KAREL CONGRESO GROUP STATISTICAL ANALYSIS
Kyle Kenneth Ruaya
2024-02-14
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, union2.1Sex
Grade Level
Strand
The tables above provides the distributions of respondents in terms of sex, year level, and strand. It can be seen that there are 54 females and 46 males; 41 of which are from grade 11 and 59 from grade 12. Moreover, 69 said that they used the Science laboratory for 1-2 days every week, 17 said they use it fro 3-4 days, and 14 said 5 days and above.
2. Is there a significant difference on the knowledge and skill, actual application of concepts and alternative learning strategy when grouped according to:
Sex
Call:
lm(formula = `Knowledge and Skill` ~ `Actual Application of Concepts` +
`Alternative Learning Strategy`, data = Data)
Coefficients:
(Intercept) `Actual Application of Concepts`
1.2208 0.2767
`Alternative Learning Strategy`
0.2990 
From this, we may deduce that the data fail to satisfy the two assumptions – Linearity and Homogeneity of Variance.
2.1.1 Sex and “Knowledge and Skill”
`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.
Attaching package: 'rstatix'The following object is masked from 'package:stats':
filterThe mean for male and female is 2.870 and 2.907, respectively.
The above graph shows the plotting of data by sex, which contains two sexes – male and female.
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ℹ This can happen when ggplot fails to infer the correct grouping structure in
the data.
ℹ Did you forget to specify a `group` aesthetic or to convert a numerical
variable into a factor?
It clearly shows that there is no significant difference between the social behavior of the student in terms of mental aspect when grouped according to their sex.
Loading required package: carData
Attaching package: 'car'The following object is masked from 'package:purrr':
someThe following object is masked from 'package:dplyr':
recode
The histogram does not resemble a bell curve as seen above, means that the residuals do not have a normal distribution. Moreover, the points in the QQ-plots roughly follow the straight line, with the majority of them falling within the confidence bands. However, this does not guarantee that residuals follow a normal distribution since when based on the diagram on the left, it is the exact opposite of it. Thus, it is more convenient to observe the two.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.96187, p-value = 0.005479The Shapiro-Wilk p-value = 0.005479 on the residuals is less than the usual significance level of 0.05. Thus, we reject the hypothesis that residuals have a normal distribution.
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 0.001 0.9748
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Wilcoxon Rank Sum Test
Wilcoxon rank sum test
data: a and b
W = 1160, p-value = 0.565
alternative hypothesis: true location shift is not equal to 0Since the p-value is larger than 0.05, we fail to reject the null hypothesis, that is, there is no significant difference on the knowledge and skills of the students when grouped according to sex.
2.1.2 Sex and Actual Application of Concepts
`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.# A tibble: 16 × 3
# Groups: Sex [2]
Sex `Actual Application of Concepts` count
<fct> <dbl> <int>
1 Female 2.6 2
2 Female 2.8 1
3 Female 3 20
4 Female 3.2 11
5 Female 3.4 5
6 Female 3.6 5
7 Female 3.8 7
8 Female 4 3
9 Male 2.6 2
10 Male 2.8 2
11 Male 3 12
12 Male 3.2 12
13 Male 3.4 9
14 Male 3.6 4
15 Male 3.8 2
16 Male 4 3The mean for male and female is 3.257 and 3.274, respectively.
The above graph shows the plotting of data by sex, which contains two sexes – male and female.
Warning: The following aesthetics were dropped during statistical transformation: fill
ℹ This can happen when ggplot fails to infer the correct grouping structure in
the data.
ℹ Did you forget to specify a `group` aesthetic or to convert a numerical
variable into a factor?
The histogram does not resemble a bell curve as seen above, means that the residuals do not have a normal distribution. Moreover, the points in the QQ-plots roughly follow the straight line, with the majority of them falling within the confidence bands. However, this does not guarantee that residuals follow a normal distribution since when based on the diagram on the left, it is the exact opposite of it. Thus, it is more convenient to observe the two.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.92654, p-value = 3.184e-05The Shapiro-Wilk p-value = 3.184e-05 on the residuals is less than the usual significance level of 0.05. Thus, we reject the hypothesis that residuals have a normal distribution.
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 0.5327 0.4672
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Wilcoxon Rank Sum Test
Wilcoxon rank sum test with continuity correction
data: c and d
W = 1252, p-value = 0.9463
alternative hypothesis: true location shift is not equal to 0Since the p-value= 0.9463 is greater than 0.05, we fail to reject the null hypothesis. Hence, there is no significant difference on the actual application of concepts when grouped according to their sex.
2.1.3 Sex and Alternative Learning Strategy
`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.# A tibble: 15 × 3
# Groups: Sex [2]
Sex `Alternative Learning Strategy` count
<fct> <dbl> <int>
1 Female 2 6
2 Female 2.2 5
3 Female 2.4 12
4 Female 2.6 13
5 Female 2.8 6
6 Female 3 10
7 Female 3.2 2
8 Male 2 6
9 Male 2.2 8
10 Male 2.4 6
11 Male 2.6 10
12 Male 2.8 9
13 Male 3 4
14 Male 3.2 1
15 Male 3.4 2The mean for male and female is 2.548 and 2.570, respectively.
The above graph shows the plotting of data by sex, which contains two sexes – male and female.
The histogram does not resemble a bell curve as seen above, means that the residuals do not have a normal distribution. Moreover, the points in the QQ-plots roughly follow the straight line, with the majority of them falling within the confidence bands. However, this does not guarantee that residuals follow a normal distribution since when based on the diagram on the left, it is the exact opposite of it. Thus, it is more convenient to observe the two.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.95986, p-value = 0.003933The Shapiro-Wilk p-value = 0.003933 on the residuals is less than the usual significance level of 0.05. Thus, we reject the hypothesis that residuals have a normal distribution.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.4587 0.4998
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Wilcoxon Rank Sum Test
Wilcoxon rank sum test with continuity correction
data: e and f
W = 1190, p-value = 0.718
alternative hypothesis: true location shift is not equal to 0Since the p-value= 0.718 is greater than 0.05, we fail to reject the null hypothesis. Hence, there is no significant difference alternative learning strategy when grouped according to their sex.
2.2 Days
2.2.1 Days and “Knowledge and Skill”
Normality Test
Shapiro-Wilk normality test
data: Data$`Knowledge and Skill`
W = 0.95223, p-value = 0.001167Since p-value = 0.001167 < 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 0.4792 0.6207
97 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: 3 × 11
Days variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1-2 Days Knowled… 69 2 4 2.8 0.4 2.89 0.409 0.049 0.098
2 3-4 Days Knowled… 17 2.4 4 3 0.6 2.98 0.435 0.106 0.224
3 5 Days and A… Knowled… 14 2 3.6 2.8 0.3 2.8 0.392 0.105 0.226The mean of 1-2 Days, 3-4 Days, and 5 Days and Above is 2.887, 2.976, and 2.800, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Knowledge and Skill 100 0.785 2 0.675 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 Knowledge and Sk… 1-2 D… 3-4 D… 69 17 0.657 0.511 1 ns
2 Knowledge and Sk… 1-2 D… 5 Day… 69 14 -0.465 0.642 1 ns
3 Knowledge and Sk… 3-4 D… 5 Day… 17 14 -0.870 0.384 1 ns 2.2.2 Days and Actual Application of Concepts
Normality Test
Shapiro-Wilk normality test
data: Data$`Actual Application of Concepts`
W = 0.91771, p-value = 1.066e-05Since p-value = 1.066e-05 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 2 0.5549 0.5759
97 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# A tibble: 3 × 11
Days variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1-2 Days Actual … 69 2.6 4 3.2 0.4 3.27 0.324 0.039 0.078
2 3-4 Days Actual … 17 2.6 4 3.2 0.6 3.38 0.418 0.101 0.215
3 5 Days and A… Actual … 14 2.6 3.8 3.1 0.2 3.1 0.321 0.086 0.185The mean of 1-2 Days, 3-4 Days, and 5 Days and Above is 3.272, 3.376, and 3.100, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Actual Application of Concepts 100 4.08 2 0.13 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 3 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Actual Applicat… 1-2 D… 3-4 D… 69 17 1.05 0.292 0.876 ns
2 Actual Applicat… 1-2 D… 5 Day… 69 14 -1.50 0.133 0.400 ns
3 Actual Applicat… 3-4 D… 5 Day… 17 14 -2.01 0.0445 0.134 ns There is a significant difference between 3-4 Days and 5 Days and Above.
2.2.3 Strand and Alternative Learning Strategy
Normality Test
Shapiro-Wilk normality test
data: Data$`Alternative Learning Strategy`
W = 0.95127, p-value = 0.001006Since p-value = 0.001006 < 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 0.5479 0.5799
97 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
Days variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 1-2 Days Alterna… 69 2 3.4 2.6 0.4 2.55 0.331 0.04 0.08
2 3-4 Days Alterna… 17 2 3.4 2.6 0.6 2.6 0.424 0.103 0.218
3 5 Days and A… Alterna… 14 2 3.2 2.6 0.4 2.56 0.369 0.099 0.213The mean of 1-2 Days, 3-4 Days, and 5 Days and Above is 2.551, 2.600, and 2.557, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Alternative Learning Strategy 100 0.239 2 0.887 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 Alternative Lear… 1-2 D… 3-4 D… 69 17 0.481 0.631 1 ns
2 Alternative Lear… 1-2 D… 5 Day… 69 14 0.177 0.860 1 ns
3 Alternative Lear… 3-4 D… 5 Day… 17 14 -0.217 0.828 1 ns 3. Is there a significant relationship between knowledge and skill, actual application of concepts, and alternative learning strategy?
Normality Test
Shapiro-Wilk normality test
data: Data1$Scores
W = 0.97114, p-value = 1.008e-05Since p-value = 1.008e-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 2 1.0458 0.3527
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 Knowledge and… Scores 100 2 4 2.8 0.6 2.89 0.41 0.041 0.081
2 Actual Applic… Scores 100 2.6 4 3.2 0.4 3.27 0.346 0.035 0.069
3 Alternative L… Scores 100 2 3.4 2.6 0.45 2.56 0.35 0.035 0.07 The mean of knowledge and skill, actual application of concepts, and alternative learning strategy is 2.890, 3.266, and 2.560, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Scores 300 122. 2 3.62e-27 Kruskal-WallisBased on the p-value, there is a 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 Knowledge … Actua… 100 100 6.09 1.12e- 9 3.37e- 9 ****
2 Scores Knowledge … Alter… 100 100 -4.92 8.52e- 7 2.56e- 6 ****
3 Scores Actual App… Alter… 100 100 -11.0 3.27e-28 9.80e-28 **** Pairwaise, there is significant difference.
4. Which have the most significant impact?
Based on the provided output above, it can be seen that actual application of concepts is the most challenging factor.