Question 1 (A)

Statement of the Appropriate Hypothesis:

Let \(\mu_A\) and \(\mu_B\) be the popultion mean 1 hr during contectrations for aspirin A and B.

The Hypothesis are:

\[ H_0: \mu\_A = \mu\_B\]

\[H_1: \mu\_A \neq \mu\_B \]

Question 1 (B)- Paired - T Test

#Aspirin Urin Samples T - Types

A <- c(15,26,13,28,17,20,7,36,12,18)
B <- c(13,20,10,21,17,22,5,30,7,11)

## Paired- T Test 
  
?t.test
t.test(A, B , alternative = c("two.sided"), paired = TRUE)
## 
##  Paired t-test
## 
## data:  A and B
## t = 3.6742, df = 9, p-value = 0.005121
## alternative hypothesis: true mean difference is not equal to 0
## 95 percent confidence interval:
##  1.383548 5.816452
## sample estimates:
## mean difference 
##             3.6

Comments: In this observation the P-Value is 0.005121 and compare to significance level \(\alpha\)  = 0.05. From the calculation we can state that out P-Value is lesser than the given \(\alpha\)  = 0.05. So, we can reject the null hypothesis \(H_0\)

##Question 1(C) - Two - Sample T Test

t.test(A, B , alternative = c("two.sided"), paired = FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  A and B
## t = 0.9802, df = 17.811, p-value = 0.3401
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -4.12199 11.32199
## sample estimates:
## mean of x mean of y 
##      19.2      15.6

Comments: In this observation the P-Value is 0.3401

Question 2

Statement of the null and alternative hypothesis:

Let \(\mu_A\) and \(\mu_N\) be the time for active and Non active exercise.

The Hypothesis are:

\[ H_0: \mu\_A = \mu\_N\]

\[H_a: \mu\_A < \mu\_N \]

The Reasoning of using Non Parametric Test

E <- c(9.5, 10.0, 9.75, 9.75, 9.0, 13.0)
N <- c(11.50, 12.0, 13.25, 11.50, 13.0, 9.0)

qqnorm(E, main = "Normal Q-Q Plot For Active Exercise")

qqnorm(N, main = "Normal Q-Q Plot for No- Active Exercise")

boxplot(E,N)

Comments: Due to our Data is not normally distributed so we need to do Non- Parametric T-Test

Analyze the Man-Whitney U -Test

wilcox.test(E,N,alternative = c("less"),conf.int = FALSE, conf.level = 0.95)
## 
##  Wilcoxon rank sum test with continuity correction
## 
## data:  E and N
## W = 9, p-value = 0.08523
## alternative hypothesis: true location shift is less than 0

Comment: Comments: In this observation the P-Value is 0.08523 and compare to significance level \(\alpha\)  = 0.05. From the calculation we can state that out P-Value is greater than the given \(\alpha\)  = 0.05. So, we can NOT reject the null hypothesis \(H_0\)

#Conclusions:

#Aspirin Urin Samples T - Types

A <- c(15,26,13,28,17,20,7,36,12,18)
B <- c(13,20,10,21,17,22,5,30,7,11)

## Paired- T Test 
  
t.test(A, B , alternative = c("two.sided"), paired = TRUE)

##Question 1(C) - Two - Sample T Test


t.test(A, B , alternative = c("two.sided"), paired = FALSE)


## The Reasoning of using Non Parametric Test


E <- c(9.5, 10.0, 9.75, 9.75, 9.0, 13.0)
N <- c(11.50, 12.0, 13.25, 11.50, 13.0, 9.0)

qqnorm(E, main = "Normal Q-Q Plot For Active Exercise")
qqnorm(N, main = "Normal Q-Q Plot for No- Active Exercise")
boxplot(E,N)

## Analyze the Man-Whitney U -Test

wilcox.test(E,N,alternative = c("less"),conf.int = FALSE, conf.level = 0.95)