2. Is there a significant difference on the variables financial status, job opportunities, and personal preference when grouped according to:

2.1 Sex


Call:
lm(formula = `Financial Status` ~ `Job Opportunities` + `Personal Preference`, 
    data = Data)

Coefficients:
          (Intercept)    `Job Opportunities`  `Personal Preference`  
               1.2518                 0.4169                 0.1868

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 Financial Status

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Financial Status`
W = 0.96438, p-value = 0.008361

Since p-value = 0.008361 < 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  1.6529 0.2016
      98

The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.


Attaching package: 'gplots'

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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    55   2.2   4      3.2   0.6  3.09 0.42  0.057 0.113
2 Female Financial Status    45   1.8   3.8    3     0.4  3.04 0.357 0.053 0.107

The mean of male and female is 3.091 and 3.036, 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     0.751     1 0.386 Kruskal-Wallis

Based on the p-value, there is no significant difference was observed between the group pairs.

2.1.2 Sex and Job Opportunities

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Job Opportunities`
W = 0.97029, p-value = 0.02327

Since p-value = 0.002327 < 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  0.2825 0.5963
      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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Warning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
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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   Job Opportuniti…    55   1.8   3.8      3   0.4  2.98 0.459 0.062 0.124
2 Female Job Opportuniti…    45   1.8   3.8      3   0.4  2.95 0.405 0.06  0.122

The mean of male and female is 2.982 and 2.951, respectively.

Mann Whitney U Test

# A tibble: 1 × 6
  .y.                   n statistic    df     p method        
* <chr>             <int>     <dbl> <int> <dbl> <chr>         
1 Job Opportunities   100     0.393     1 0.531 Kruskal-Wallis

Based on the p-value, there is no significant difference was observed between the group pairs.

2.1.3 Sex and Personal Preference

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Personal Preference`
W = 0.94866, p-value = 0.0006774

Since p-value = 0.0006774 < 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  0.9866  0.323
      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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Warning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
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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   Personal Prefer…    55   2.2     4      3   0.4  3.14 0.396 0.053 0.107
2 Female Personal Prefer…    45   1.8     4      3   0.6  3.03 0.478 0.071 0.144

The mean of female and male is 3.138 and 3.027, respectively.

Mann Whitney U Test

# A tibble: 1 × 6
  .y.                     n statistic    df     p method        
* <chr>               <int>     <dbl> <int> <dbl> <chr>         
1 Personal Preference   100      1.05     1 0.307 Kruskal-Wallis

Based on the p-value, there is no significant difference was observed between the group pairs.

2.2 Strand

2.2.1 Strand and Financial Status

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Financial Status`
W = 0.96438, p-value = 0.008361

Since p-value = 0.008361 < 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  1.3541 0.2615
      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 parameter

Warning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
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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   Financial Status    25   2.2   3.8    3     0.4  3.02 0.341 0.068 0.141
2 ABM    Financial Status    25   2.4   4      3     0.4  3.05 0.425 0.085 0.176
3 HUMSS  Financial Status    25   2.4   3.8    3.2   0.4  3.15 0.323 0.065 0.133
4 GAS    Financial Status    25   1.8   3.8    3     0.6  3.05 0.47  0.094 0.194

The mean of STEM, ABM, HUMSS, and GAS is 3.016, 3.048, 3.152, and 3.048, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.                  n statistic    df     p method        
* <chr>            <int>     <dbl> <int> <dbl> <chr>         
1 Financial Status   100      2.91     3 0.405 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 Financial Status STEM   ABM       25    25     0.270 0.787 1     ns          
2 Financial Status STEM   HUMSS     25    25     1.58  0.113 0.680 ns          
3 Financial Status STEM   GAS       25    25     0.446 0.656 1     ns          
4 Financial Status ABM    HUMSS     25    25     1.31  0.189 1     ns          
5 Financial Status ABM    GAS       25    25     0.176 0.860 1     ns          
6 Financial Status HUMSS  GAS       25    25    -1.14  0.255 1     ns

2.2.2 Strand and Job Opportunities

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Job Opportunities`
W = 0.97029, p-value = 0.02327

Since p-value = 0.02327 < 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  0.7774 0.5094
      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 parameter

Warning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameter

Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter

# 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   Job Opportuniti…    25   2.2   3.6    2.8   0.4  2.90 0.361 0.072 0.149
2 ABM    Job Opportuniti…    25   2     3.6    3     0.6  2.90 0.405 0.081 0.167
3 HUMSS  Job Opportuniti…    25   1.8   3.8    3.2   0.6  3.06 0.555 0.111 0.229
4 GAS    Job Opportuniti…    25   2.2   3.8    3     0.6  3.02 0.391 0.078 0.161

The mean of STEM, ABM, HUMSS, and GAS is 2.904, 2.896, 3.056, and 3.016, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.                   n statistic    df     p method        
* <chr>             <int>     <dbl> <int> <dbl> <chr>         
1 Job Opportunities   100      4.00     3 0.261 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 Job Opportuniti… STEM   ABM       25    25     0.288 0.773  1     ns          
2 Job Opportuniti… STEM   HUMSS     25    25     1.78  0.0744 0.446 ns          
3 Job Opportuniti… STEM   GAS       25    25     1.16  0.246  1     ns          
4 Job Opportuniti… ABM    HUMSS     25    25     1.50  0.135  0.808 ns          
5 Job Opportuniti… ABM    GAS       25    25     0.872 0.383  1     ns          
6 Job Opportuniti… HUMSS  GAS       25    25    -0.623 0.533  1     ns

2.2.3 Strand and Personal Preference

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Personal Preference`
W = 0.94866, p-value = 0.0006774

Since p-value = 0.0006774 < 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  3.0839 0.03096 *
      96                  
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

The p-value is less than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is not met.

Warning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter

Warning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameter

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   Personal Prefer…    25   2.2     4      3   0.2  3.07 0.369 0.074 0.152
2 ABM    Personal Prefer…    25   2.6     4      3   0.4  3.14 0.394 0.079 0.163
3 HUMSS  Personal Prefer…    25   2.6     4      3   0.2  3.16 0.342 0.068 0.141
4 GAS    Personal Prefer…    25   1.8     4      3   0.8  2.98 0.595 0.119 0.246

The mean of STEM, ABM, HUMSS, and GAS is 3.072, 3.144, 3.160, and 2.976, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.                     n statistic    df     p method        
* <chr>               <int>     <dbl> <int> <dbl> <chr>         
1 Personal Preference   100      1.72     3 0.633 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 Personal Prefere… STEM   ABM       25    25     0.152 0.879     1 ns          
2 Personal Prefere… STEM   HUMSS     25    25     0.582 0.560     1 ns          
3 Personal Prefere… STEM   GAS       25    25    -0.704 0.481     1 ns          
4 Personal Prefere… ABM    HUMSS     25    25     0.431 0.667     1 ns          
5 Personal Prefere… ABM    GAS       25    25    -0.856 0.392     1 ns          
6 Personal Prefere… HUMSS  GAS       25    25    -1.29  0.198     1 ns

3. Is there a significant difference between financial status, job opportunities, and personal preference?

Normality Test


    Shapiro-Wilk normality test

data:  Data1$Scores
W = 0.96953, p-value = 5.648e-06

Since p-value = 5.648e-06 < 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  0.3566 0.7004
      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 parameter

Warning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameter

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   1.8   4        3  0.4   3.07 0.392 0.039 0.078
2 Job Opportuni… Scores     100   1.8   3.8      3  0.4   2.97 0.434 0.043 0.086
3 Personal Pref… Scores     100   1.8   4        3  0.45  3.09 0.436 0.044 0.087

The mean of Financial Status, Job Opportunities, and Personal Preference is 3.066, 2.968 and 3.088, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.        n statistic    df     p method        
* <chr>  <int>     <dbl> <int> <dbl> <chr>         
1 Scores   300      3.86     2 0.145 Kruskal-Wallis

Based 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 Scores Financial Status Job O…   100   100   -1.68   0.0932 0.280 ns          
2 Scores Financial Status Perso…   100   100    0.0467 0.963  1     ns          
3 Scores Job Opportuniti… Perso…   100   100    1.73   0.0845 0.253 ns

4. Which variable have the most significant?

Based on the provided output above, we can say that it is the personal preference.

DELJEN GONZALES GROUP STATISTICAL ANALYSIS

Kyle Kenneth Ruaya

2024-02-20

Data

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

Sex

Strand

2. Is there a significant difference on the variables financial status, job opportunities, and personal preference when grouped according to:

2.1 Sex

2.1.1 Sex and Financial Status

Normality Test

Equality of Variance

Mann Whitney U Test

2.1.2 Sex and Job Opportunities

Normality Test

Equality of Variance

Mann Whitney U Test

2.1.3 Sex and Personal Preference

Normality Test

Equality of Variance

Mann Whitney U Test

2.2 Strand

2.2.1 Strand and Financial Status

Normality Test

Equality of Variance

Kruskal-wallis Test

Pairwise Comparisons

2.2.2 Strand and Job Opportunities

Normality Test

Equality of Variance

Kruskal-wallis Test

Pairwise Comparisons

2.2.3 Strand and Personal Preference

Normality Test

Equality of Variance

Kruskal-wallis Test

Pairwise Comparisons

3. Is there a significant difference between financial status, job opportunities, and personal preference?

Normality Test

Equality of Variance

Kruskal-wallis Test

Pairwise Comparisons

4. Which variable have the most significant?