BRENT MORILLA GROUP STATISTICAL ANALYSIS
Kyle Kenneth Ruaya
2024-02-23
Data
1. What is the demographic profile of the respondents in terms of:
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Strand
Daily Screen Time
The tables above provides the distributions of respondents in terms of sex, strand and daily screen time. It can be seen that there are 72 females and 28 males; 20 of which are from ABM, 20 from GAS, 40 from HUMSS, and 20 from STEM. Moreover, 26 students have a daily screen time of 1-2 hours, 24 for 3-4 hours, 22 for 5-6 hours, and 28 for more than 6 hours.
2. Is there a significant difference on the variables academic proficiency, physical and mental health, and social interactions when grouped according to:
2.1 Sex
Call:
lm(formula = `Academic Proficiency` ~ `Mental Health` + `Physical Health` +
`Social Interactions`, data = Data)
Coefficients:
(Intercept) `Mental Health` `Physical Health`
1.31602 0.24408 0.18613
`Social Interactions`
0.05491 
From this, we may deduce that the data fail to satisfy two assumptions – Linearity and Homogeneity of Variance.
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filter2.1.1 Sex and Physical Health
Normality Test
Shapiro-Wilk normality test
data: Data$`Physical Health`
W = 0.92897, p-value = 4.352e-05Since p-value = 4.352e-05 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
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Df F value Pr(>F)
group 1 0.5427 0.4631
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
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# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Female Physical Health 72 1.6 3.4 2.8 0.4 2.73 0.329 0.039 0.077
2 Male Physical Health 28 1.6 3.4 2.7 0.4 2.69 0.387 0.073 0.15 The mean of female and male is 2.731 and 2.686, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Physical Health 100 0.471 1 0.492 Kruskal-WallisBased on the p-value, there is no significant difference on the physical health when grouped according to sex.
2.1.2 Sex and Mental Health
Normality Test
Shapiro-Wilk normality test
data: Data$`Mental Health`
W = 0.9454, p-value = 0.000418Since p-value = 0.000418 < 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 0.1865 0.6668
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
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# 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 Female Mental Health 72 2 3.2 2.6 0.4 2.65 0.297 0.035 0.07
2 Male Mental Health 28 2 3.2 2.6 0.45 2.61 0.333 0.063 0.129The mean of female and male is 2.647 and 2.607, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Mental Health 100 0.212 1 0.645 Kruskal-WallisBased on the p-value, there is no significant difference on the mental health when grouped according to sex.
2.1.3 Sex and Academic Proficiency
Normality Test
Shapiro-Wilk normality test
data: Data$`Academic Proficiency`
W = 0.94945, p-value = 0.0007636Since p-value = 0.0007636 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.0236 0.8783
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
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# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Female Academic Profic… 72 1.6 3.2 2.6 0.4 2.60 0.305 0.036 0.072
2 Male Academic Profic… 28 1.8 3.2 2.6 0.45 2.63 0.325 0.061 0.126The mean of female and male is 2.597 and 2.629, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Academic Proficiency 100 0.225 1 0.635 Kruskal-WallisBased on the p-value, there is no significant difference on the physical health when grouped according to sex.
Normality Test
Shapiro-Wilk normality test
data: Data$`Social Interactions`
W = 0.9488, p-value = 0.0006918Since p-value = 0.0006918 < 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 5.35 0.02281 *
98
---
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.
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# 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 Female Social Interact… 72 2 3.2 2.6 0.4 2.57 0.323 0.038 0.076
2 Male Social Interact… 28 1.2 3.8 2.6 0.6 2.54 0.547 0.103 0.212The mean of female and male is 2.572 and 2.536, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Social Interactions 100 0.280 1 0.596 Kruskal-WallisBased on the p-value, there is no significant difference on the physical health when grouped according to sex.
2.2 Daily Screen Time
2.2.1 Daily Screen Time and Physical Health
Normality Test
Shapiro-Wilk normality test
data: Data$`Physical Health`
W = 0.92897, p-value = 4.352e-05Since p-value = 4.352e-05 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 3 1.4341 0.2376
96 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
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# A tibble: 4 × 11
`Daily Screen Time` variable n min max median iqr mean sd se
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1-2 hours Physical… 26 1.6 3.4 2.8 0.6 2.71 0.413 0.081
2 3-4 hours Physical… 24 2 3 2.8 0.4 2.78 0.272 0.056
3 5-6 hours Physical… 22 2 3.4 2.8 0.55 2.73 0.363 0.077
4 More Physical… 28 1.6 3.4 2.6 0.2 2.67 0.328 0.062
# ℹ 1 more variable: ci <dbl>The mean of 1-2 hours, 3-4 hours, 5-6 hours, and more is 2.708, 2.775, 2.727, and 2.671, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Physical Health 100 2.00 3 0.571 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.2.2 Daily Screen Time and Mental Health
Normality Test
Shapiro-Wilk normality test
data: Data$`Mental Health`
W = 0.9454, p-value = 0.000418Since p-value = 0.000418 < 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 0.7506 0.5246
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
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# A tibble: 4 × 11
`Daily Screen Time` variable n min max median iqr mean sd se
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1-2 hours Mental H… 26 2 3.2 2.6 0.4 2.65 0.318 0.062
2 3-4 hours Mental H… 24 2.2 3 2.8 0.4 2.68 0.256 0.052
3 5-6 hours Mental H… 22 2 3.2 2.6 0.55 2.64 0.324 0.069
4 More Mental H… 28 2 3.2 2.6 0.45 2.59 0.331 0.063
# ℹ 1 more variable: ci <dbl>The mean of 1-2 hours, 3-4 hours, 5-6 hours, and more is 2.654, 2.675, 2.636, and 2.586, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Mental Health 100 1.15 3 0.765 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.2.3 Daily Screen Time and Academic Proficiency
Normality Test
Shapiro-Wilk normality test
data: Data$`Academic Proficiency`
W = 0.94945, p-value = 0.0007636Since p-value = 0.0007636 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 3 0.5001 0.6831
96 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
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# A tibble: 4 × 11
`Daily Screen Time` variable n min max median iqr mean sd se
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1-2 hours Academic… 26 2 3.2 2.6 0.35 2.67 0.315 0.062
2 3-4 hours Academic… 24 2 3.2 2.5 0.4 2.53 0.281 0.057
3 5-6 hours Academic… 22 2.2 3 2.8 0.2 2.7 0.26 0.055
4 More Academic… 28 1.6 3.2 2.6 0.4 2.54 0.344 0.065
# ℹ 1 more variable: ci <dbl>The mean of 1-2 hours, 3-4 hours, 5-6 hours, and more is 2.669, 2.533, 2.700, and 2.536, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Academic Proficiency 100 6.75 3 0.0804 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
Normality Test
Shapiro-Wilk normality test
data: Data$`Social Interactions`
W = 0.9488, p-value = 0.0006918Since p-value = 0.0006918 < 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 1.7959 0.1531
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
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# A tibble: 4 × 11
`Daily Screen Time` variable n min max median iqr mean sd se
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1-2 hours Social I… 26 2 3 2.6 0.6 2.53 0.325 0.064
2 3-4 hours Social I… 24 2.2 3 2.6 0.4 2.57 0.255 0.052
3 5-6 hours Social I… 22 2 3.8 2.4 0.75 2.56 0.46 0.098
4 More Social I… 28 1.2 3.8 2.6 0.4 2.59 0.505 0.095
# ℹ 1 more variable: ci <dbl>The mean of 1-2 hours, 3-4 hours, 5-6 hours, and more is 2.531, 2.567, 2.564, and 2.586, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Social Interactions 100 0.766 3 0.858 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
Normality Test
Shapiro-Wilk normality test
data: Data1$Scores
W = 0.95675, p-value = 1.956e-09Since p-value = 1.956e-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 3 1.2869 0.2785
395 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning: Removed 1 rows containing non-finite values (`stat_boxplot()`).
Warning: Removed 1 rows containing non-finite values (`stat_summary()`).Warning: Removed 1 rows containing missing values (`geom_point()`).
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# A tibble: 4 × 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 Prof… Scores 100 1.6 3.2 2.6 0.4 2.61 0.309 0.031 0.061
2 Physical Heal… Scores 100 1.6 3.4 2.8 0.4 2.72 0.345 0.034 0.068
3 Mental Health Scores 100 2 3.2 2.6 0.4 2.64 0.307 0.031 0.061
4 Social Intera… Scores 99 1.2 3.8 2.6 0.4 2.56 0.395 0.04 0.079The mean of Academic Proficiency, Physical Health, Mental Health, and Social Interactions is 2.606, 2.718, 2.636, and 2.558, respectively.
Warning: Removed 1 rows containing non-finite values (`stat_boxplot()`).
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Scores 400 14.2 3 0.00264 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 Scores Academic Pro… Physi… 100 100 2.59 9.52e-3 0.0571 ns
2 Scores Academic Pro… Menta… 100 100 0.531 5.96e-1 1 ns
3 Scores Academic Pro… Socia… 100 99 -1.08 2.81e-1 1 ns
4 Scores Physical Hea… Menta… 100 100 -2.06 3.92e-2 0.235 ns
5 Scores Physical Hea… Socia… 100 99 -3.66 2.48e-4 0.00149 **
6 Scores Mental Health Socia… 100 99 -1.61 1.08e-1 0.648 ns There is significant difference between academic proficiency and physical health, physical and mental health, and physical health and social interactions
4. On which variable have the most significant impact?
Based on the provided output above, we can say that it is the physical health.