NICOLE EGUNA GROUP STATISTICAL ANALYSIS
Kyle Kenneth Ruaya
2024-02-20
Data
1. What is the demographic profile of the respondents in terms of:
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intersect, setdiff, setequal, unionAge
Sex
Strand
Frequency
The tables above provides the distributions of respondents in terms of age, sex, strand and frequency. It can be seen that there are 33 students ages 15-16 years old and 67 students ages 17-18 years old; 25 of which were females and 75 were males. Moreover, there are 21 students each from ABM, GAS, and STEM, and 37 students from HUMSS.
2. Is there a significant difference on the technological aspects, critical thinking, and vocabulary when grouped according to:
2.1 Sex
Call:
lm(formula = `Technological Aspects` ~ `Critical Thinking` +
Vocabulary, data = Data)
Coefficients:
(Intercept) `Critical Thinking` Vocabulary
1.4348 0.2650 0.2349 
From this, we may deduce that the data fail to satisfy the two assumptions – Linearity and Homogeneity of Variance.
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filter2.1.1 Sex and Technological Aspects
Normality Test
Shapiro-Wilk normality test
data: Data$`Technological Aspects`
W = 0.91148, p-value = 5.111e-06Since p-value = 5.111e-06 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.7565 0.3865
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
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# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Technological A… 75 1.4 3.6 3 0.2 2.89 0.418 0.048 0.096
2 Female Technological A… 25 1.8 3.6 2.8 0.4 2.78 0.432 0.086 0.178The mean of male and female is 2.888 and 2.784, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Technological Aspects 100 1.63 1 0.202 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.1.2 Sex and Critical Thinking
Normality Test
Shapiro-Wilk normality test
data: Data$`Critical Thinking`
W = 0.96261, p-value = 0.006197Since p-value = 0.006197 < 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 0.0604 0.8063
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
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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 Male Critical Thinki… 75 1.8 3.6 2.8 0.6 2.70 0.45 0.052 0.104
2 Female Critical Thinki… 25 1.8 3.6 2.8 0.6 2.81 0.471 0.094 0.194The mean of male and female is 2.701 and 2.808, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Critical Thinking 100 1.23 1 0.268 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.1.3 Sex and Vocabulary
Normality Test
Shapiro-Wilk normality test
data: Data$Vocabulary
W = 0.9378, p-value = 0.0001419Since p-value = 0.0001419 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.0261 0.8719
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
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# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Vocabulary 75 1.8 4 3 0.5 3.01 0.497 0.057 0.114
2 Female Vocabulary 25 1.8 3.6 3 0.4 2.95 0.539 0.108 0.223The mean of male and female is 3.013 and 2.952, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Vocabulary 100 0.0289 1 0.865 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.2 Frequency
2.2.1 Frequency and Technological Aspects
Normality Test
Shapiro-Wilk normality test
data: Data$`Technological Aspects`
W = 0.91148, p-value = 5.111e-06Since p-value = 5.111e-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 0.8691 0.46
96 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in stats::qt(ci/2 + 0.5, data_sum$length - 1): NaNs produced
Warning in qt((1 + p)/2, ns - 1): NaNs producedWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
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Warning: There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning:
! There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning in `stats::qt()`:
! NaNs produced# A tibble: 4 × 11
Frequency variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Always Technolo… 26 1.4 3.6 3 0.4 2.82 0.497 0.098 0.201
2 Sometimes Technolo… 70 1.8 3.6 3 0.35 2.90 0.386 0.046 0.092
3 Seldom Technolo… 3 2.2 2.6 2.2 0.2 2.33 0.231 0.133 0.574
4 Never Technolo… 1 3 3 3 0 3 NA NA NaN The mean of always, sometimes, seldom, and never is 2.823, 2.897, 2.333, and 3.000, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Technological Aspects 100 5.40 3 0.145 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 6 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Technological A… Always Somet… 26 70 0.496 0.620 1 ns
2 Technological A… Always Seldom 26 3 -2.03 0.0423 0.254 ns
3 Technological A… Always Never 26 1 0.306 0.759 1 ns
4 Technological A… Somet… Seldom 70 3 -2.29 0.0218 0.131 ns
5 Technological A… Somet… Never 70 1 0.197 0.844 1 ns
6 Technological A… Seldom Never 3 1 1.34 0.179 1 ns There is significant difference between always and seldom so with sometimes and seldom.
2.2.2 Frequency and Critical Thinking
Normality Test
Shapiro-Wilk normality test
data: Data$`Critical Thinking`
W = 0.96261, p-value = 0.006197Since p-value = 0.006197 < 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 0.6342 0.5947
96 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in stats::qt(ci/2 + 0.5, data_sum$length - 1): NaNs produced
Warning in qt((1 + p)/2, ns - 1): NaNs producedWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
Warning: There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning:
! There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning in `stats::qt()`:
! NaNs produced# A tibble: 4 × 11
Frequency variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Always Critical… 26 1.8 3.4 2.8 0.6 2.77 0.472 0.093 0.191
2 Sometimes Critical… 70 1.8 3.6 2.8 0.6 2.73 0.449 0.054 0.107
3 Seldom Critical… 3 1.8 2.6 2.6 0.4 2.33 0.462 0.267 1.15
4 Never Critical… 1 2.4 2.4 2.4 0 2.4 NA NA NaN The mean of always, sometimes, seldom, and never is 2.769, 2.734, 2.333, and 2.400, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Critical Thinking 100 3.45 3 0.327 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 6 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Critical Thinking Always Somet… 26 70 -0.605 0.545 1 ns
2 Critical Thinking Always Seldom 26 3 -1.63 0.103 0.616 ns
3 Critical Thinking Always Never 26 1 -0.988 0.323 1 ns
4 Critical Thinking Somet… Seldom 70 3 -1.45 0.146 0.878 ns
5 Critical Thinking Somet… Never 70 1 -0.862 0.389 1 ns
6 Critical Thinking Seldom Never 3 1 -0.0100 0.992 1 ns 2.2.3 Frequency and Vocabulary
Normality Test
Shapiro-Wilk normality test
data: Data$Vocabulary
W = 0.9378, p-value = 0.0001419Since p-value = 0.0001419 < 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 0.7744 0.5111
96 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in stats::qt(ci/2 + 0.5, data_sum$length - 1): NaNs produced
Warning in qt((1 + p)/2, ns - 1): NaNs producedWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
Warning: There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning:
! There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning in `stats::qt()`:
! NaNs produced# A tibble: 4 × 11
Frequency variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Always Vocabula… 26 2.2 4 3 0.35 3.07 0.404 0.079 0.163
2 Sometimes Vocabula… 70 1.8 4 3 0.6 2.98 0.54 0.064 0.129
3 Seldom Vocabula… 3 2.2 3.2 2.6 0.5 2.67 0.503 0.291 1.25
4 Never Vocabula… 1 3.4 3.4 3.4 0 3.4 NA NA NaN The mean of always, sometimes, seldom, and never is 3.069, 2.980, 2.667, and 3.400, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Vocabulary 100 2.71 3 0.439 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 6 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Vocabulary Always Sometimes 26 70 -0.481 0.630 1 ns
2 Vocabulary Always Seldom 26 3 -1.22 0.222 1 ns
3 Vocabulary Always Never 26 1 0.979 0.328 1 ns
4 Vocabulary Sometimes Seldom 70 3 -1.08 0.282 1 ns
5 Vocabulary Sometimes Never 70 1 1.10 0.271 1 ns
6 Vocabulary Seldom Never 3 1 1.51 0.131 0.789 ns There is significant difference between always and seldom so with sometimes and seldom.
3. Is there a significant relationship between technological aspects, critical thinking, and vocabulary?
Normality Test
Shapiro-Wilk normality test
data: Data1$Scores
W = 0.95572, p-value = 6.888e-08Since p-value = 6.888e-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 1.4362 0.2395
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
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# 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 Technological… Scores 100 1.4 3.6 3 0.4 2.86 0.421 0.042 0.084
2 Critical Thin… Scores 100 1.8 3.6 2.8 0.6 2.73 0.455 0.046 0.09
3 Vocabulary Scores 100 1.8 4 3 0.6 3.00 0.506 0.051 0.1 The mean of Technological Aspects, Critical Thinking and Vocabulary is 2.862, 2.728, and 2.998,respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Scores 300 18.3 2 0.000107 Kruskal-WallisBased on the p-value, there is 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 Technologica… Criti… 100 100 -2.18 2.95e-2 8.86e-2 ns
2 Scores Technologica… Vocab… 100 100 2.10 3.57e-2 1.07e-1 ns
3 Scores Critical Thi… Vocab… 100 100 4.28 1.90e-5 5.69e-5 **** There is significant difference on all pairs of groups.
4. Which have the most significant impact?
Based on the provided output above, we can say that it is the vocabulary.