RAMIL GANADOS GROUP STATISTICAL ANALYSIS
Kyle Kenneth Ruaya
2024-02-18
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, year level, and strand. It can be seen that there are 73 females and 27 males; 20 of which are from ABM, 20 from GAS, 40 from HUMSS, and 20 from STEM.
2. Is there a significant difference on the variables physical and mental stability, time management, personal growth, skill development, and academic pressure when grouped according to:
2.1 Sex
Call:
lm(formula = `Physical and Mental Stability` ~ `Time Management` +
`Personal Growth` + `Skill Development` + `Academic Pressure`,
data = Data)
Coefficients:
(Intercept) `Time Management` `Personal Growth`
0.99369 0.31874 0.17161
`Skill Development` `Academic Pressure`
0.09234 0.06344 
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 Physical and Mental Stability
Normality Test
Shapiro-Wilk normality test
data: Data$`Physical and Mental Stability`
W = 0.92915, p-value = 4.456e-05Since p-value = 4.456e-05 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.941 0.3344
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Attaching package: 'gplots'The following object is masked from 'package:stats':
lowessWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Physical and Me… 27 1.4 3.2 2.6 0.4 2.69 0.424 0.082 0.168
2 Female Physical and Me… 73 2.2 3.6 3 0.6 2.91 0.309 0.036 0.072The mean of female and male is 2.689 and 2.910, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Physical and Mental Stability 100 5.79 1 0.0161 Kruskal-WallisBased on the p-value, there is significant difference was observed between the group pairs.
2.1.2 Sex and Time Management
Normality Test
Shapiro-Wilk normality test
data: Data$`Time Management`
W = 0.96454, p-value = 0.008593Since p-value = 0.008593 < 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 1.1232 0.2918
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 27 1.8 3 2.6 0.6 2.49 0.343 0.066 0.136
2 Female Time Management 73 1 4 3 0.4 2.84 0.516 0.06 0.12 The mean of female and male is 2.489 and 2.841, 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 14.3 1 0.000159 Kruskal-WallisBased on the p-value, there is significant difference was observed between the group pairs.
2.1.3 Sex and Personal Growth
Normality Test
Shapiro-Wilk normality test
data: Data$`Personal Growth`
W = 0.967, p-value = 0.01308Since p-value = 0.01308 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.5735 0.4507
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Personal Growth 27 1.8 3.8 2.8 0.6 2.89 0.485 0.093 0.192
2 Female Personal Growth 73 2.2 4 3 0.6 3.08 0.409 0.048 0.095The mean of female and male is 2.889 and 3.077, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Personal Growth 100 2.81 1 0.0937 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.1.4 Sex and Skill Development
Normality Test
Shapiro-Wilk normality test
data: Data$`Skill Development`
W = 0.90646, p-value = 2.883e-06Since p-value = 2.883e-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 0.0041 0.9488
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 Skill Developme… 27 1.8 3.8 3 0.5 3.02 0.468 0.09 0.185
2 Female Skill Developme… 73 1.4 4 3.2 0.6 3.23 0.453 0.053 0.106The mean of female and male is 3.022 and 3.233, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Skill Development 100 3.90 1 0.0482 Kruskal-WallisBased on the p-value, there is significant difference was observed between the group pairs.
2.1.5 Sex and Academic Pressure
Normality Test
Shapiro-Wilk normality test
data: Data$`Academic Pressure`
W = 0.96531, p-value = 0.009783Since p-value = 0.009783 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 1.5186 0.2208
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Academic Pressu… 27 1 3.2 2.6 0.4 2.52 0.478 0.092 0.189
2 Female Academic Pressu… 73 1.4 3.8 2.8 0.6 2.71 0.495 0.058 0.115The mean of female and male is 2.519 and 2.707, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Academic Pressure 100 2.38 1 0.123 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.2 Strand
2.2.1 Strand and Physical and Mental Stability
Normality Test
Shapiro-Wilk normality test
data: Data$`Physical and Mental Stability`
W = 0.92915, p-value = 4.456e-05Since p-value = 4.456e-05 < 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 0.9753 0.4077
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 Physical and Me… 20 2.4 3.6 3 0.4 2.98 0.297 0.066 0.139
2 ABM Physical and Me… 20 2.2 3.4 3 0.3 2.9 0.321 0.072 0.15
3 HUMSS Physical and Me… 40 1.4 3.4 2.8 0.4 2.76 0.354 0.056 0.113
4 GAS Physical and Me… 20 1.8 3.4 2.8 0.6 2.85 0.415 0.093 0.194The mean of STEM, ABM, HUMSS, and GAS is 2.98, 2.90, 2.76, and 2.85, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Physical and Mental Stability 100 5.76 3 0.124 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 Physical and Me… STEM ABM 20 20 -0.569 0.569 1 ns
2 Physical and Me… STEM HUMSS 20 40 -2.23 0.0257 0.154 ns
3 Physical and Me… STEM GAS 20 20 -1.04 0.299 1 ns
4 Physical and Me… ABM HUMSS 20 40 -1.57 0.116 0.694 ns
5 Physical and Me… ABM GAS 20 20 -0.469 0.639 1 ns
6 Physical and Me… HUMSS GAS 40 20 1.03 0.302 1 ns There is significant difference between STEM and HUMSS.
2.2.2 Strand and Time Management
Normality Test
Shapiro-Wilk normality test
data: Data$`Time Management`
W = 0.96454, p-value = 0.008593Since p-value = 0.008593 < 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 0.6287 0.5983
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 Time Management 20 1.8 3.8 3 0.6 2.99 0.549 0.123 0.257
2 ABM Time Management 20 1 4 3 0.3 2.82 0.649 0.145 0.304
3 HUMSS Time Management 40 1.8 3.2 2.6 0.6 2.62 0.368 0.058 0.118
4 GAS Time Management 20 1.6 3.4 2.7 0.6 2.67 0.441 0.099 0.207The mean of STEM, ABM, HUMSS, and GAS is 2.990, 2.820, 2.625, and 2.70, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Time Management 100 8.88 3 0.0309 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 Time Manageme… STEM ABM 20 20 -0.737 0.461 1 ns
2 Time Manageme… STEM HUMSS 20 40 -2.70 0.00691 0.0415 *
3 Time Manageme… STEM GAS 20 20 -1.95 0.0511 0.307 ns
4 Time Manageme… ABM HUMSS 20 40 -1.85 0.0642 0.385 ns
5 Time Manageme… ABM GAS 20 20 -1.21 0.225 1 ns
6 Time Manageme… HUMSS GAS 40 20 0.449 0.653 1 ns There is significant difference between STEM and HUMSS.
2.2.3 Strand and Personal Growth
Normality Test
Shapiro-Wilk normality test
data: Data$`Personal Growth`
W = 0.967, p-value = 0.01308Since p-value = 0.01308 < 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 1.0224 0.3863
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 Personal Growth 20 1.8 4 3 0.45 2.99 0.509 0.114 0.238
2 ABM Personal Growth 20 2.4 3.8 3 0.4 3.01 0.34 0.076 0.159
3 HUMSS Personal Growth 40 1.8 3.8 2.9 0.65 2.99 0.479 0.076 0.153
4 GAS Personal Growth 20 2.4 3.8 3.1 0.4 3.15 0.355 0.079 0.166The mean of STEM, ABM, HUMSS, and GAS is 2.99, 3.01, 2.99, and 3.15, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Personal Growth 100 2.66 3 0.446 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 Personal Growth STEM ABM 20 20 -0.00276 0.998 1 ns
2 Personal Growth STEM HUMSS 20 40 -0.212 0.832 1 ns
3 Personal Growth STEM GAS 20 20 1.18 0.237 1 ns
4 Personal Growth ABM HUMSS 20 40 -0.208 0.835 1 ns
5 Personal Growth ABM GAS 20 20 1.18 0.236 1 ns
6 Personal Growth HUMSS GAS 40 20 1.58 0.115 0.689 ns 2.2.4 Strand and Skill Development
Normality Test
Shapiro-Wilk normality test
data: Data$`Skill Development`
W = 0.90646, p-value = 2.883e-06Since p-value = 2.883e-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 0.8412 0.4746
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 Skill Developme… 20 1.4 3.8 3.1 0.45 3.13 0.532 0.119 0.249
2 ABM Skill Developme… 20 2.6 4 3.2 0.4 3.2 0.367 0.082 0.172
3 HUMSS Skill Developme… 40 1.8 4 3.2 0.6 3.22 0.532 0.084 0.17
4 GAS Skill Developme… 20 2.4 3.8 3 0.25 3.11 0.34 0.076 0.159The mean of STEM, ABM, HUMSS, and GAS is 3.13, 3.20, 3.22, and 3.11, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Skill Development 100 2.68 3 0.444 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 Skill Development STEM ABM 20 20 0.141 0.888 1 ns
2 Skill Development STEM HUMSS 20 40 0.861 0.389 1 ns
3 Skill Development STEM GAS 20 20 -0.637 0.524 1 ns
4 Skill Development ABM HUMSS 20 40 0.698 0.485 1 ns
5 Skill Development ABM GAS 20 20 -0.778 0.437 1 ns
6 Skill Development HUMSS GAS 40 20 -1.60 0.110 0.662 ns 2.2.5 Strand and Academic Pressure
Normality Test
Shapiro-Wilk normality test
data: Data$`Academic Pressure`
W = 0.96531, p-value = 0.009783Since p-value = 0.009783 < 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 0.6562 0.581
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 Academic Pressu… 20 1.8 3.2 2.8 0.65 2.65 0.425 0.095 0.199
2 ABM Academic Pressu… 20 1.4 3.8 2.8 0.8 2.8 0.602 0.135 0.282
3 HUMSS Academic Pressu… 40 1 3.2 2.6 0.8 2.56 0.459 0.072 0.147
4 GAS Academic Pressu… 20 1.4 3.6 2.6 0.45 2.72 0.504 0.113 0.236The mean of STEM, ABM, HUMSS, and GAS is 2.650, 2.800, 2.555, and 2.720, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Academic Pressure 100 3.48 3 0.323 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 Academic Pressu… STEM ABM 20 20 0.880 0.379 1 ns
2 Academic Pressu… STEM HUMSS 20 40 -0.701 0.484 1 ns
3 Academic Pressu… STEM GAS 20 20 0.484 0.628 1 ns
4 Academic Pressu… ABM HUMSS 20 40 -1.72 0.0859 0.516 ns
5 Academic Pressu… ABM GAS 20 20 -0.396 0.692 1 ns
6 Academic Pressu… HUMSS GAS 40 20 1.26 0.208 1 ns 3. Is there a significant relationship between physical and mental stability, time management, personal growth, skill development, and academic pressure?
Normality Test
Shapiro-Wilk normality test
data: Data1$Scores
W = 0.96783, p-value = 5.184e-09Since p-value = 5.184e-09 < 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 4 1.8806 0.1126
495 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: 5 × 11
Variables variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Physical and … Scores 100 1.4 3.6 2.9 0.45 2.85 0.355 0.036 0.071
2 Time Manageme… Scores 100 1 4 2.8 0.6 2.75 0.5 0.05 0.099
3 Personal Grow… Scores 100 1.8 4 3 0.6 3.03 0.436 0.044 0.087
4 Skill Develop… Scores 100 1.4 4 3.2 0.45 3.18 0.465 0.046 0.092
5 Academic Pres… Scores 100 1 3.8 2.6 0.6 2.66 0.495 0.049 0.098The mean of “Physical and Mental Stability”, “Time Management”, “Personal Growth”, “Skill Development”, “Academic Pressure” is 2.850, 2.746, 3.026, 3.176, and 2.656, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Scores 500 81.1 4 1.03e-16 Kruskal-WallisBased on the p-value, there is significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 10 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Scores Physical … Time … 100 100 -1.50 1.33e- 1 1 e+ 0 ns
2 Scores Physical … Perso… 100 100 2.56 1.04e- 2 1.04e- 1 ns
3 Scores Physical … Skill… 100 100 5.28 1.28e- 7 1.28e- 6 ****
4 Scores Physical … Acade… 100 100 -2.58 9.96e- 3 9.96e- 2 ns
5 Scores Time Mana… Perso… 100 100 4.07 4.78e- 5 4.78e- 4 ***
6 Scores Time Mana… Skill… 100 100 6.79 1.16e-11 1.16e-10 ****
7 Scores Time Mana… Acade… 100 100 -1.07 2.83e- 1 1 e+ 0 ns
8 Scores Personal … Skill… 100 100 2.72 6.54e- 3 6.54e- 2 ns
9 Scores Personal … Acade… 100 100 -5.14 2.75e- 7 2.75e- 6 ****
10 Scores Skill Dev… Acade… 100 100 -7.86 3.87e-15 3.87e-14 **** There is significant difference on the following pairs of groups: Physical and Mental Stability and Personal Growth; Physical and Mental Stability and Skill Development; Physical and Mental Stability and Academic Pressure; Time Management and Personal Growth; Time Management and Skill Development; Personal Growth and Skill Development; Personal Growth and Academic Pressure; and Skill Development and Academic Pressure.
4. Which have the most significant impact?
Based on the provided output above, we can say that it is the variable skill development.