VANESSA LUTCHA GROUP STATISTICAL ANALYSIS
Kyle Kenneth Ruaya
2024-02-13
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 of Reading
The tables above provides the distributions of respondents in terms of sex, strand, and frequency of reading. It can be seen that there are 52 females and 48 males; 10 of which are from ABM, 35 from GAS, and 55 from HUMSS. Moreover, 44 said that they always read, 53 said sometimes, 2 said they seldom read, and 1 said never.
2. Is there a significant difference on the students’ reading comprehension (technology distractions and teacher support) when grouped according to:
Sex
Call:
lm(formula = `Technology Distractions` ~ `Teacher Support`, data = Data)
Coefficients:
(Intercept) `Teacher Support`
1.8618 0.2892 
From this, we may deduce that the data fail to satisfy the two assumptions – Linearity and Homogeneity of Variance.
2.1.1 Sex and Technology Distractions
`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 2.654 and 2.738, respectively.
The above graph shows the plotting of data by sex, which contains two sexes – male and female.
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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 reading comprehension of the student considering the effects of technology distractions when grouped according to their sex. However, illustration do not give exact results to see if the difference is significant. Thus, we have the following.
Loading required package: carData
Attaching package: 'car'The following object is masked from 'package:purrr':
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recode
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. This also indicates that residuals have a normal distribution.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.98076, p-value = 0.1524The Shapiro-Wilk p-value = 0.1524 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 1.5767 0.2122
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: a and b
t = -1.4304, df = 97.797, p-value = 0.1558
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.2012475 0.0326578
sample estimates:
mean of x mean of y
2.654167 2.738462 Since the p-value is larger than 0.05, we fail to reject the null hypothesis, that is, there is no significant difference between the reading comprehension of the students in consideration to the effects of technology distractions when grouped according to sex.
2.1.2 Sex and Teacher Support
`summarise()` has grouped output by 'Sex'. You can override using the `.groups`
argument.# A tibble: 19 × 3
# Groups: Sex [2]
Sex `Teacher Support` count
<fct> <dbl> <int>
1 Female 2.2 1
2 Female 2.4 3
3 Female 2.6 6
4 Female 2.8 8
5 Female 3 24
6 Female 3.2 3
7 Female 3.4 4
8 Female 3.6 2
9 Female 3.8 1
10 Male 1.4 1
11 Male 1.8 1
12 Male 2.2 1
13 Male 2.4 3
14 Male 2.6 9
15 Male 2.8 9
16 Male 3 17
17 Male 3.2 4
18 Male 3.4 1
19 Male 3.6 2The mean for male and female is 2.825 and 2.954, 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 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 = 0.94019, p-value = 0.0001979The Shapiro-Wilk p-value = 0.0001979 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 1.6773 0.1983
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 with continuity correction
data: c and d
W = 1028, p-value = 0.1148
alternative hypothesis: true location shift is not equal to 0Since the p-value= 0.1148 is greater than 0.05, we fail to reject the null hypothesis. Hence, there is no significant difference between the reading comprehension of the students in considering the effects of teacher support when grouped according to their sex.
2.2 Frequency of reading
2.2.1 Frequency of reading and Technology Distractions
`summarise()` has grouped output by 'Frequency of reading'. You can override
using the `.groups` argument.# A tibble: 18 × 3
# Groups: Frequency of reading [4]
`Frequency of reading` `Technology Distractions` count
<fct> <dbl> <int>
1 Always 1.8 1
2 Always 2.4 6
3 Always 2.6 14
4 Always 2.8 11
5 Always 3 8
6 Always 3.2 2
7 Always 3.4 1
8 Always 3.6 1
9 Never 2.6 1
10 Seldom 2.4 1
11 Seldom 2.8 1
12 Sometimes 2 2
13 Sometimes 2.2 5
14 Sometimes 2.4 6
15 Sometimes 2.6 16
16 Sometimes 2.8 11
17 Sometimes 3 11
18 Sometimes 3.2 2Warning: There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning:
! There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning in `stats::qt()`:
! NaNs producedThe mean for always, never, seldom, and sometimes is 2.745, 2.600, 2.600, and 2.664, respectively.
The above graph shows the plotting of data by sex, which contains two sexes – male and female.
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$residuals
W = 0.9818, p-value = 0.183The Shapiro-Wilk p-value = 0.183 on the residuals is greater than the usual significance level of 0.05. Thus, we fail to reject the hypothesis that residuals have a normal distribution.
Equality of Variance
Warning in leveneTest.default(y = y, group = group, ...): group coerced to
factor.Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 3 0.4548 0.7145
96 The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
One-way ANOVA
Df Sum Sq Mean Sq F value Pr(>F)
`Frequency of reading` 3 0.086 0.02874 0.222 0.881
Residuals 96 12.427 0.12945 Since p-value = 0.881 > 0.05, we fail to reject the null hypothesis, that is, the reading comprehension in consideration of the effects of technology distractions do not differ when grouped according to the frequency of reading.
2.2.2 Frequency of reading and Teacher Support
`summarise()` has grouped output by 'Frequency of reading'. You can override
using the `.groups` argument.# A tibble: 22 × 3
# Groups: Frequency of reading [4]
`Frequency of reading` `Teacher Support` count
<fct> <dbl> <int>
1 Always 1.4 1
2 Always 1.8 1
3 Always 2.2 1
4 Always 2.4 4
5 Always 2.6 4
6 Always 2.8 8
7 Always 3 15
8 Always 3.2 2
9 Always 3.4 4
10 Always 3.6 3
# ℹ 12 more rowsWarning: There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning:
! There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning in `stats::qt()`:
! NaNs producedThe mean for always, never, seldom, sometimes is 2.895, 2.600, 2.900, and 2.894, respectively.
The above graph shows the plotting of data by frequency of reading.
Warning: The following aesthetics were dropped during statistical transformation: fill
ℹ This can happen when ggplot fails to infer the correct grouping structure in
the data.
ℹ Did you forget to specify a `group` aesthetic or to convert a numerical
variable into a factor?
It clearly shows that there is a difference between the reading comprehension of the student cosidering teacher support when grouped according to their frequency of reading. However, this does not assure anyone that the difference is significant.
The histogram does not resemble a bell curve as seen above, means that the residuals does not have a normal distribution. Moreover, the points in the QQ-plots does not follow the straight line, with the majority of them falling outside the confidence bands. This also indicates that residuals does not have normal distribution.
Normality Test
Shapiro-Wilk normality test
data: res_aov$residuals
W = 0.90319, p-value = 2.003e-06The Shapiro-Wilk p-value = 2.003e-06 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 3 2.2337 0.08923 .
96
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1The p-value is greater than the 0.05 level of significance. Thus, the homogeneity assumption of the variance is met.
Kruskal-wallis Test
Warning: There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning:
! There was 1 warning in `mutate()`.
ℹ In argument: `ci = abs(stats::qt(alpha/2, .data$n - 1) * .data$se)`.
Caused by warning in `stats::qt()`:
! NaNs produced# A tibble: 4 × 11
`Frequency of reading` variable n min max median iqr mean sd
<fct> <fct> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 Always Teacher Su… 44 1.4 3.8 3 0.25 2.90 0.457
2 Sometimes Teacher Su… 53 2.2 3.6 3 0.2 2.89 0.256
3 Seldom Teacher Su… 2 2.8 3 2.9 0.1 2.9 0.141
4 Never Teacher Su… 1 2.6 2.6 2.6 0 2.6 NA
# ℹ 2 more variables: se <dbl>, ci <dbl>
# A tibble: 1 × 6
.y. n statistic df p method
* <chr> <int> <dbl> <int> <dbl> <chr>
1 Teacher Support 100 1.42 3 0.7 Kruskal-WallisBased on the p-value, no significant difference was observed between the group pairs.
3. Is there a significant difference on the reading comprehension of the students between the two factors, technology distractions and teacher support? Which of the two have the most significant impact on students’ reading comprehension?
Normality Test
Shapiro-Wilk normality test
data: Data1$`Scores in terms of reading comprehension`
W = 0.94273, p-value = 3.992e-07Since p-value = 3.992e-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 1 0.0802 0.7774
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 Technology Di… Scores … 100 1.8 3.6 2.6 0.25 2.70 0.297 0.03 0.059
2 Teacher Suppo… Scores … 100 1.4 3.8 3 0.25 2.89 0.356 0.036 0.071The mean of technology distractions and teacher support is 2.698 and 2.892, 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 reading comprehension 200 21.5 1 3.49e-6 Krusk…Based on the p-value, there is a significant difference was observed between the group pairs.
Lastly, it is the teacher support that has been the main factor that affects the reading comprehension of the students.