3.1 Bootstrap

Each group is resampled with replacement to create 1000 samples of 10 members. The sample means are then calculated. This creates a bootstrapped sample mean distribution from 1000 means of each group. The code used is displayed below.

# set random number generator seed for reproducibility
set.seed(1234)

bootstrap <- function(supp_str, dose, n_samp) {
    
    data <- group_extract(supp_str, dose)
    
    resampled <- replicate(n_samp, mean(sample(data, replace = TRUE)))
    
    return(resampled)
}

4.1 Effect of Dosage

It may be of interest to limit the deciding factor to just the dosage. In this case the groups are only differentiated by the dosage factor. The following hypothesis is formed:

\[H_0: \bar{X}_{2} - \bar{X}_{1} = \bar{X}_{2} - \bar{X}_{0.5} = \bar{X}_{1} - \bar{X}_{0.5} = 0 \] \[H_A: \bar{X}_{2} > \bar{X}_{1}, \bar{X}_{2} > \bar{X}_{0.5}, \bar{X}_{1} > \bar{X}_{0.5}\] In other words, the null hypothesis is that dosage has no significant effect on odontoblast lengths. The alternative hypothesis is that 2mg/day results in the greatest average length while 0.5mg/day results in the shortest average length. One-sided two-sample bootstrap tests are carried out to check if 2 mg/day average is greater than other groups. The p-values are listed below:

##   (X2 > X1) (X2 > X0.5) (X1 > X0.5) 
##           0           0           0

With such small p-values it is safe to reject the null hypothesis. Therefore the alternative hypothesis, that is 2mg/day of vitamin C produces longer lengths than both 1mg/day and 0.5mg/day and 1mg/day produces longer lengths than 0.5mg/day, can be accepted.

4.2 Effect of Supplement Type

The supplement type is now used as the sole grouping factor. The hypothesis to be tested is:

\[H_0: \bar{X}_{OJ} - \bar{X}_{VC} = 0\] \[H_A: \bar{X}_{OJ} > \bar{X}_{VC}\]

In other words, the alternative hypothesis is giving orange juice is more effective at increasing teeth length compared to giving ascorbic acid. The computed p-value is 0.029. Since it is once again smaller than the traditional cutoff value at 0.05, the null hypothesis is rejected. Orange juice is found to increase teeth length more effectively compared to ascorbic acid in general.

4.3 All combinations

Finally both supplement type and dosage are used to define the groups. The following hypothesis is tested first

\[H_0: \bar{X}_{VC 2} - \bar{X}_{OJ 0.5} = 0\] \[H_A: \bar{X}_{VC 2} - \bar{X}_{OJ 0.5} > 0\]

In other words, the null hypothesis is there is no significant mean difference in teeth length between VC 2 and OJ 0.5. The alternative hypothesis is VC 2 produces significantly longer teeth lengths compared to OJ 0.5 on average. The p-value for the hypothesis test above is 0. Since it is less than the customery cutoff p = 0.05 the null hypothesis is rejected. VC 2 produces significantly longer teeth on average compared to OJ 0.5.

The same process can be repeated multiple times with other groups to produce the following table:

This table has to be read as The P value for the alternative hypothesis (row name) > (column name). The following alternative hypothesis should therefore be accepted for making the cutoff at P = 0.05:

## [1] OJ_0.5 > VC_0.5, P-val = 0.002
## [1] VC_1 > OJ_0.5, P-val = 0.006
## [1] VC_1 > VC_0.5, P-val = 0
## [1] OJ_2 > OJ_0.5, P-val = 0
## [1] OJ_2 > VC_1, P-val = 0
## [1] OJ_2 > VC_0.5, P-val = 0
## [1] OJ_2 > OJ_1, P-val = 0.021
## [1] OJ_1 > OJ_0.5, P-val = 0
## [1] OJ_1 > VC_1, P-val = 0
## [1] OJ_1 > VC_0.5, P-val = 0
## [1] VC_2 > OJ_0.5, P-val = 0
## [1] VC_2 > VC_1, P-val = 0
## [1] VC_2 > VC_0.5, P-val = 0
## [1] VC_2 > OJ_1, P-val = 0.043

Almost everything in this table agrees with the assessment so far from analyzing the effect of dosage and supplement type. However, notice that a break in pattern occurs when the hypothesis test fails to reject the null for VC 2 > OJ 2 and OJ 2 > VC 2. This means for the 2mg/day case the lengths resulted from different supplement types are not signifcantly different from each other.