MARIELLE BELOSILLO GROUP STATISTICAL ANALYSIS
Kyle Kenneth Ruaya
2024-02-20
Data
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 and strand. It can be seen that there are 66 females and 34 males; 22 of which are from ABM, 22 from GAS, 34 from HUMSS, and 22 from STEM.
2. Is there a significant difference on the variables financial status, students behavior and time management when grouped according to:
2.1 Sex
Call:
lm(formula = `Financial Status` ~ `Students Behavior` + `Time Management`,
data = Data)
Coefficients:
(Intercept) `Students Behavior` `Time Management`
2.7685 -0.1273 0.1324 
From this, we may deduce that the data fail to satisfy two assumptions – Linearity and Homogeneity of Variance.
Loading required package: carData
Attaching package: 'car'The following object is masked from 'package:dplyr':
recode── Attaching core tidyverse packages ──────────────────────── tidyverse 2.0.0 ──
✔ forcats 1.0.0 ✔ readr 2.1.5
✔ ggplot2 3.4.4 ✔ stringr 1.5.1
✔ lubridate 1.9.3 ✔ tibble 3.2.1
✔ purrr 1.0.2 ✔ tidyr 1.3.1
── Conflicts ────────────────────────────────────────── tidyverse_conflicts() ──
✖ dplyr::filter() masks stats::filter()
✖ dplyr::lag() masks stats::lag()
✖ car::recode() masks dplyr::recode()
✖ purrr::some() masks car::some()
ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors
Attaching package: 'rstatix'
The following object is masked from 'package:stats':
filter2.1.1 Sex and Financial Status
Normality Test
Shapiro-Wilk normality test
data: Data$`Financial Status`
W = 0.96701, p-value = 0.01311Since p-value = 0.01311 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.0028 0.958
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Attaching package: 'gplots'The following object is masked from 'package:stats':
lowessWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Financial Status 34 1.8 3.6 2.8 0.4 2.77 0.394 0.068 0.138
2 Female Financial Status 66 1.6 3.6 2.8 0.4 2.78 0.415 0.051 0.102The mean of male and female is 2.771 and 2.782, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Financial Status 100 0.0985 1 0.754 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.1.2 Sex and Students Behavior
Normality Test
Shapiro-Wilk normality test
data: Data$`Students Behavior`
W = 0.91318, p-value = 6.228e-06Since p-value = 6.228e-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 1 1.2538 0.2656
98 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: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Students Behavi… 34 1.2 3.4 2.8 0.4 2.72 0.487 0.084 0.17
2 Female Students Behavi… 66 1.2 3.8 2.8 0.4 2.74 0.375 0.046 0.092The mean of male and female is 2.724 and 2.745, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Students Behavior 100 0.190 1 0.663 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.1.3 Sex and Time Management
Normality Test
Shapiro-Wilk normality test
data: Data$`Time Management`
W = 0.93998, p-value = 0.0001922Since p-value = 0.0001922 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.0889 0.7662
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Time Management 34 1.6 3.4 2.8 0.4 2.71 0.421 0.072 0.147
2 Female Time Management 66 1.6 3.6 2.6 0.6 2.7 0.414 0.051 0.102The mean of female and male is 2.712 and 2.700, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Time Management 100 0.0524 1 0.819 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.2 Strand
2.2.1 Strand and Financial Status
Normality Test
Shapiro-Wilk normality test
data: Data$`Financial Status`
W = 0.96701, p-value = 0.01311Since p-value = 0.01311 < 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 0.1012 0.9592
96 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 Financial Status 22 1.8 3.6 2.8 0.3 2.84 0.435 0.093 0.193
2 ABM Financial Status 22 2.2 3.4 2.8 0.55 2.81 0.362 0.077 0.161
3 HUMSS Financial Status 34 1.6 3.2 2.6 0.55 2.62 0.361 0.062 0.126
4 GAS Financial Status 22 1.8 3.6 3 0.4 2.93 0.43 0.092 0.191The mean of STEM, ABM, HUMSS, and GAS is 2.836, 2.809, 2.624, and 2.927, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Financial Status 100 9.83 3 0.0201 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 Financial Sta… STEM ABM 22 22 -0.417 0.677 1 ns
2 Financial Sta… STEM HUMSS 22 34 -1.99 0.0470 0.282 ns
3 Financial Sta… STEM GAS 22 22 0.926 0.355 1 ns
4 Financial Sta… ABM HUMSS 22 34 -1.53 0.127 0.760 ns
5 Financial Sta… ABM GAS 22 22 1.34 0.180 1 ns
6 Financial Sta… HUMSS GAS 34 22 3.01 0.00264 0.0158 * There is significant difference between STEM and HUMSS so with HUMSS and GAS.
2.2.2 Strand and Students Behavior
Normality Test
Shapiro-Wilk normality test
data: Data$`Students Behavior`
W = 0.91318, p-value = 6.228e-06Since p-value = 6.228e-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 1.3916 0.2501
96 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 Students Behavi… 22 2.4 3.8 2.7 0.2 2.74 0.317 0.068 0.141
2 ABM Students Behavi… 22 2 3.2 2.6 0.4 2.72 0.325 0.069 0.144
3 HUMSS Students Behavi… 34 2 3.4 2.9 0.55 2.83 0.366 0.063 0.128
4 GAS Students Behavi… 22 1.2 3.4 2.8 0.55 2.62 0.604 0.129 0.268The mean of STEM, ABM, HUMSS, and GAS is 2.736, 2.718, 2.829, and 2.618, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Students Behavior 100 2.97 3 0.396 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 Students Behavior STEM ABM 22 22 0.283 0.778 1 ns
2 Students Behavior STEM HUMSS 22 34 1.54 0.123 0.738 ns
3 Students Behavior STEM GAS 22 22 0.399 0.690 1 ns
4 Students Behavior ABM HUMSS 22 34 1.23 0.218 1 ns
5 Students Behavior ABM GAS 22 22 0.116 0.908 1 ns
6 Students Behavior HUMSS GAS 34 22 -1.10 0.270 1 ns 2.2.3 Strand and Time Management
Normality Test
Shapiro-Wilk normality test
data: Data$`Time Management`
W = 0.93998, p-value = 0.0001922Since p-value = 0.0001922 < 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 1.7821 0.1557
96 The 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 Time Management 22 2 3.6 2.9 0.6 2.77 0.401 0.086 0.178
2 ABM Time Management 22 2 3.2 2.6 0.2 2.66 0.265 0.056 0.117
3 HUMSS Time Management 34 1.6 3.4 2.6 0.55 2.67 0.412 0.071 0.144
4 GAS Time Management 22 1.6 3.4 2.8 0.4 2.73 0.55 0.117 0.244The mean of STEM, ABM, HUMSS, and GAS is 2.773, 2.664, 2.671 and 2.727, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Time Management 100 1.83 3 0.608 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 Time Management STEM ABM 22 22 -0.981 0.327 1 ns
2 Time Management STEM HUMSS 22 34 -0.707 0.480 1 ns
3 Time Management STEM GAS 22 22 0.174 0.862 1 ns
4 Time Management ABM HUMSS 22 34 0.374 0.708 1 ns
5 Time Management ABM GAS 22 22 1.15 0.248 1 ns
6 Time Management HUMSS GAS 34 22 0.898 0.369 1 ns 3. Is there a significant difference between financial status, students behavior, and time management?
Normality Test
Shapiro-Wilk normality test
data: Data1$Scores
W = 0.95216, p-value = 2.522e-08Since p-value = 2.522e-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 2 0.0827 0.9206
297 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: 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 Financial Sta… Scores 100 1.6 3.6 2.8 0.4 2.78 0.406 0.041 0.081
2 Students Beha… Scores 100 1.2 3.8 2.8 0.4 2.74 0.415 0.041 0.082
3 Time Manageme… Scores 100 1.6 3.6 2.8 0.6 2.70 0.414 0.041 0.082The mean of Financial Status, Students Behavior, and Time Management is 2.778, 2.738 and 2.704, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Scores 300 1.24 2 0.537 Kruskal-WallisBased on the p-value, there is no 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 Financial Status Stude… 100 100 -0.318 0.750 1 ns
2 Scores Financial Status Time … 100 100 -1.08 0.278 0.835 ns
3 Scores Students Behavior Time … 100 100 -0.766 0.444 1 ns 4. Which variable have the most significant?
Based on the provided output above, we can say that it is the financial status.