MIDTERM
Ian Duhaylungsod
2022-11-29
Question 1:
A COVID frontliner wants to estimate the average amount that a resident in Maramag would donate to COVID affected families in Maramag. Twenty residents were randomly selected from the Municipality of Maramag. The 20 randomly residents were contacted by telephone and asked how much they would be willing to donate. Their responses are given below.
30 20 15 8 10 40 20 25 20 28 20 25 50 40 20 10 25 25 20 15
1.1 Test at 0.05 level of significance using R for a two-tailed
test.
1.2 Test at 0.05 level of significance using R for a an appropriate
one-tailed test.
Solution:
x <- c(30, 20, 15, 8, 10, 40, 20, 25, 20, 28, 20, 25, 50, 40, 20, 10, 25, 25, 20, 15)
x [1] 30 20 15 8 10 40 20 25 20 28 20 25 50 40 20 10 25 25 20 15mean(x)[1] 23.31.1
t.test(x, alternative = c("two.sided"), mu = 20, conf.level = 0.95)
One Sample t-test
data: x
t = 1.3905, df = 19, p-value = 0.1804
alternative hypothesis: true mean is not equal to 20
95 percent confidence interval:
18.33282 28.26718
sample estimates:
mean of x
23.3 Interpretation:
Based on the result, the p-value is 0.1804. Clearly, it is greater than 0.05 level of significance. With this, we fail to reject the null hypothesis that the mean of x is equal to 20, that is, we do not have evidence that it is different from 20.
1.2
t.test(x, alternative = c("greater"), mu = 20, conf.level = 0.95)
One Sample t-test
data: x
t = 1.3905, df = 19, p-value = 0.09022
alternative hypothesis: true mean is greater than 20
95 percent confidence interval:
19.19641 Inf
sample estimates:
mean of x
23.3 Interpretation:
Since the p-value is greater than the significance level 0.05, then we fail to reject the null hypothesis, that is, there is no enough evidence to prove that Maramag residents would donate more than 20.
t.test(x, alternative = c("less"), mu = 20, conf.level = 0.95)
One Sample t-test
data: x
t = 1.3905, df = 19, p-value = 0.9098
alternative hypothesis: true mean is less than 20
95 percent confidence interval:
-Inf 27.40359
sample estimates:
mean of x
23.3 Interpretation:
Since p-value is greater than the significance level 0.05, we fail to reject the null hypothesis, that is, there is no enough eveidence to prove that Maramag residents would donate less than 20.
Question 2:
Refer to the pgviews data, answer the following:
2.1 Is the distribution of the data normally distributed?
2.2 Are the variances equal?
2.3 At 0.05 level of significance, does Site A and Site B differ
statistically?
Solution:
library(readxl)
pgviews <- read_excel("D:/COLLEGE 4TH YEAR/1st SEMESTER/STAT 55 EXPERIMENTAL DESIGN/MIDTERM/pgviews.xlsx")
pgviewsian <- head(pgviews, 20)
paged_table(ian)2.1
shapiro.test(ian$Pages)
Shapiro-Wilk normality test
data: ian$Pages
W = 0.92449, p-value = 0.1209Interpretation:
Since p-value = 0.1209 which is greater than 0.05 level of significance, it is conclusive that we assume normality.
2.2
str(ian)tibble [20 × 3] (S3: tbl_df/tbl/data.frame)
$ Subject: num [1:20] 1 2 3 4 5 6 7 8 9 10 ...
$ Site : chr [1:20] "B" "B" "A" "B" ...
$ Pages : num [1:20] 2 6 5 7 3 2 6 1 3 4 ...var.test(Pages ~ Site, ian, alternative = "two.sided")
F test to compare two variances
data: Pages by Site
F = 0.60957, num df = 8, denom df = 10, p-value = 0.4948
alternative hypothesis: true ratio of variances is not equal to 1
95 percent confidence interval:
0.1581284 2.6181715
sample estimates:
ratio of variances
0.6095679 Interpretation:
Clearly, the p-value = 0.4948 is greater than 0.05 level of significance. With this, we can conclude that there is no significant difference between two variances, meaning the variances are equal.
2.3
t.test(Pages ~ Site, data = ian, var.equal = TRUE)
Two Sample t-test
data: Pages by Site
t = -1.5765, df = 18, p-value = 0.1323
alternative hypothesis: true difference in means between group A and group B is not equal to 0
95 percent confidence interval:
-2.8510432 0.4065987
sample estimates:
mean in group A mean in group B
3.777778 5.000000 Interpretation:
The p-value = 0.1323 is greater than 0.05 level of significance. Hence, there is no significant difference between Site A and Site B.
Question 3:
3-1. Two medications for the treatment of panic disorder, Medication1 and Medication 2, were compared with a placebo. A random sample of 36 patients was obtained from a listing of about 4,000 panic disorder patients who volunteered to participate in clinical trials. The 36 patients were randomly divided into three groups of equal size. The first group received Medication 1 for 10 weeks, the second group received Medication 2 for 10 weeks, and the third group received a placebo pill for 10 weeks. On week 11, all 36 patients were given the 7-item Panic Disorder Severity Scale (PDSC) which is scored on a 0 to 28 scale (lower scores are better). The PDSC scores are given below.
Group 1: 12 10 11 14 15 9 11 12 13 10 15 10
Group 2: 12 13 17 11 16 13 12 14 17 12 16 18
Group 3: 14 21 17 16 17 22 16 22 19 20 18 16
3.1 Is the distribution of the data normally distributed?
3.2 Are the variances equal?
3.3 At 0.05 level of significance, does the PDSC scores differ
significantly among the different treatments?
3.4 Use R for the conduct of multiple comparison test provided that 3.3
is significant.
Solution:
library(readxl)
Ian_Data_for_Question_3_Midterm_Exam <- read_excel("D:/COLLEGE 4TH YEAR/1st SEMESTER/STAT 55 EXPERIMENTAL DESIGN/MIDTERM/Ian Data for Question 3 Midterm Exam.xlsx")
Ian_Data_for_Question_3_Midterm_Exam3.1
str(Ian_Data_for_Question_3_Midterm_Exam)tibble [36 × 2] (S3: tbl_df/tbl/data.frame)
$ Treatment : chr [1:36] "Group 1" "Group 1" "Group 1" "Group 1" ...
$ Observation: num [1:36] 12 10 11 14 15 9 11 12 13 10 ...shapiro.test(Ian_Data_for_Question_3_Midterm_Exam$Observation)
Shapiro-Wilk normality test
data: Ian_Data_for_Question_3_Midterm_Exam$Observation
W = 0.95851, p-value = 0.1933Interpretation:
It can be seen that the p-value = 0.1933 is greater than 0.05 significance level. This implies that we can assume normality.
3.2
res <- bartlett.test(Observation ~ Treatment, data = Ian_Data_for_Question_3_Midterm_Exam)
res
Bartlett test of homogeneity of variances
data: Observation by Treatment
Bartlett's K-squared = 0.67864, df = 2, p-value = 0.7123Interpretation:
Since the p-value = 0.7123 is greater than 0.05 level of significance, we can say that there is no significant difference between the variances.
3.3
ian1 <- head(Ian_Data_for_Question_3_Midterm_Exam, 36)
paged_table(ian1)library(dplyr)
Attaching package: 'dplyr'The following objects are masked from 'package:stats':
filter, lagThe following objects are masked from 'package:base':
intersect, setdiff, setequal, unionlibrary(ggpubr)Warning: package 'ggpubr' was built under R version 4.2.1Loading required package: ggplot2Warning: package 'ggplot2' was built under R version 4.2.1library(gplots)Warning: package 'gplots' was built under R version 4.2.1
Attaching package: 'gplots'The following object is masked from 'package:stats':
lowessset.seed(123)
ian2 <- dplyr::sample_n(Ian_Data_for_Question_3_Midterm_Exam, 10)
paged_table(ian2)ian3 <- group_by(ian1, Treatment)%>%
summarise(count = n(),
mean = mean(Observation, na.rm = TRUE),
sd = sd(Observation, na.rm = TRUE)
)
paged_table(ian3)ian1$Treatment <- ordered(ian1$Treatment, levels = c("Group 1", "Group 2", "Group 3"))levels(ian1$Treatment)[1] "Group 1" "Group 2" "Group 3"library("ggpubr")
ggboxplot(ian1, x = "Treatment", y = "Observation",
color = "Treatment", palette = c("#00AFBB", "#E7B800", "#FC4E07"),
order = c("Group 1", "Group 2", "Group 3"),
ylab = "Observation", xlab = "Treatment")
ggline(ian1, x = "Treatment", y = "Observation",
add = c("mean_se", "jitter"),
order = c("Group 1", "Group 2", "Group 3"),
ylab = "Observation", xlab = "Treatment")
plotmeans(Observation ~ Treatment, data = ian1, frame = FALSE,
xlab = "Treatment", ylab = "Observation",
main="Mean Plot with 95% CI")Warning 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
res.aov <- aov(Observation ~ Treatment, data = Ian_Data_for_Question_3_Midterm_Exam)summary(res.aov) Df Sum Sq Mean Sq F value Pr(>F)
Treatment 2 245.2 122.58 21.8 9.25e-07 ***
Residuals 33 185.6 5.62
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1Interpretation:
Based on the results, there are significant differences between the treatments highlighted with “***” in the model summary since the p-value = 9.25e-07 is less than the significance level of 0.05.
3.4
TukeyHSD(res.aov) Tukey multiple comparisons of means
95% family-wise confidence level
Fit: aov(formula = Observation ~ Treatment, data = Ian_Data_for_Question_3_Midterm_Exam)
$Treatment
diff lwr upr p adj
Group 2-Group 1 2.416667 0.04105701 4.792276 0.0454940
Group 3-Group 1 6.333333 3.95772367 8.708943 0.0000006
Group 3-Group 2 3.916667 1.54105701 6.292276 0.0008425Interpretation:
It is visible in the results that the difference between Group 2 and Group 1 is significant with an adjusted p-value = 0.0454940. Also, the difference between Group 3 and Group 1 is significant with an adjusted p-value = 0.0000006. Further, the difference between Group 3 and Group 2 is also significant since the adjusted p-value = 0.0008425.