2. Is there a significant difference on the impact of watching Korean dramas in terms of sleep patterns, language skills and style and behavior when grouped according to:

2.1 Sex


Call:
lm(formula = `Sleep Patterns` ~ `Language Skills` + `Style and Behavior`, 
    data = Data)

Coefficients:
         (Intercept)     `Language Skills`  `Style and Behavior`  
              1.0576                0.1215                0.4454

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

2.1.1 Sex and Sleep Patterns

`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.


Attaching package: 'rstatix'

The following object is masked from 'package:stats':

    filter

The mean for male and female is 2.156 and 2.740, respectively.

The above graph shows the plotting of data by sex, which contains two sexes – male and female.

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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?

It clearly shows that there is a significant difference between the impact of watching Korean dramas in terms of sleep patterns when grouped according to their sex.

Loading required package: carData


Attaching package: 'car'

The following object is masked from 'package:purrr':

    some

The 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 = 0.96178, p-value = 0.005394

The Shapiro-Wilk p-value = 0.005394 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  0.7814 0.3789
      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 = 414.5, p-value = 7.925e-06
alternative hypothesis: true location shift is not equal to 0

Since the p-value is less than 0.05, we reject the null hypothesis, that is, there is significant difference on the impact of watching Korean dramas in terms of sleep patterns when grouped according to sex.

2.1.2 Sex and Language Skills

`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.

The mean for male and female is 2.356 and 2.795, 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?

It clearly shows that there is a significant difference between the impact of watching Korean dramas in terms of language skills when grouped according to their sex.

Normality Test


    Shapiro-Wilk normality test

data:  res_aov$residuals
W = 0.95309, p-value = 0.001333

The Shapiro-Wilk p-value = 0.001333 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  24.203 3.497e-06 ***
      98                      
---
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.

Wilcoxon Rank Sum Test


    Wilcoxon rank sum test

data:  c and d
W = 708.5, p-value = 0.02923
alternative hypothesis: true location shift is not equal to 0

Since the p-value is less than 0.05, we reject the null hypothesis, that is, there is significant difference on the impact of watching Korean dramas in terms of language skills when grouped according to sex.

2.1.3 Sex and Style and Behavior

`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.

The mean for male and female is 2.407 and 2.797, 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?

It clearly shows that there is a significant difference between the impact of watching Korean dramas in terms of style and behavior when grouped according to their sex.

Normality Test


    Shapiro-Wilk normality test

data:  res_aov$residuals
W = 0.97385, p-value = 0.04386

The Shapiro-Wilk p-value = 0.04386 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  4.2625 0.0416 *
      98                 
---
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.

Wilcoxon Rank Sum Test


    Wilcoxon rank sum test

data:  e and f
W = 717.5, p-value = 0.03549
alternative hypothesis: true location shift is not equal to 0

Since the p-value is less than 0.05, we reject the null hypothesis, that is, there is significant difference on the impact of watching Korean dramas in terms of style and behavior when grouped according to sex.

2.2 Strand

2.2.1 Strand and Sleep Patterns

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Sleep Patterns`
W = 0.96487, p-value = 0.009087

Since p-value = 0.009087 < 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  0.9947 0.3988
      96

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':

    lowess

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 GAS    Sleep Patterns    40   1     3.8    2.8   0.6  2.70 0.631 0.1   0.202
2 ABM    Sleep Patterns    20   1.4   3.4    2.8   0.3  2.62 0.494 0.11  0.231
3 HUMSS  Sleep Patterns    15   2.2   3.4    2.8   0.7  2.77 0.413 0.107 0.229
4 STEM   Sleep Patterns    25   1.2   3.4    2     0.8  2.26 0.576 0.115 0.238

The mean of GAS, ABM, HUMSS, and STEM is 2.695, 2.620, 2.773, and 2.256, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.                n statistic    df       p method        
* <chr>          <int>     <dbl> <int>   <dbl> <chr>         
1 Sleep Patterns   100      11.8     3 0.00809 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 Sleep Patterns GAS    ABM       40    20    -0.441 0.659   1      ns          
2 Sleep Patterns GAS    HUMSS     40    15     0.505 0.614   1      ns          
3 Sleep Patterns GAS    STEM      40    25    -3.03  0.00242 0.0145 *           
4 Sleep Patterns ABM    HUMSS     20    15     0.801 0.423   1      ns          
5 Sleep Patterns ABM    STEM      20    25    -2.17  0.0296  0.178  ns          
6 Sleep Patterns HUMSS  STEM      15    25    -2.84  0.00458 0.0275 *

There is a significant difference between GAS and STEM, ABM and STEM, so with HUMSS and STEM.

2.2.2 Strand and Language Skills

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Language Skills`
W = 0.88638, p-value = 3.414e-07

Since p-value = 3.414e-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  3  1.6545  0.182
      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
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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 GAS    Language Skills    40     1   3.8    2.8   0.4  2.76 0.48  0.076 0.153
2 ABM    Language Skills    20     1   3.6    3     0.3  2.8  0.632 0.141 0.296
3 HUMSS  Language Skills    15     2   3.4    2.8   0.4  2.75 0.389 0.1   0.215
4 STEM   Language Skills    25     1   3.4    2.4   0.8  2.4  0.683 0.137 0.282

The mean of GAS, ABM, HUMSS, and STEM is 2.760, 2.800, 2.747, and 2.400, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.                 n statistic    df      p method        
* <chr>           <int>     <dbl> <int>  <dbl> <chr>         
1 Language Skills   100      7.75     3 0.0514 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 Language Skil… GAS    ABM       40    20     0.992 0.321   1      ns          
2 Language Skil… GAS    HUMSS     40    15    -0.406 0.685   1      ns          
3 Language Skil… GAS    STEM      40    25    -2.07  0.0383  0.230  ns          
4 Language Skil… ABM    HUMSS     20    15    -1.16  0.248   1      ns          
5 Language Skil… ABM    STEM      20    25    -2.67  0.00766 0.0460 *           
6 Language Skil… HUMSS  STEM      15    25    -1.24  0.215   1      ns

There is a significant difference between GAS and STEM, ABM and STEM.

2.2.3 Strand and Style and Behavior

Normality Test


    Shapiro-Wilk normality test

data:  Data$`Style and Behavior`
W = 0.95697, p-value = 0.002461

Since p-value = 0.002461 < 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   0.985 0.4033
      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
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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 GAS    Style and Behav…    40   1.4   4      3     0.4  2.84 0.509 0.081 0.163
2 ABM    Style and Behav…    20   1.8   3.2    2.8   0.5  2.7  0.47  0.105 0.22 
3 HUMSS  Style and Behav…    15   2     3.8    3     0.6  2.89 0.501 0.129 0.277
4 STEM   Style and Behav…    25   1     3.8    2.2   0.8  2.32 0.698 0.14  0.288

The mean of GAS, ABM, HUMSS, and STEM is 2.845, 2.700, 2.893, and 2.320, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.                    n statistic    df       p method        
* <chr>              <int>     <dbl> <int>   <dbl> <chr>         
1 Style and Behavior   100      12.6     3 0.00561 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 Style and Be… GAS    ABM       40    20    -0.817 4.14e-1 1       ns          
2 Style and Be… GAS    HUMSS     40    15     0.104 9.17e-1 1       ns          
3 Style and Be… GAS    STEM      40    25    -3.32  9.02e-4 0.00541 **          
4 Style and Be… ABM    HUMSS     20    15     0.748 4.55e-1 1       ns          
5 Style and Be… ABM    STEM      20    25    -2.08  3.80e-2 0.228   ns          
6 Style and Be… HUMSS  STEM      15    25    -2.69  7.19e-3 0.0431  *

There is a significant difference between GAS and STEM, ABM and STEM, so with HUMSS and STEM.

3. Is there a significant difference between sleep patterns, language skills, and style and behavior in terms of the impact of watching Korean dramas?

Normality Test


    Shapiro-Wilk normality test

data:  Data1$`Scores in terms of the impact of watching Korean dramas`
W = 0.94386, p-value = 2.852e-09

Since p-value = 2.852e-09 < 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  0.6816 0.5066
      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
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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 Sleep Patterns Scores …   100     1   3.8    2.6   0.8  2.58 0.588 0.059 0.117
2 Language Skil… Scores …   100     1   3.8    2.8   0.6  2.68 0.573 0.057 0.114
3 Style and Beh… Scores …   100     1   4      2.8   0.6  2.69 0.59  0.059 0.117

The mean of sleep patterns, language skills, and style and behavior is 2.582, 2.676, and 2.692, respectively.

Kruskal-wallis Test

# A tibble: 1 × 6
  .y.                                             n statistic    df     p method
* <chr>                                       <int>     <dbl> <int> <dbl> <chr> 
1 Scores in terms of the impact of watching …   300      2.82     2 0.244 Krusk…

It is conclusive that there is no significant difference between 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 in terms … Sleep… Langu…   100   100     1.50  0.132 0.397 ns          
2 Scores in terms … Sleep… Style…   100   100     1.40  0.162 0.486 ns          
3 Scores in terms … Langu… Style…   100   100    -0.107 0.915 1     ns

Pairwaise, there is no significant difference.

Which of the three: sleep patterns, language skills, style and behavior, does watching Korean drama have the most significant impact?

Based on the results, it is the style and behavior.

CHONA MADORA GROUP STATISTICAL ANALYSIS

Kyle Kenneth Ruaya

2024-02-16

Data

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

Sex

Strand

2. Is there a significant difference on the impact of watching Korean dramas in terms of sleep patterns, language skills and style and behavior when grouped according to:

2.1 Sex

2.1.1 Sex and Sleep Patterns

Normality Test

Equality of Variance

Wilcoxon Rank Sum Test

2.1.2 Sex and Language Skills

Normality Test

Equality of Variance

Wilcoxon Rank Sum Test

2.1.3 Sex and Style and Behavior

Normality Test

Equality of Variance

Wilcoxon Rank Sum Test

2.2 Strand

2.2.1 Strand and Sleep Patterns

Normality Test

Equality of Variance

Kruskal-wallis Test

Pairwise Comparisons

2.2.2 Strand and Language Skills

Normality Test

Equality of Variance

Kruskal-wallis Test

Pairwise Comparisons

2.2.3 Strand and Style and Behavior

Normality Test

Equality of Variance

Kruskal-wallis Test

Pairwise Comparisons

3. Is there a significant difference between sleep patterns, language skills, and style and behavior in terms of the impact of watching Korean dramas?

Normality Test

Equality of Variance

Kruskal-wallis Test

Pairwise Comparisons

Which of the three: sleep patterns, language skills, style and behavior, does watching Korean drama have the most significant impact?