question1

1a

\(H_a: \mu\_1 - \mu\_2 \neq 0\)

\(H_a: \mu\_1 - \mu\_2 \ = 0\)

Where μ1 & μ2 = mean concentrations of Aspirin A & B in a subject’s specimen

 AspirinA<- c(15,26,13,28,17,20,7,36,12,18)
 AspirinB<- c(13,20,10,21,17,22,5,30,7,11)
t.test(AspirinA,AspirinB,paired = TRUE)
## 
##  Paired t-test
## 
## data:  AspirinA and AspirinB
## 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

1b The p-value of paired t-test is 0.005121

Conclusion: Since p-value (0.005121) < significance level (0.05), we can reject the Null hypothesis

t.test(AspirinA,AspirinB,paired = FALSE)
## 
##  Welch Two Sample t-test
## 
## data:  AspirinA and AspirinB
## 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

1c

The p-value of Two sample t-test is 0.3401

Conclusion: Since p-value (0.3401) > significance level (0.05), we fail to reject Null Hypothesis

Question 2

2a

\(H_a: \mu\_1 - \mu\_2 \neq 0\)

\(H_a: \mu\_1 - \mu\_2 \ < 0\)

Where μ1 = mean time to walk in months for the ‘active exercise group’ & μ2= mean time to walk in months for the ‘non-stimulated Ones’

A<- c(9.5, 10, 9.75, 9.75, 9, 13) 
B<- c(11.5, 12, 13.25, 11.5, 13, 9)

qqnorm(A)
qqline(A)

qqnorm(B)
qqline(B)

2b

Conclusion:From the plot above we can see that the data are skewed which means the data are not normally distributed so we need to perform the non-parametric test

?wilcox.test
## starting httpd help server ... done
wilcox.test(A,B, alternative = "less")
## Warning in wilcox.test.default(A, B, alternative = "less"): cannot compute
## exact p-value with ties
## 
##  Wilcoxon rank sum test with continuity correction
## 
## data:  A and B
## W = 9, p-value = 0.08523
## alternative hypothesis: true location shift is less than 0

2c

Conclusion: p=0.08523 and Test Statistic W=9. The p value is greater than the significance level which means it failed to reject the null hypothesis

complete code

AspirinA<- c(15,26,13,28,17,20,7,36,12,18)
 AspirinB<- c(13,20,10,21,17,22,5,30,7,11)
 
t.test(AspirinA,AspirinB,paired = TRUE)
t.test(AspirinA,AspirinB,paired = FALSE)

A<- c(9.5, 10, 9.75, 9.75, 9, 13) 
B<- c(11.5, 12, 13.25, 11.5, 13, 9)

qqnorm(A)
qqline(A)
qqnorm(B)
qqline(B)
?wilcox.test
wilcox.test(A,B, alternative = "less")