BREN NASAYAO GROUP STATISTICAL ANALYSIS
Kyle Kenneth Ruaya
2024-02-18
Data
1. What is the demographic profile of the respondents in terms of:
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Strand
The tables above provides the distributions of respondents in terms of sex, year level, and strand. It can be seen that there are 48 females and 52 males; 25 of which are from each strand.
2. Is there a significant difference on the variables behavioral changes, college readiness, and academic performance when grouped according to:
2.1 Sex
Call:
lm(formula = `Behavioral Changes` ~ `College Readiness` + `Academic Performance`,
data = Data)
Coefficients:
(Intercept) `College Readiness` `Academic Performance`
0.1190 0.4656 0.3849 
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 Behavioral Changes
Normality Test
Shapiro-Wilk normality test
data: Data$`Behavioral Changes`
W = 0.95905, p-value = 0.003444Since p-value = 0.003444 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 5.3162 0.02324 *
98
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The p-value is less than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is not met.
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# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Behavioral Chan… 52 1 2.4 1.8 0.25 1.70 0.349 0.048 0.097
2 Female Behavioral Chan… 48 1 3 1.7 0.8 1.70 0.458 0.066 0.133The mean of female and male is 1.696 and 1.704, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Behavioral Changes 100 0.0226 1 0.881 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.1.2 Sex and College Readiness
Normality Test
Shapiro-Wilk normality test
data: Data$`College Readiness`
W = 0.95669, p-value = 0.002351Since p-value = 0.002351 < 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 0.1055 0.746
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 College Readine… 52 1 2.8 1.8 0.4 1.82 0.373 0.052 0.104
2 Female College Readine… 48 1.2 3 2 0.45 1.84 0.376 0.054 0.109The mean of female and male is 1.823 and 1.842, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 College Readiness 100 0.0606 1 0.806 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.1.3 Sex and Academic Performance
Normality Test
Shapiro-Wilk normality test
data: Data$`Academic Performance`
W = 0.94744, p-value = 0.000565Since p-value = 0.000565 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.287 0.5933
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
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# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Academic Perfor… 52 1 3.2 2 0.4 1.92 0.357 0.05 0.099
2 Female Academic Perfor… 48 1 2.6 2 0.4 1.86 0.372 0.054 0.108The mean of female and male is 1.919 and 1.862, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Academic Performance 100 0.239 1 0.625 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.2 Strand
2.2.1 Strand and Behavioral Changes
Normality Test
Shapiro-Wilk normality test
data: Data$`Behavioral Changes`
W = 0.95905, p-value = 0.003444Since p-value = 0.003444 < 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 0.2209 0.8817
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 Behavioral Chan… 25 1.4 3 2 0.4 1.99 0.381 0.076 0.157
2 ABM Behavioral Chan… 25 1 2.2 1.4 0.6 1.49 0.361 0.072 0.149
3 HUMSS Behavioral Chan… 25 1 2.2 1.8 0.4 1.71 0.337 0.067 0.139
4 GAS Behavioral Chan… 25 1 2.2 1.6 0.6 1.61 0.367 0.073 0.152The mean of STEM, ABM, HUMSS, and GAS is 1.992, 1.488, 1.712, and 1.608, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Behavioral Changes 100 18.3 3 0.000389 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 Behavioral C… STEM ABM 25 25 -4.11 3.89e-5 2.33e-4 ***
2 Behavioral C… STEM HUMSS 25 25 -2.11 3.52e-2 2.11e-1 ns
3 Behavioral C… STEM GAS 25 25 -3.02 2.57e-3 1.54e-2 *
4 Behavioral C… ABM HUMSS 25 25 2.01 4.47e-2 2.68e-1 ns
5 Behavioral C… ABM GAS 25 25 1.10 2.72e-1 1 e+0 ns
6 Behavioral C… HUMSS GAS 25 25 -0.909 3.64e-1 1 e+0 ns There is significant difference on the first four group pairs.
2.2.2 Strand and College Readiness
Normality Test
Shapiro-Wilk normality test
data: Data$`College Readiness`
W = 0.95669, p-value = 0.002351Since p-value = 0.002351 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 3 0.6172 0.6055
96 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
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# A tibble: 4 × 11
Strand variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 STEM College Readine… 25 1.2 3 2 0.6 1.94 0.398 0.08 0.164
2 ABM College Readine… 25 1.2 2.6 1.6 0.4 1.73 0.326 0.065 0.135
3 HUMSS College Readine… 25 1 2.6 2 0.4 1.88 0.351 0.07 0.145
4 GAS College Readine… 25 1 2.8 1.8 0.4 1.78 0.393 0.079 0.162The mean of STEM, ABM, HUMSS, and GAS is 1.994, 1.728, 1.880, and 1.776, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 College Readiness 100 6.24 3 0.101 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 6 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 College Readine… STEM ABM 25 25 -2.16 0.0306 0.184 ns
2 College Readine… STEM HUMSS 25 25 -0.333 0.739 1 ns
3 College Readine… STEM GAS 25 25 -1.57 0.116 0.698 ns
4 College Readine… ABM HUMSS 25 25 1.83 0.0674 0.405 ns
5 College Readine… ABM GAS 25 25 0.591 0.554 1 ns
6 College Readine… HUMSS GAS 25 25 -1.24 0.216 1 ns There is significant difference between STEM and ABM.
2.2.3 Strand and Academic Performance
Normality Test
Shapiro-Wilk normality test
data: Data$`Academic Performance`
W = 0.94744, p-value = 0.000565Since p-value = 0.000565 < 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 0.4868 0.6923
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
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 Perfor… 25 1.2 3.2 2 0.4 2.03 0.435 0.087 0.179
2 ABM Academic Perfor… 25 1 2.4 1.8 0.4 1.81 0.358 0.072 0.148
3 HUMSS Academic Perfor… 25 1.4 2.6 2 0.2 1.93 0.294 0.059 0.121
4 GAS Academic Perfor… 25 1 2.4 1.8 0.4 1.8 0.321 0.064 0.133The mean of STEM, ABM, HUMSS, and GAS is 2.032, 1.808, 1.928, and 1.800, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Academic Performance 100 5.61 3 0.132 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 Perfor… STEM ABM 25 25 -1.89 0.0594 0.356 ns
2 Academic Perfor… STEM HUMSS 25 25 -0.809 0.419 1 ns
3 Academic Perfor… STEM GAS 25 25 -2.06 0.0391 0.235 ns
4 Academic Perfor… ABM HUMSS 25 25 1.08 0.282 1 ns
5 Academic Perfor… ABM GAS 25 25 -0.178 0.859 1 ns
6 Academic Perfor… HUMSS GAS 25 25 -1.25 0.210 1 ns There is significant difference between STEM and GAS.
3. Is there a significant relationship between behavioral changes, college readiness, and academic performance?
Normality Test
Shapiro-Wilk normality test
data: Data1$Scores
W = 0.96224, p-value = 4.934e-07Since p-value = 4.934e-07 < 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 1.3236 0.2677
297 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: 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 Behavioral Ch… Scores 100 1 3 1.8 0.6 1.7 0.403 0.04 0.08
2 College Readi… Scores 100 1 3 1.8 0.4 1.83 0.373 0.037 0.074
3 Academic Perf… Scores 100 1 3.2 2 0.4 1.89 0.363 0.036 0.072The mean of Behavioral Changes, College Readiness, and Academic Performance is 1.700, 1.832, and 1.892, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Scores 300 12.7 2 0.00176 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 Behavioral C… Colle… 100 100 2.27 2.32e-2 0.0695 ns
2 Scores Behavioral C… Acade… 100 100 3.51 4.45e-4 0.00134 **
3 Scores College Read… Acade… 100 100 1.24 2.14e-1 0.643 ns There is significant performance between behavioral changes and college readiness so with behavioral changes and academic performance.
4. On which variable does additional two year program on career development for SHS have the most significant impact?
Based on the provided output above, we can say that it is the academic performance.