1 Hypothesis of Drug mean Concentrations

  • We want to see if the mean concentrations of the two drugs are the same in urine concentrations, this give us μ₁ = μ₂ as our null hypothesis that needs to be tested with the given sample data.

1.1 Paired T-Test

## 
##  Paired t-test
## 
## data:  AsprinA and AsprinB
## 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
  • Running our Paired T-Test we get t=3.6742, df=9, and a p-value of .005121 as our values. By looking at our p-value being less that .05 we are able to reject our hypothesis previously stated above. Being able to reject our hypothesis give us evidence to say that the two mean concentrations of the two drugs are the same in urine concentrations is false and that the mean concentrations are not equal to each other. We can conclude that using a paired t-test gives us a p-value less than .05, which allows us to say that μ₁ ≠ μ₂ and is greater than or less than 0.

1.2 Two Sample T-Test

## 
##  Welch Two Sample t-test
## 
## data:  AsprinA and AsprinB
## 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
  • Running our Two Sample T-Test we get t=.9802, df=18 rounding up, and a p-value of .3401 as our values. By looking at our p-value being greater than .05 we can not reject the hypothesis previously stated above. By not being able to reject our hypothesis, this gives us evidence to say that the two mean concentrations of the two drugs are the same in urine concentrations is true. By running the two sample t-test we can conclude that our p-value being greater than .05 allows to say that μ₁ = μ₂ with the statistical evidence provided.

2 Null and Alternative Hypothesis of Infants on Exercise vs. Non-Exercise

  • Our Null hypothesis in would be that there is evidence that infants that are given exercise do not have a shorten time to start walking when compared to infants that do not get exercise. This gives us an equation of μ₁ = μ₂ where μ is the mean time it takes for infants to start walking after eight weeks. Our alternative hypothesis would be that there is evidence that infants that exercise do have a shortened time to start walking when compared to infants that do not get exercise. This would give us an equation that looks like this μ₁ less than μ.

2.1 Non-Parametric Test

  • For the set of data that is given for the infants on time it takes for them to start walking I’d like to do a non-parametric t-test. The reason I would run this type of test is because of the small data size that we have for each category that we are testing. And if a normal probability plot was to be done, there wouldn’t be enough data points in the plot itself to determine normality of the data or not. Running non-parametric t-test will still give us accurate results which can be used to either reject or fail to reject the null hypothesis.

2.2 Results from Non-Parametric test

## 
##  Wilcoxon rank sum exact test
## 
## data:  active and non_active
## W = 9, p-value = 0.09524
## alternative hypothesis: true location shift is less than 0
  • From running our Wilcox test which is also known as a Mann-Whitney Test which is a form of a non-parametric test that can be done on a set of data. Looking at a p-value of .09524 which is greater than our confidence level of .05 means that we fail to reject our null hypothesis. This means that there is no objective evidence to say that exercise shortens the time it takes for infants to start walking as opposed to no exercise. Gathering a bigger set of data may allows for us to see if the null hypothesis could stand agian with more data, it would also allow for us to see if the data is normally distributed and is not skewed or wide-tailed on either side of the distribution from a NPP.

3 Code used for testing

AsprinA <- c(15,26,13,28,17,20,7,36,12,18)

AsprinB <- c(13,20,10,21,17,22,5,30,7,11)

result1 <- t.test(AsprinA, AsprinB, paired = TRUE, conf.level = .95)

result1

result2 <-t.test(AsprinA, AsprinB, conf.level = .95)

result2

active <- c(9.50,10.00,9.75,9.75,9.00,13.00)

non_active<-c(11.50,12.00,13.25,11.50,13.00,9.00)

wilcox.test(active,non_active, alternative = c(“less”))