JAIRA JOY GROUP STATISTICAL ANALYSIS
Kyle Kenneth Ruaya
2024-02-12
Data
1. What is the demographic profile of the respondents in terms of:
Attaching package: 'dplyr'The following objects are masked from 'package:stats':
filter, lagThe following objects are masked from 'package:base':
intersect, setdiff, setequal, unionSex
Strand
Frequency in using Facebook
The tables above provides the distributions of respondents in terms of sex, strand, and frequency in using facebook. It can be seen that there are 55 females and 45 males; 20 of whch are from ABM, 20 from GAS, 20 from STEM, and 40 from HUMSS. Moreover, 64 students who always use facebook, 34 said they sometimes use facebook, and two said seldom
2. Is there a significant difference on the students’ academic performance (study habit and academic productivity) when grouped according to:
Sex
Call:
lm(formula = `Study Habit` ~ `Academic Productivity`, data = Data)
Coefficients:
(Intercept) `Academic Productivity`
0.5490 0.7649 
From this, we may deduce that the data fail to satisfy the two assumptions – Linearity and Homogeneity of Variance.
2.1.1 Sex and Study Habit
`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.
Attaching package: 'rstatix'The following object is masked from 'package:stats':
filterThe mean for male and female is 2.538 and 2.524, respectively.
The above graph shows the plotting of data by sex, which contains two sexes – male and female.
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ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errorsWarning: 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 no significant difference between the academic performance of the student in terms of study habit when grouped according to their sex.
Loading required package: carData
Attaching package: 'car'The following object is masked from 'package:purrr':
someThe following object is masked from 'package:dplyr':
recode
The histogram does not resembles 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. This indicates that residuals have a normal distribution. However, we may only get the true interpretation when both illustrations shows the same answer in which, in this case, failed to do so. Thus, we have the following.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.97406, p-value = 0.04553The Shapiro-Wilk p-value = 0.04553 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 0.2016 0.6545
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 = 1164, p-value = 0.9473
alternative hypothesis: true location shift is not equal to 0Since the p-value is larger than 0.05, we fail to reject the null hypothesis, that is, there is no significant difference between the academic performance of the students in terms of study habit when grouped according to sex.
2.1.2 Sex and Academic Productivity
`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.# A tibble: 19 × 3
# Groups: Sex [2]
Sex `Academic Productivity` count
<fct> <dbl> <int>
1 Female 1.6 2
2 Female 2 6
3 Female 2.2 6
4 Female 2.4 4
5 Female 2.6 12
6 Female 2.8 10
7 Female 3 7
8 Female 3.2 4
9 Female 3.4 4
10 Male 1.4 2
11 Male 1.8 2
12 Male 2 4
13 Male 2.2 4
14 Male 2.4 5
15 Male 2.6 11
16 Male 2.8 8
17 Male 3 6
18 Male 3.2 2
19 Male 3.8 1The mean for male and female is 2.542 and 2.629, 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?
The histogram resembles a bell curve as seen above, means that the residuals 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.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.98053, p-value = 0.1461The Shapiro-Wilk p-value = 0.1461 on the residuals is greater than the usual significance level of 0.05. Thus, we fail to 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 0.0131 0.909
98 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Two Sample T-test
Welch Two Sample t-test
data: c and d
t = -1.1006, df = 91.401, p-value = 0.274
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.28225721 0.08098737
sample estimates:
mean of x mean of y
2.542222 2.642857 Since the p-value= 0.274 is greater than 0.05, we fail to reject the null hypothesis. Hence, there is no significant difference between the academic performance of the students in terms of academic productivity when grouped according to their sex.
2.2 Frequency in using Facebook
2.2.1 Frequency in using Facebook and Study Habit
`summarise()` has grouped output by 'Frequency in using Facebook'. You can
override using the `.groups` argument.# A tibble: 25 × 3
# Groups: Frequency in using Facebook [3]
`Frequency in using Facebook` `Study Habit` count
<fct> <dbl> <int>
1 Always 1.4 1
2 Always 1.6 2
3 Always 1.8 1
4 Always 2 9
5 Always 2.2 8
6 Always 2.4 7
7 Always 2.6 9
8 Always 2.8 13
9 Always 3 8
10 Always 3.2 2
# ℹ 15 more rowsThe mean for always, seldom, and sometimes is 2.553, 2.900, and 2.465, respectively.
The above graph shows the plotting of data by frequency in using Facebook: Always, Sometimes, and Seldom
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 difference between the academic performance of the student in terms of study habit when grouped according to their frequency in using Facebook. However, this does not assure anyone that the difference is significant.
The histogram resemble a bell curve as seen above, means that the residuals 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. This also indicates that residuals have normal distribution.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.98597, p-value = 0.372The Shapiro-Wilk p-value = 0.372 on the residuals is greater than the usual significance level of 0.05. Thus, we fail to 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 2 1.2098 0.3027
97 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
One-way ANOVA
Df Sum Sq Mean Sq F value Pr(>F)
`Frequency in using Facebook` 2 0.453 0.2265 0.901 0.409
Residuals 97 24.377 0.2513 Since p-value = 0.409 > 0.05, we fail to reject the null hypothesis, that is, the academic performance in terms of study habit do not differ when grouped according to frequency in using Facebook.
2.2.2 Frequency in using Facebook and Academic Productivity
`summarise()` has grouped output by 'Frequency in using Facebook'. You can
override using the `.groups` argument.# A tibble: 23 × 3
# Groups: Frequency in using Facebook [3]
`Frequency in using Facebook` `Academic Productivity` count
<fct> <dbl> <int>
1 Always 1.4 1
2 Always 1.6 1
3 Always 1.8 2
4 Always 2 5
5 Always 2.2 7
6 Always 2.4 4
7 Always 2.6 15
8 Always 2.8 12
9 Always 3 9
10 Always 3.2 4
# ℹ 13 more rowsThe mean for always, seldom, and sometimes is 2.625, 2.700, and 2.518, respectively.
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 difference between the social behavior of the student in terms of emotional aspect when grouped according to their strand. However, illustrations does only give overviews of the data. It does not guarantee the exact result.
The histogram resembles a bell curve as seen above, means that the residuals 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.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.98155, p-value = 0.1751The Shapiro-Wilk p-value = 0.1751 on the residuals is greater than the usual significance level of 0.05. Thus, we fail to 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 2 1.609 0.2054
97 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
One-way ANOVA
Df Sum Sq Mean Sq F value Pr(>F)
`Frequency in using Facebook` 2 0.281 0.1403 0.667 0.515
Residuals 97 20.389 0.2102 Since p-value = 0.515 > 0.05, we fail to reject the null hypothesis, that is, the academic performance in terms of academic productivity do not differ when grouped according to frequency in using Facebook.
Based on the pairwise comparison, no significant difference was observed.
3. Is there a significant difference between the study habit and academic productivity in terms of the effects of frequent use of Facebook?
Normality Test
Shapiro-Wilk normality test
data: Data1$`Scores in terms of the effects of frequent use of Facebook to the students' academic performance`
W = 0.97409, p-value = 0.0009325Since p-value = 0.0009325 < 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 2.0314 0.1557
198 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':
lowessWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameterWarning in axis(1, at = 1:length(means), labels = legends, ...): "frame" is not
a graphical parameterWarning in plot.xy(xy.coords(x, y), type = type, ...): "frame" is not a
graphical parameter
# 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 Study Habit Scores … 100 1.2 3.8 2.6 0.6 2.53 0.501 0.05 0.099
2 Academic Prod… Scores … 100 1.4 3.8 2.6 0.6 2.59 0.457 0.046 0.091The mean of study habit and academic productivity is 2.53 and 2.59, respectively.
Wilcoxon Rank Sum Test
Wilcoxon rank sum test
data: x and y
W = 4670.5, p-value = 0.4163
alternative hypothesis: true location shift is not equal to 0Since the p-value is larger than 0.05, we fail to reject the null hypothesis, that is, there is no significant difference between the study habit and academic productivity of the students in terms of academic performance.
4. Which of the two: study habit and academic productivity, does frequent use of Facebook have the most significant impact?
Based on the provided output above, it can be seen that frequent use of Facebook have the most significant impact to the students’ academic performace in terms of academic productivity.