Data

1. What is the demographic profile of the respondents in terms of:


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Sex

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; 23 of which are from ABM, 22 from GAS, 31 from HUMSS, and 24 from STEM.

2. Is there a significant difference on the variables physical health and mental health when grouped according to:

2.1 Sex


Call:
lm(formula = `Physical Health` ~ `Mental Health`, data = Data)

Coefficients:
    (Intercept)  `Mental Health`  
         1.1373           0.6103  

From this, we may deduce that the data fail to satisfy two assumptions – Linearity and Homogeneity of Variance.

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2.1.1 Sex and Physical Health

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Physical Health`
W = 0.95947, p-value = 0.003686

Since p-value = 0.003686 < 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  0.2379 0.6268
      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 Physical Health    58   2     3.8    3     0.4  2.92 0.4   0.053 0.105
2 Male   Physical Health    42   2.2   4      2.8   0.2  2.90 0.383 0.059 0.119

The mean of female and male is 2.917 and 2.905, respectively.

Mann Whitney U Test

# A tibble: 1 × 6
  .y.                 n statistic    df     p method        
* <chr>           <int>     <dbl> <int> <dbl> <chr>         
1 Physical Health   100    0.0830     1 0.773 Kruskal-Wallis

Based on the p-value, there is no significant difference on the physical health when grouped according to sex.

2.1.2 Sex and Mental Health

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Mental Health`
W = 0.95781, p-value = 0.002816

Since p-value = 0.002816 < 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  0.8262 0.3656
      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 Mental Health    58   2     3.8    3    0.4   2.91 0.352 0.046 0.092
2 Male   Mental Health    42   1.8   4      2.9  0.35  2.90 0.44  0.068 0.137

The mean of female and male is 2.910 and 2.905, respectively.

Mann Whitney U Test

# A tibble: 1 × 6
  .y.               n statistic    df     p method        
* <chr>         <int>     <dbl> <int> <dbl> <chr>         
1 Mental Health   100   0.00325     1 0.955 Kruskal-Wallis

Based on the p-value, there is no significant difference on the physical health when grouped according to sex.

2.2 Strand

2.2.1 Strand and Physical Health

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Physical Health`
W = 0.95947, p-value = 0.003686

Since p-value = 0.003686 < 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   0.072 0.9748
      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   Physical Health    24   2.6   4      3    0.3   3.13 0.371 0.076 0.157
2 ABM    Physical Health    23   2.2   3.6    2.8  0.4   2.78 0.366 0.076 0.158
3 HUMSS  Physical Health    31   2     3.6    2.8  0.4   2.86 0.384 0.069 0.141
4 GAS    Physical Health    22   2.2   3.6    3    0.35  2.88 0.374 0.08  0.166

The mean of STEM, ABM, HUMSS, and GAS is 3.133, 2.783, 2.858, and 2.882, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.                 n statistic    df      p method        
* <chr>           <int>     <dbl> <int>  <dbl> <chr>         
1 Physical Health   100      11.1     3 0.0114 Kruskal-Wallis

Based 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 Physical Heal… STEM   ABM       24    23    -3.14  0.00170 0.0102 *           
2 Physical Heal… STEM   HUMSS     24    31    -2.58  0.00984 0.0591 ns          
3 Physical Heal… STEM   GAS       24    22    -1.91  0.0559  0.335  ns          
4 Physical Heal… ABM    HUMSS     23    31     0.777 0.437   1      ns          
5 Physical Heal… ABM    GAS       23    22     1.18  0.239   1      ns          
6 Physical Heal… HUMSS  GAS       31    22     0.493 0.622   1      ns          

There is significant difference between STEM and ABM, STEM and HUMSS.

2.2.2 Strand and Mental Health

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Mental Health`
W = 0.95781, p-value = 0.002816

Since p-value = 0.002816 < 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  0.3016 0.8242
      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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Warning 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   Mental Health    24   2.4   4      3    0.4   3.02 0.377 0.077 0.159
2 ABM    Mental Health    23   2     3.8    2.8  0.4   2.87 0.416 0.087 0.18 
3 HUMSS  Mental Health    31   1.8   3.6    2.8  0.4   2.81 0.363 0.065 0.133
4 GAS    Mental Health    22   2.2   3.8    3    0.35  2.97 0.392 0.084 0.174

The mean of STEM, ABM, HUMSS, and GAS is 3.017, 2.870, 2.806, and 2.973, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.               n statistic    df     p method        
* <chr>         <int>     <dbl> <int> <dbl> <chr>         
1 Mental Health   100      5.62     3 0.132 Kruskal-Wallis

Based 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 Mental Health STEM   ABM       24    23    -1.42  0.154  0.926 ns          
2 Mental Health STEM   HUMSS     24    31    -2.15  0.0314 0.188 ns          
3 Mental Health STEM   GAS       24    22    -0.439 0.660  1     ns          
4 Mental Health ABM    HUMSS     23    31    -0.616 0.538  1     ns          
5 Mental Health ABM    GAS       23    22     0.959 0.338  1     ns          
6 Mental Health HUMSS  GAS       31    22     1.63  0.102  0.614 ns          

There is significant difference between STEM and HUMSS.

3. Is there a significant relationship between physical health and mental health?

Normality Test


    Shapiro-Wilk normality test

data:  Data1$Scores
W = 0.96029, p-value = 2.126e-05

Since p-value = 2.126e-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   1   0.012  0.913
      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
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Warning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
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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 Physical Heal… Scores     100   2       4    2.9   0.4  2.91 0.391 0.039 0.078
2 Mental Health  Scores     100   1.8     4    3     0.4  2.91 0.389 0.039 0.077

The mean of physical health and mental health is 2.912 and 2.908, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.        n statistic    df     p method        
* <chr>  <int>     <dbl> <int> <dbl> <chr>         
1 Scores   200   0.00675     1 0.935 Kruskal-Wallis

Based on the p-value, there is no significant difference between physical health and mental health.

4. On which variable does academic workload have the most significant impact?

Based on the provided output above, we can say that it is the physical health.