSS 30 30 0.845 0.398 1 ns
3 Facilities and R… STEM GAS 30 31 0.863 0.388 1 ns
4 Facilities and R… ABM HUMSS 29 30 0.564 0.573 1 ns
5 Facilities and R… ABM GAS 29 31 0.579 0.562 1 ns
6 Facilities and R… HUMSS GAS 30 31 0.0109 0.991 1 ns 2.2.3 Strand and Community and Governance
Normality Test
Shapiro-Wilk normality test
data: Data$`Community and Governance`
W = 0.91716, p-value = 1.654e-06Since p-value = 1.654e-06 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 3 1.2218 0.305
116 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# A tibble: 4 × 11
Strand variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 STEM Community and G… 30 2 4 3.1 0.4 3.13 0.447 0.082 0.167
2 ABM Community and G… 29 2.2 4 3 0.2 3.08 0.353 0.065 0.134
3 HUMSS Community and G… 30 2.4 4 3 0.2 3.09 0.343 0.063 0.128
4 GAS Community and G… 31 2 4 3 0.1 3.03 0.345 0.062 0.127The mean of STEM, ABM, HUMSS, and GAS is 3.127, 3.083, 3.087, 3.026, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Community and Governance 120 2.18 3 0.535 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 6 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Community and Go… STEM ABM 30 29 -0.754 0.451 1 ns
2 Community and Go… STEM HUMSS 30 30 -0.760 0.447 1 ns
3 Community and Go… STEM GAS 30 31 -1.48 0.139 0.836 ns
4 Community and Go… ABM HUMSS 29 30 0.000265 1.00 1 ns
5 Community and Go… ABM GAS 29 31 -0.705 0.481 1 ns
6 Community and Go… HUMSS GAS 30 31 -0.711 0.477 1 ns 3. Is there a significant relationship between academic track specialization, facilities and resources, and community and governance?
Normality Test
Shapiro-Wilk normality test
data: Data1$Scores
W = 0.94222, p-value = 1.242e-10Since p-value = 1.242e-10 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 2 3.3326 0.03681 *
357
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The p-value is less than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is not met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# A tibble: 3 × 11
Variables variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Academic Trac… Scores 120 2.4 4 3.2 0.6 3.30 0.397 0.036 0.072
2 Facilities an… Scores 120 1.4 3.8 3 0.45 2.91 0.372 0.034 0.067
3 Community and… Scores 120 2 4 3 0.2 3.08 0.372 0.034 0.067The mean of Academic Track Specialization, Facilities and Resources, and Community and Governance is 3.303, 2.910, and 3.080, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Scores 360 52.1 2 4.89e-12 Kruskal-WallisBased on the p-value, there is a significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 3 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Scores Academic T… Facil… 120 120 -7.22 5.31e-13 1.59e-12 ****
2 Scores Academic T… Commu… 120 120 -3.62 2.94e- 4 8.82e- 4 ***
3 Scores Facilities… Commu… 120 120 3.60 3.22e- 4 9.67e- 4 *** Pairwaise, there is significant difference.
4. Which have the most significant impact?
Based on the provided output above, we can say that it is the academic track specialization.