CHEGIE CORDITA 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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Year Level
Program
The tables above provides the distributions of respondents in terms of sex, year level, and program It can be seen that there are 48 females and 52 males; 28 of which are 1st year students, 20 are 2nd year, 38 are 3rd year, and 14 are 4th year students. Moreover, there are 6 BEED students, 33 BSCrim, 18 BSED, 12 BSIT, 22 BSOA, and 9 BSTM.
2. Is there a significant difference on the variables financial status, physical and mental health, and time management when grouped according to:
2.1 Sex
Call:
lm(formula = `Financial Status` ~ `Time Management` + `Physical and Mental Health`,
data = Data)
Coefficients:
(Intercept) `Time Management`
1.80782 -0.05985
`Physical and Mental Health`
0.38955 
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 Financial Status
Normality Test
Shapiro-Wilk normality test
data: Data$`Financial Status`
W = 0.97266, p-value = 0.03545Since p-value = 0.03545 < 0.05, it is conclusive that we reject the null hypothesis. That is, we cannot assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.0804 0.7773
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
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# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Financial Status 52 1.4 3.8 2.6 0.6 2.73 0.527 0.073 0.147
2 Female Financial Status 48 1.4 4 3 0.6 2.85 0.559 0.081 0.162The mean of male and female is 2.727 and 2.850, respectively.
Mann Whitney U Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Financial Status 100 1.98 1 0.159 Kruskal-WallisBased on the p-value, there is no significant difference was observed between the group pairs.
2.1.2 Sex and Physical and Mental Health
Normality Test
Shapiro-Wilk normality test
data: Data$`Physical and Mental Health`
W = 0.9775, p-value = 0.0848Since p-value = 0.0848 > 0.05, it is conclusive that we fail to reject the null hypothesis. That is, we 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 0.7688 0.3827
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 Physical and Me… 52 1.6 4 2.7 0.8 2.82 0.587 0.081 0.163
2 Female Physical and Me… 48 1.4 4 3 0.6 2.95 0.559 0.081 0.162The mean of female and male is 2.823 and 2.946, respectively.
Two Sample T-test
Welch Two Sample t-test
data: a and b
t = -1.0708, df = 97.897, p-value = 0.2869
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.3502528 0.1047400
sample estimates:
mean of x mean of y
2.823077 2.945833 Based on the p-value, there is no significant difference was observed between the group pairs.
2.1.3 Sex and Time Management
Normality Test
Shapiro-Wilk normality test
data: Data$`Time Management`
W = 0.98001, p-value = 0.1331Since p-value = 0.1331 > 0.05, it is conclusive that we fail to reject the null hypothesis. That is, we assume normality.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 1 0.5501 0.46
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
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# A tibble: 2 × 11
Sex variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Male Time Management 52 1.2 4 2.4 0.65 2.40 0.584 0.081 0.163
2 Female Time Management 48 1 4 2.4 0.8 2.43 0.608 0.088 0.177The mean of female and male is 2.396 and 2.433, respectively.
Two Sample T-test
Welch Two Sample t-test
data: c and d
t = -0.24913, df = 96.946, p-value = 0.8038
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.2650514 0.2059319
sample estimates:
mean of x mean of y
2.403774 2.433333 Based on the p-value, there is no significant difference was observed between the group pairs.
2.2 Strand
2.2.1 Program and Physical and Financial Status
Normality Test
Shapiro-Wilk normality test
data: Data$`Financial Status`
W = 0.97266, p-value = 0.03545Since p-value = 0.03545 < 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 5 1.0768 0.3783
94 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: 6 × 11
Program variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 BEED Financial Stat… 6 2.2 3 2.7 0.5 2.67 0.327 0.133 0.343
2 BSED Financial Stat… 18 2.6 4 3.2 0.7 3.2 0.423 0.1 0.21
3 BSOA Financial Stat… 22 1.4 3.8 2.6 0.8 2.66 0.625 0.133 0.277
4 BSIT Financial Stat… 12 1.8 3.2 2.6 0.35 2.6 0.391 0.113 0.248
5 BSTM Financial Stat… 9 2.4 3.6 3 0.4 3.09 0.362 0.121 0.278
6 BSCRIM Financial Stat… 33 1.4 3.8 2.6 0.8 2.65 0.541 0.094 0.192The mean of BEED, BSED, BSOA, BSIT, BSTM, and BSCRIM is 2.667, 3.200, 2.664, 2.600, 3.089, and 2.648 respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Financial Status 100 19.6 5 0.00147 Kruskal-WallisBased on the p-value, there is significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 15 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Financial S… BEED BSED 6 18 2.26 2.39e-2 0.359 ns
2 Financial S… BEED BSOA 6 22 0.281 7.79e-1 1 ns
3 Financial S… BEED BSIT 6 12 -0.180 8.57e-1 1 ns
4 Financial S… BEED BSTM 6 9 1.75 7.98e-2 1 ns
5 Financial S… BEED BSCRIM 6 33 0.0434 9.65e-1 1 ns
6 Financial S… BSED BSOA 18 22 -2.94 3.26e-3 0.0489 *
7 Financial S… BSED BSIT 18 12 -3.10 1.95e-3 0.0293 *
8 Financial S… BSED BSTM 18 9 -0.346 7.30e-1 1 ns
9 Financial S… BSED BSCRIM 18 33 -3.57 3.61e-4 0.00541 **
10 Financial S… BSOA BSIT 22 12 -0.611 5.41e-1 1 ns
11 Financial S… BSOA BSTM 22 9 2.01 4.48e-2 0.672 ns
12 Financial S… BSOA BSCRIM 22 33 -0.400 6.89e-1 1 ns
13 Financial S… BSIT BSTM 12 9 2.30 2.16e-2 0.324 ns
14 Financial S… BSIT BSCRIM 12 33 0.324 7.46e-1 1 ns
15 Financial S… BSTM BSCRIM 9 33 -2.40 1.62e-2 0.243 ns There is significant difference between BEED and BSED, BSED and BSOA, BSED and BSIT, and BSED and BSCRIM.
2.2.2 Program and Physical and Physical and Mental Health
Normality Test
Shapiro-Wilk normality test
data: Data$`Physical and Mental Health`
W = 0.9775, p-value = 0.0848Since p-value = 0.0848 > 0.05, it is conclusive that we reject the null hypothesis. That is, we 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 5 1.0928 0.3696
94 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: 6 × 11
Program variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 BEED Physical and M… 6 2.4 3.4 2.7 0.35 2.8 0.358 0.146 0.375
2 BSED Physical and M… 18 2.2 3.8 3 0.55 3.02 0.46 0.108 0.229
3 BSOA Physical and M… 22 1.8 4 3 0.85 3.04 0.644 0.137 0.286
4 BSIT Physical and M… 12 1.4 3.6 2.8 1.05 2.72 0.695 0.201 0.442
5 BSTM Physical and M… 9 2.2 3.6 3.2 1 3.04 0.555 0.185 0.426
6 BSCRIM Physical and M… 33 1.6 4 2.6 0.6 2.73 0.547 0.095 0.194The mean of BEED, BSED, BSOA, BSIT, BSTM, and BSCRIM is 2.800, 3.022, 3.045, 2.717, 3.044, and 2.727 respectively.
One Way ANOVA
Df Sum Sq Mean Sq F value Pr(>F)
Program 5 2.338 0.4675 1.45 0.214
Residuals 94 30.310 0.3224 Based on the p-value, there is no significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 15 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Physical and M… BEED BSED 6 18 0.953 0.341 1 ns
2 Physical and M… BEED BSOA 6 22 1.00 0.317 1 ns
3 Physical and M… BEED BSIT 6 12 -0.0694 0.945 1 ns
4 Physical and M… BEED BSTM 6 9 0.858 0.391 1 ns
5 Physical and M… BEED BSCRIM 6 33 -0.290 0.772 1 ns
6 Physical and M… BSED BSOA 18 22 0.0375 0.970 1 ns
7 Physical and M… BSED BSIT 18 12 -1.30 0.194 1 ns
8 Physical and M… BSED BSTM 18 9 0.00708 0.994 1 ns
9 Physical and M… BSED BSCRIM 18 33 -1.97 0.0486 0.729 ns
10 Physical and M… BSOA BSIT 22 12 -1.38 0.167 1 ns
11 Physical and M… BSOA BSTM 22 9 -0.0228 0.982 1 ns
12 Physical and M… BSOA BSCRIM 22 33 -2.14 0.0321 0.482 ns
13 Physical and M… BSIT BSTM 12 9 1.10 0.270 1 ns
14 Physical and M… BSIT BSCRIM 12 33 -0.278 0.781 1 ns
15 Physical and M… BSTM BSCRIM 9 33 -1.54 0.123 1 ns There is significant difference between BSED and BSCRIM.
2.2.2 Program and Time Management
Normality Test
Shapiro-Wilk normality test
data: Data$`Time Management`
W = 0.98001, p-value = 0.1331Since p-value = 0.1331 > 0.05, it is conclusive that we reject the null hypothesis. That is, we 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 5 1.5052 0.1956
94 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: 6 × 11
Program variable n min max median iqr mean sd se ci
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 BEED Time Management 6 1.6 3 2 0.45 2.13 0.501 0.204 0.525
2 BSED Time Management 18 1.8 3 2.3 0.75 2.37 0.396 0.093 0.197
3 BSOA Time Management 22 1 3.2 2.4 0.55 2.44 0.526 0.112 0.233
4 BSIT Time Management 12 1.2 4 2.6 1.05 2.6 0.795 0.23 0.505
5 BSTM Time Management 9 1 3.2 2.4 1.2 2.31 0.807 0.269 0.62
6 BSCRIM Time Management 33 1.2 4 2.4 0.6 2.44 0.609 0.106 0.216The mean of BEED, BSED, BSOA, BSIT, BSTM, and BSCRIM is 2.133, 2.367, 2.436, 2.600, 2.311 and 2.436 respectively.
One Way ANOVA
Df Sum Sq Mean Sq F value Pr(>F)
Program 5 1.05 0.2102 0.585 0.711
Residuals 94 33.77 0.3592 Based on the p-value, there is no significant difference was observed between the group pairs.
Pairwise Comparisons
# A tibble: 15 × 9
.y. group1 group2 n1 n2 statistic p p.adj p.adj.signif
* <chr> <chr> <chr> <int> <int> <dbl> <dbl> <dbl> <chr>
1 Time Management BEED BSED 6 18 0.945 0.345 1 ns
2 Time Management BEED BSOA 6 22 1.43 0.153 1 ns
3 Time Management BEED BSIT 6 12 1.70 0.0901 1 ns
4 Time Management BEED BSTM 6 9 1.09 0.276 1 ns
5 Time Management BEED BSCRIM 6 33 1.31 0.191 1 ns
6 Time Management BSED BSOA 18 22 0.669 0.503 1 ns
7 Time Management BSED BSIT 18 12 1.08 0.281 1 ns
8 Time Management BSED BSTM 18 9 0.314 0.753 1 ns
9 Time Management BSED BSCRIM 18 33 0.460 0.645 1 ns
10 Time Management BSOA BSIT 22 12 0.528 0.598 1 ns
11 Time Management BSOA BSTM 22 9 -0.214 0.831 1 ns
12 Time Management BSOA BSCRIM 22 33 -0.283 0.777 1 ns
13 Time Management BSIT BSTM 12 9 -0.621 0.535 1 ns
14 Time Management BSIT BSCRIM 12 33 -0.793 0.428 1 ns
15 Time Management BSTM BSCRIM 9 33 0.0177 0.986 1 ns 3. Is there a significant difference between financial status, physical and mental health, and time management?
Normality Test
Shapiro-Wilk normality test
data: Data1$Scores
W = 0.98394, p-value = 0.001941Since p-value = 0.001941 < 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 0.2397 0.787
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
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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 Financial Sta… Scores 100 1.4 4 2.8 0.65 2.79 0.543 0.054 0.108
2 Physical and … Scores 100 1.4 4 2.8 0.6 2.88 0.574 0.057 0.114
3 Time Manageme… Scores 100 1 4 2.4 0.8 2.41 0.593 0.059 0.118The mean of Financial Status, Physical and Mental Health, and Time Management is 2.786, 2.882, and 2.414, respectively.
Kruskal-wallis Test
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Scores 300 32.8 2 0.0000000746 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 Financial St… Physi… 100 100 0.866 3.87e-1 1 e+0 ns
2 Scores Financial St… Time … 100 100 -4.47 7.76e-6 2.33e-5 ****
3 Scores Physical and… Time … 100 100 -5.34 9.43e-8 2.83e-7 **** There is significant difference between financial status and time management so with physical and mental health and time management
4. On which variable have the most significant impact?
Based on the provided output above, we can say that it is the physical and mental health.