Assignment_6

Question 1

alpha= 0.05
\(h_o\)= \(u_1\) - \(u_2\) = 0
\(h_a\)= \(u_1\) != \(u_2\)

Paired t-test

dat <- read.csv(file.choose())
t.test(dat$A,dat$B,paired=TRUE)

## 
##  Paired t-test
## 
## data:  dat$A and dat$B
## t = 3.6742, df = 9, p-value = 0.005121
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  1.383548 5.816452
## sample estimates:
## mean of the differences 
##                     3.6

The p.value 0.005121 is less than alpha, so We have enough evidence to reject Ho. The aspirine A has a different effect than aspirine B.

two-sample t-test

t.test(dat$A,dat$B,var.equal = TRUE)

## 
##  Two Sample t-test
## 
## data:  dat$A and dat$B
## t = 0.9802, df = 18, p-value = 0.34
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -4.116103 11.316103
## sample estimates:
## mean of x mean of y 
##      19.2      15.6

The p.value 0.34 is greater than alpha, so We dont have enough evidence to reject Ho. The aspirine A has the same effect than aspirine B.

Question 2

Boxplot

dat2<- read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/main/normtemp.csv",na.strings = "") 
 
males<- dat2 %>% filter(Sex == 1)
females <- dat2 %>% filter(Sex == 2)

boxplot(males$ï..Temp,females$ï..Temp, main="Resting temparatures", names=c("Males","Females"),ylab = "Temperature")

The variances don´t seem to be constant because the median are different and the range is wider for the males.

Hypothesis criteria

\(H_0\) = \(u_1\) - \(u_2\) = 0
\(H_a\) = \(u_1\) ≠ \(u_2\)

Pooled variance

t.test(males$ï..Temp,females$ï..Temp,var.equal = TRUE)

## 
##  Two Sample t-test
## 
## data:  males$ï..Temp and females$ï..Temp
## t = -2.2854, df = 128, p-value = 0.02393
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.53963938 -0.03882216
## sample estimates:
## mean of x mean of y 
##  98.10462  98.39385

Unequal variance

t.test(males$ï..Temp,females$ï..Temp,var.equal = FALSE)

## 
##  Welch Two Sample t-test
## 
## data:  males$ï..Temp and females$ï..Temp
## t = -2.2854, df = 127.51, p-value = 0.02394
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.53964856 -0.03881298
## sample estimates:
## mean of x mean of y 
##  98.10462  98.39385

As we can see, the p values for both tests are almost equal. Therefore, the variances appear to be constant. In addition, as the p value is less than alpha, we have enough evidence to reject H0. Hence, the resting temperatures appear to be different for males and females

CODE USE

library(dplyr)
dat <- read.csv(file.choose())
qqnorm(dat$A)
qqnorm(dat$B)
t.test(dat$A,dat$B,paired=TRUE)
t.test(dat$A,dat$B,var.equal = TRUE)
dat2<- read.csv("https://raw.githubusercontent.com/tmatis12/datafiles/main/normtemp.csv",na.strings = "") 
males<- dat2 %>% filter(Sex == 1)
females <- dat2 %>% filter(Sex == 2)
boxplot(males$ï..Temp,females$ï..Temp, main="Resting temparatures", names=c("Males","Females"),ylab = "Temperature")
t.test(males$ï..Temp,females$ï..Temp,var.equal = TRUE)
t.test(males$ï..Temp,females$ï..Temp,var.equal = FALSE)