SHEINA CORDITA GROUP STATISTICAL ANALYSIS
Kyle Kenneth Ruaya
2024-02-19
Data
1. What is the demographic profile of the respondents in terms of:
Attaching package: 'dplyr'The following objects are masked from 'package:stats':
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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 58 females and 42 males; 24 of which are from ABM, 23 from GAS, 37 from HUMSS, and 16 from STEM.
2. Is there a significant difference on the variables background knowledge and vocabulary when grouped according to:
2.1 Sex
Call:
lm(formula = `Background Knowledge` ~ Vocabulary, data = Data)
Coefficients:
(Intercept) Vocabulary
1.4028 0.5336 
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 Background Knowledge
Normality Test
Shapiro-Wilk normality test
data: Data$`Background Knowledge`
W = 0.97339, p-value = 0.04043Since p-value = 0.04043 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
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Df F value Pr(>F)
group 1 0.2384 0.6265
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
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# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 FEMALE Background Know… 58 2 4 3 0.938 3 0.56 0.074 0.147
2 MALE Background Know… 42 1.5 4 2.88 0.75 2.83 0.604 0.093 0.188The mean of female and male is 3.000 and 2.827, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Background Knowledge 100 1.40 1 0.237 Kruskal-WallisBased on the p-value, there is no significant difference on the Background Knowledge when grouped according to sex.
2.1.2 Sex and Vocabulary
Normality Test
Shapiro-Wilk normality test
data: Data$Vocabulary
W = 0.96332, p-value = 0.006982Since p-value = 0.006982 < 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 0.5562 0.4576
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 Vocabulary 58 1.75 4 2.75 0.75 2.86 0.554 0.073 0.146
2 MALE Vocabulary 42 1.75 4 2.75 0.75 2.85 0.595 0.092 0.185The mean of female and male is 2.862 and 2.851, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Vocabulary 100 0.000609 1 0.98 Kruskal-WallisBased on the p-value, there is no significant difference on the Vocabulary when grouped according to sex.
2.2 Strand
2.2.1 Strand and Background Knowledge
Normality Test
Shapiro-Wilk normality test
data: Data$`Background Knowledge`
W = 0.97339, p-value = 0.04043Since p-value = 0.04043 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 3 1.233 0.302
96 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
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# A tibble: 4 × 11
Strand variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 STEM Background Know… 16 1.5 4 3.12 1 2.95 0.759 0.19 0.405
2 ABM Background Know… 24 2 4 3 0.625 2.98 0.566 0.116 0.239
3 HUMSS Background Know… 37 1.75 4 2.75 1 2.78 0.555 0.091 0.185
4 GAS Background Know… 23 2.25 4 3 0.625 3.10 0.469 0.098 0.203The mean of STEM, ABM, HUMSS, and GAS is 2.953, 2.979, 2.777, and 3.098, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Background Knowledge 100 4.99 3 0.173 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 6 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Background Know… STEM ABM 16 24 -0.285 0.776 1 ns
2 Background Know… STEM HUMSS 16 37 -1.47 0.143 0.855 ns
3 Background Know… STEM GAS 16 23 0.316 0.752 1 ns
4 Background Know… ABM HUMSS 24 37 -1.32 0.186 1 ns
5 Background Know… ABM GAS 24 23 0.668 0.504 1 ns
6 Background Know… HUMSS GAS 37 23 2.04 0.0413 0.248 ns There is significant difference between HUMMS and GAS.
2.2.2 Strand and Vocabulary
Normality Test
Shapiro-Wilk normality test
data: Data$Vocabulary
W = 0.96332, p-value = 0.006982Since p-value = 0.006982 < 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 0.5704 0.6359
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
Strand variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 STEM Vocabulary 16 1.75 3.5 3.25 0.812 2.97 0.531 0.133 0.283
2 ABM Vocabulary 24 2 3.75 2.75 0.562 2.78 0.496 0.101 0.21
3 HUMSS Vocabulary 37 1.75 4 2.75 0.75 2.82 0.609 0.1 0.203
4 GAS Vocabulary 23 2 4 2.75 0.75 2.91 0.615 0.128 0.266The mean of STEM, ABM, HUMSS, and GAS is 2.969, 2.781, 2.824, and 2.913, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Vocabulary 100 1.49 3 0.685 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 STEM ABM 16 24 -1.18 0.240 1 ns
2 Vocabulary STEM HUMSS 16 37 -0.921 0.357 1 ns
3 Vocabulary STEM GAS 16 23 -0.589 0.556 1 ns
4 Vocabulary ABM HUMSS 24 37 0.395 0.693 1 ns
5 Vocabulary ABM GAS 24 23 0.643 0.520 1 ns
6 Vocabulary HUMSS GAS 37 23 0.316 0.752 1 ns 3. Is there a significant relationship between background knowledge and vocabulary?
Normality Test
Shapiro-Wilk normality test
data: Data1$Scores
W = 0.97258, p-value = 0.0005945Since p-value = 0.0005945 < 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 0.0966 0.7563
198 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: 2 × 11
Variables variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Background Kn… Scores 100 1.5 4 3 0.75 2.93 0.582 0.058 0.115
2 Vocabulary Scores 100 1.75 4 2.75 0.75 2.86 0.569 0.057 0.113The mean of physical health and mental health is 2.928 and 2.857, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Scores 200 0.820 1 0.365 Kruskal-WallisBased on the p-value, there is no significant difference between physical health and mental health.
4. Which variable have the most significant impact towards comprehension in English language?
Based on the provided output above, we can say that it is the Background Knowledge.