• There are multiple types of t-tests, two of which will be used in this presentation.
• The data being analyzed will be from “sleep”, which is
• The objectives will be to determine if the drug administered yielded improved sleep, and also to determine if one drug was more effective than the other.

• As can be seen, both groups average above the 0 hour mark, though outliers under the 0 hour mark do exist.

• By using a t-test, the efficacy of the drug administered can be determined; that is, it can be determined with confidence that the average extra sleep is not less than nor equal to 0.
• The confidence will be the standard 0.95, so we will be looking for a p-value of 0.05 or lower.

## 
##  One Sample t-test
## 
## data:  sleep$extra[1:10]
## t = 1.3257, df = 9, p-value = 0.1088
## alternative hypothesis: true mean is greater than 0
## 95 percent confidence interval:
##  -0.2870553        Inf
## sample estimates:
## mean of x 
##      0.75

• Because the p-value is not below 0.05, the efficacy of drug 1 cannot be determined as it does not deviate enough from 0.

## 
##  One Sample t-test
## 
## data:  sleep$extra[11:20]
## t = 3.6799, df = 9, p-value = 0.002538
## alternative hypothesis: true mean is greater than 0
## 95 percent confidence interval:
##  1.169334      Inf
## sample estimates:
## mean of x 
##      2.33

• On the other hand, the p-value of group 2 is well below 0.05, so that means that the drug had an actual effect on the improvement of the subjects’ sleep.

• Though we have already seen that only one drug was determined to have improved the subjects’ sleep, a paired t-test can see if there is a difference between the two drugs.
• If there is no difference, the mean difference should be equal to 0.
• If there is a difference, the mean difference will not be equal to 0.
• We will use the paired t-test to determine if drug 2 produced significantly more sleep than drug 1.

## 
##  Paired t-test
## 
## data:  (filter(.data = sleep, group == 2))$extra and (filter(.data = sleep, group == 1))$extra
## t = 4.0621, df = 9, p-value = 0.001416
## alternative hypothesis: true mean difference is greater than 0
## 95 percent confidence interval:
##  0.8669947       Inf
## sample estimates:
## mean difference 
##            1.58

• As can be seen with the p-value of 0.001416, the group 2 drug produced significantly more sleep than group 1.

• This barplot will show the visual of the difference of extra hours of sleep from drug 2 to drug 1.