Experimental Design (STAT 55) Midterm Exam
Kyle Kenneth Ruaya
2022-11-27
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.
mar = c(30, 20, 15, 8, 10, 40, 20, 25, 20, 28, 20, 25, 50, 40, 20, 10, 25, 25, 20, 15)
mean(mar)## [1] 23.31.1 Test at 0.05 level of significance using R for a two-tailed test.
t.test(mar, alternative="two.sided", mu = 20, conf.level = 0.95)##
## One Sample t-test
##
## data: mar
## 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
## 23.3Given the high p-value = 0.1804, we fail to reject the null hypothesis that the mean of mar is equal to 20. That is, we don’t have evidence that it is different from 20.
1.2 Test at 0.05 level of significance using R for a an appropriate one-tailed test.
t.test(mar, alternative="greater", mu = 20, conf.level = 0.95)##
## One Sample t-test
##
## data: mar
## 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
## 23.3Since the p-value is greater than the signficance level 0.05, 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(mar, alternative="less", mu = 20, conf.level = 0.95)##
## One Sample t-test
##
## data: mar
## 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
## 23.3Since the p-value is greater than the signficance level 0.05, we fail to reject the null hypothesis. That is, there is no enough evidence 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?
library(readxl)
pgviews<- read_excel("C:/Users/63966/Downloads/pgviews.xlsx")
View(pgviews)ken <- head(pgviews, 20)
ken## # A tibble: 20 × 3
## Subject Site Pages
## <dbl> <chr> <dbl>
## 1 1 B 2
## 2 2 B 6
## 3 3 A 5
## 4 4 B 7
## 5 5 A 3
## 6 6 B 2
## 7 7 B 6
## 8 8 A 1
## 9 9 A 3
## 10 10 A 4
## 11 11 B 6
## 12 12 B 6
## 13 13 B 4
## 14 14 A 5
## 15 15 A 3
## 16 16 A 6
## 17 17 B 6
## 18 18 B 3
## 19 19 A 4
## 20 20 B 72.1 Is the distribution of the data normally distributed?
Here, Null Hypothesis: The data is normally distributed; Alternative Hypothesis: The data is not normally distributed.
shapiro.test(ken$Pages)##
## Shapiro-Wilk normality test
##
## data: ken$Pages
## W = 0.92449, p-value = 0.1209Since p-value = 0.1209 > 0.05, it is conclusive that we fail to reject the null hypothesis. That is, we can assume normality.
2.2 Are the variances equal?
str(ken)## 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, ken, 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
## 0.6095679The p-value of 0.4948 is greater than the significance level of 0.05. We can conclude that there is no significant difference between the two variances.
2.3 At 0.05 level of significance, does Site A and Site B differ statistically?
kyle <- t.test(Pages ~ Site, data = ken, var.equal = TRUE,conf.level = 0.95)
kyle##
## 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 5.000000The p-value of the test is 0.1323, which is greater than the significance level alpha = 0.05. Hence we do not have enough evidences to prove that Site A and Site B differ statistically.
Question 3
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.
K1 <- c(12, 10, 11, 14, 15, 9, 11, 12, 13, 10, 15, 10)
K1## [1] 12 10 11 14 15 9 11 12 13 10 15 10K2 <- c(12, 13, 17, 11, 16, 13, 12, 14, 17, 12, 16, 18)
K2## [1] 12 13 17 11 16 13 12 14 17 12 16 18K3 <- c(14, 21, 17, 16, 17, 22, 16, 22, 19, 20, 18, 16)
K3## [1] 14 21 17 16 17 22 16 22 19 20 18 16K <- c(K1, K2, K3)
K## [1] 12 10 11 14 15 9 11 12 13 10 15 10 12 13 17 11 16 13 12 14 17 12 16 18 14
## [26] 21 17 16 17 22 16 22 19 20 18 16str(K)## num [1:36] 12 10 11 14 15 9 11 12 13 10 ...shapiro.test(K)##
## Shapiro-Wilk normality test
##
## data: K
## W = 0.95851, p-value = 0.1933Since p-value = 0.1933 > 0.05, it is conclusive that we fail to reject the null hypothesis. That is, we can assume normality.
3.2 Are the variances equal?
library(readxl)
med<- read_excel("D:/Regression Analysis/Book1.xlsx")
View(med)kenneth <- head(med, 37)
kenneth## # A tibble: 36 × 2
## Group Observation
## <chr> <dbl>
## 1 Group1 12
## 2 Group1 10
## 3 Group1 11
## 4 Group1 14
## 5 Group1 15
## 6 Group1 9
## 7 Group1 11
## 8 Group1 12
## 9 Group1 13
## 10 Group1 10
## # … with 26 more rowsstr(kenneth)## tibble [36 × 2] (S3: tbl_df/tbl/data.frame)
## $ Group : chr [1:36] "Group1" "Group1" "Group1" "Group1" ...
## $ Observation: num [1:36] 12 10 11 14 15 9 11 12 13 10 ...res <- bartlett.test(Observation ~ Group, data = kenneth)
res##
## Bartlett test of homogeneity of variances
##
## data: Observation by Group
## Bartlett's K-squared = 0.67864, df = 2, p-value = 0.7123The p-value is 0.7123 is greater than the significance level of 0.05. We can conclude that there is no significant difference between the tested sample variances.
3.3 At 0.05 level of significance, does the PDSC scores differ significantly among the different treatments?
set.seed(123)
dplyr::sample_n(kenneth, 36)## # A tibble: 36 × 2
## Group Observation
## <chr> <dbl>
## 1 Group3 16
## 2 Group2 17
## 3 Group2 13
## 4 Group1 11
## 5 Group1 10
## 6 Group2 13
## 7 Group2 12
## 8 Group1 15
## 9 Group1 15
## 10 Group2 14
## # … with 26 more rowslibrary(ggpubr)## Loading required package: ggplot2library(ggplot2)
library(dplyr)##
## Attaching package: 'dplyr'## The following objects are masked from 'package:stats':
##
## filter, lag## The following objects are masked from 'package:base':
##
## intersect, setdiff, setequal, uniongroup_by(med, Group) %>%
summarise(
count = n(),
mean = mean(Observation, na.rm = TRUE),
sd = sd(Observation, na.rm = TRUE)
)## # A tibble: 3 × 4
## Group count mean sd
## <chr> <int> <dbl> <dbl>
## 1 Group1 12 11.8 2.04
## 2 Group2 12 14.2 2.42
## 3 Group3 12 18.2 2.62kenneth$Group <- ordered(kenneth$Group,
levels = c("Group1", "Group2", "Group3"))levels(kenneth$Group)## [1] "Group1" "Group2" "Group3"library("ggpubr")
ggboxplot(med, x = "Group", y = "Observation",
color = "Group", palette = c("#00AFBB", "#E7B800", "#FC4E07"),
order = c("Group1", "Group2", "Group3"),
ylab = "Observation", xlab = "Group")
library("ggpubr")
ggline(med, x = "Group", y = "Observation",
add = c("mean_se", "jitter"),
order = c("Group1", "Group2", "Group3"),
ylab = "Observation", xlab = "Group")
library("gplots")##
## Attaching package: 'gplots'## The following object is masked from 'package:stats':
##
## lowessplotmeans(Observation ~ Group, data = med, frame = FALSE,
xlab = "Group", ylab = "Observation",
main="Mean Plot with 95% CI")## 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
res.aov <- aov(Observation ~ Group, data = kenneth)
res##
## Bartlett test of homogeneity of variances
##
## data: Observation by Group
## Bartlett's K-squared = 0.67864, df = 2, p-value = 0.7123summary(res.aov)## Df Sum Sq Mean Sq F value Pr(>F)
## Group 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 ' ' 1We can conclude that there are significant differences between the groups highlighted with “***” in the model summary because the p-value is less than the significance level of 0.05.
3.4 Use R for the conduct of multiple comparison test provided that 3.3 is significant.
TukeyHSD(res.aov)## Tukey multiple comparisons of means
## 95% family-wise confidence level
##
## Fit: aov(formula = Observation ~ Group, data = kenneth)
##
## $Group
## diff lwr upr p adj
## Group2-Group1 2.416667 0.04105701 4.792276 0.0454940
## Group3-Group1 6.333333 3.95772367 8.708943 0.0000006
## Group3-Group2 3.916667 1.54105701 6.292276 0.0008425The difference between Group2 and Group1 is significant, as shown by the output, with an adjusted p-value of 0.0454940. The difference between Group3 and Group1 is significant, as shown by the output, with an adjusted p-value of 0.0000006. The difference between Group3 and Group2 is significant, as shown by the output, with an adjusted p-value of 0.0008425.