10.3 – Paired t-test

Introduction

Good experiments include controls. Interested in a new treatment for weight loss? Define a control group to compare the weight loss by a group using the new product. In many cases, the best control is the individual.

Consider now a basic experimental design, the randomized crossover trial (Fig 1), introduced in Chapter 2.4.

Figure 1. A two group Randomized Crossover Trial.

Subjects randomly selected from population of interest, then again once recruited into one of two treatment arms: arm 1, subjects first receive the experimental treatment, then some time later the subjects receive the control treatment; arm 2, subjects first receive the control treatment, then some time later the subjects receive the experimental treatment. Note the difference between this paired or repeated measures design and the independent sample design (see Chapter 10.1). Repeated measures designs have many advantages; we discuss them further in Chapter 14.6. At the start, repeated measures designs have greater statistical power compared to cross-sectional (independent) sample designs.

Many experiments are designed so that subjects receive all treatments and responses are gauged against the initial values recorded on the subjects. Repeated measures statistical tests, like the paired t-test, are needed however to analyze the data. These types of statistical procedures are similar to the two sample independent t-test that we discussed earlier.

However, there is an important difference between these two types of statistical procedures. For the two independent sample t-test the samples are unpaired: we observed one variable on some individuals assigned to two different groups. These groups might be

• Two locations where we measure plants or animals
• A treatment (or experimental) group with a control group.
• Expression of cytokeratin genes (eg, ΔΔC_T, fold-change) from breast cancer patients compared to healthy donor subjects (Andergassen et al 2016).

The point is that samples in one group are not the same samples in the second group.

In the paired t-test we have two groups but the observations in these two groups are paired. Paired means that there is some relationship between one observation in the first sample and one observation in the second sample (every observation in one sample must be paired with one observation in another sample).

For example, weight change in humans before and after a change in diet could be performed as a paired analysis. Each subject’s weight before the diet was “paired” with the same subject’s weight after the diet.

Another example comes from genetics. Siblings or Monozygotic twins or clones, strains or varieties of plants or animals can be Paired in an experiment.

• You can give one of the twins a particular diet, or the plant or animal clones or strains can be raised in a particular environment (nutrient)
• The other twin or plant or animal clone or variety can serve as a type of control by providing a normal diet or normal environment.

Another example is a study of environmental pollution on cancer rates in many different communities.

• The researchers selects pairs of communities with similar characteristics for many socioeconomic factors.
• Each pair of communities differed with respect to the proximity to a known source of pollution: one of the pair was close to a source of pollution and one of the pair was far from a source of pollution.

The purpose of pairing in this example is to attempt to “control” for all the socioeconomic factors that might contribute to cancer but they did not want to directly measure. These other factors should be similar for each member of the pair.

Example: How repeatable is human running performance?

The subheading is misleading — we leave “repeatability” or, do individuals perform consistently to another time (see Chapter 12.3 – Fixed effects, random effects, and agreement). Here, we focus on a different question — we test whether mean performance differs across repeated trials. This may seem a subtle difference, but consistency asks the question if performance, whether it is answering questions on an exam or running times for a race, is stable across trials. Testing the means of first and second trials asks a different question: we test for a trend in performance, which may reflect learning.

Here’s an example in which a measure was taken twice for the same individuals. The data are running speed or pace during 5K race held annually on Oahu for a selected sample of female runners with repeat race times (20 – 29 years old). The race was run annually on Oahu, and the data reported are the pace for the first race and the second race, which occurred a year later (Jamba Juice – Banana Man Chase, Ala Moana Beach Park, data extracted from source, https://timelinehawaii.com).

Table 1. 5K pace times (kph) for 15 women (20 – 29 years)

Load the data into R as an unstacked data set. Data available at end of this page or click here.

Begin with description and exploration of the data. Start with histograms to get a sense of the sample distributions (hint: we’re looking to see if the data looks like it could come from a normal distribution, see Chapter 13.3 Assumptions) (Fig 2).

Figure 2. Histograms shows the distribution of 5K running times of 15 women who ran the race twice.

R code (stacked data set, then used defaults R Commander to make the histogram, then modified the code and submitted modified code to make Fig 1)

with(stackExCh10.3, Hist(obs, groups=Race, scale="frequency", breaks="Sturges", col="blue", 
xlab="Time (min)", ylab="Frequency")))

Conclusion? The histograms don’t look normally distributed so we keep this in mind as we proceed. (See Chapter 13 – Assumptions of parametric tests for additional information about assumptions.)

A box plot comparing the first and second pace times (Fig 3).

Figure 3. Box plot of  race speed (kph) for 15 women 5K in two successive years.

I added a trend line (linear regression, see Chapter 17, red line) and connected the averages (blue line)  for visual emphasis — no differences between the means — but note that one wouldn’t do this as part of an analysis (see Chapter 4 discussion).

R code for Fig 2

Boxplot(obs~Race, data=stackExCh10.3, id=list(method="y"), xlab="", ylab="Pace (kph)") #boxplot was made in Rcmdr
abline(lm(obs ~ as.numeric(Race), data=stackExCh10.3), col="red", lwd=2)
means <- tapply(obs, Race, mean)
points(1:2, means, pch=7, col="blue")
lines(1:2, means, col="blue", lwd=2)

The box plot works to show the median difference, but loses the paired information. A nice package called PairedData has several functions that work well with paired data, for example a profile plot (Fig 4). A profile plot is useful for showing an association between paired data.

Figure 4. Profile plot, PairedData package.

R commands for Figure 4.

require(PairedData)
attach(example.ch10.3) # remember to attach dataframe so you don't have to call variables like example.ch10.3$Race1
races <- paired(Race1, Race2)
plot(races, type = "profile")

Paired t-test calculation

The paired t-test is a straight-forward extension of the independent sample t-test; the key concept is that the two samples are no longer independent, they are paired. Thus, instead of mean of group 1 minus mean of group two, we test the differences between sample 1 and sample 2 for each paired observation.

1. Compute the differences between the Paired Samples (as in tables above)
2. Calculate the MEAN difference score, : in the previous example = -0.094 kmh
3. Calculate the degrees of freedom df = # pairs – 1 = n – 1, where n is the number of pairs
4. Calculate the standard error of the mean of d.
   
   where
   
5. Calculate the test statistic for paired data
   
6. Compare to the Critical Value in Table C. 4 (Appendix Table 2)
7. Find the Critical Value = t _{α (2), df}

Try as difference instead of paired

Before you answer, take a look at the box plot of the mean difference between the repeat measures of 5K pace for the 15 women (Fig 5).

Figure 5. Box plot of differences, Red dotted lines shows the null hypothesis.

Create a new variable, raceDiff, equal to Race2 minus Race1. Then, use the one sample T-test on raceDiff. I’ll leave you to complete the work (Question 2).

R code

t.test(Race1, Race2, paired = TRUE, alternative = "two.sided")

R output

Paired t-test

data: Race.1 and Race.2
t = 0.19389, df = 14, p-value = 0.849
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-0.9491017 1.1377521
sample estimates:
mean of the differences 
0.09432517

Rcmdr, paired t-test

Rcmdr: Statistics → Means → Paired t-test…

Note: your two groups must be in two different columns (unstacked!) to run this version of the test.

Figure 6. R Commander Paired t-test menu, Rcmdr version 2.7.

After selecting the variables, set null hypothesis after clicking on Options tab (Fig 6).

Figure 7. R Commander Paired t-Test options, select null hypothesis.

Click OK to proceed. Results will be the same as the output listed above.

the following text needs to be updated

Interpret the results.

So, what can we conclude about the null hypothesis? Interpret the 95% CI, the T-test statistic, and the P-value.

Do not ignore sample dependence

What if we ignored the repeated measures design and treated the first and second races as independent? The important concept here is to ask, what would have happened if we had done a two independent sample t-test instead?

Let’s run the analysis again, this time incorrectly using the independent sample t-test. We need to manipulate the data set before we do.

Manage your data: Stack the data

This is a good time to share how to Stack data in R. If you look at our active data set, the results of the two trials are in two different columns. In order to run the independent sample t-test we need the data in one column (with a label column).

stackExCh10.3 <- stack(example.ch10.3[, c("Race1","Race2")])
names(stackExCh10.3) <- c("obs", "Race")

Rcmdr: Data → Active data set → Stack variables in data set…

Figure 8. R Commander: Stack worksheet. Select the two variables, Race1 and Race2.

I entered values for name of the new data set, the new variable, and the name for the factor (label) column.

Figure 9. R Commander, select independent sample t-Test …

Figure 10. R commander, independent sample t test menu.

Figure 11. R Commander, select options for independent sample t-Test (assume equal variance).

Here are the results of the independent sample t-test from R.

t.test(obs~Race, alternative='two.sided', conf.level=.95, var.equal=TRUE, + data=stackCh10.3)

Two Sample t-test
data: obs by Race
t = 0.070645, df = 28, p-value = 0.9442
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -2.640719 2.829369
sample estimates:
mean in group Race1 mean in group Race2 
 9.886342 9.792017

End R output

In this case, we would have reached the same general conclusion, but the p-values are different. The p-value from the paired t-test was about 0.85 whereas the p-value from the independent sample t-test was higher, nearly 0.95, suggesting little difference between the two trials.

While the general conclusion holds, this time, that there were no statistically significant difference between the means for first and second trials. However, it won’t always work out that way. And besides, if you treated the paired data as independent, you’ve clearly violated one of the assumptions of the test.

Take a look at the degrees of freedom for the two analyses. By ignoring the pairing of samples we gain twice the number of degrees of freedom … that can’t both be right. The way to distinguish between the two is to go back to the experimental units.

Question: What are the sampling units in the case of repeat measures on individuals: the individuals themselves? the pairs of burst speed trials? something else?

it is important to note that the paired t-test is still the best for this situation because it accurately reflects the experiment — individuals were measured twice, therefore the two groups (trial 1 and trial 2) are not independent! Thus, the p-value from the paired t-test correctly reflect our best analyses of the test of the null hypothesis because the correct degrees of freedom were 14 and not 28.

In the case of the independent sample t-test we necessarily make the assumption that the two groups are independent — that is, that they are measured on different sampling units (eg, different individuals or subjects). In statistical terms, that means that you assume that the correlation between trial results is equal to zero. By incorrectly choosing an independent sample test in these repeated measures cases, I would make two null hypotheses: (1) that the means are the same and (2) that the correlation between repeat measures is zero. The problem? The t-test only evaluates the first hypothesis (means).

Questions

1. Refer to Figure 6 again and related data set. Were runners faster the second year or the first year running the 5k? What about the points labeled 4 and 14? What was the average difference between first and second races?
2. Complete the test of the null hypothesis of no difference between race 1 and race 2 (raceDiff) with the one sample t-test. Set up a table to compare the test statistic, df, and p-values for results from paired t test, one sample t-test, and independent sample t-test. How do these results compare?
3. I’ve called the observed value “pace,” but runners would know that pace is actually amount of time per kilometer, not the total time over 5k, which is what I called pace.
   • Create a new variable and report average pace for Race1 and Race2.
   • Redo the paired analysis, including box plot, on your new variable.
   • What is the null hypothesis for your new variable?
   • Summarize your results and add to the table you created for question 2.
4. Redo the two plots in Figure 2 so that the two histograms are on the sample plot. Instructions and examples were provided in Chapter 4.2.

Quiz Chapter 10.3

Paired t-test

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Data set

example.ch10.3 <- read.table(header=TRUE, text = "
ID Race1 Race2
1 15.28 15.61
2 11.22 11.19
3 8.80 9.14
4 8.88 5.46
5 9.81 10.50
6 6.12 5.69
7 8.31 8.71
8 6.26 7.42
9 17.16 16.41
10 16.23 15.82
11 5.90 7.12
12 8.31 10.48
13 5.93 8.64
14 10.54 5.99
15 9.53 8.69
")

Chapter 10 contents

• Introduction
• Compare two independent sample means
• Digging deeper into t-test Plus the Welch test
• Paired t-test
• References and suggested reading

5.5 – Importance of randomization in experimental design

Introduction

If the goal of the research is to make general, evidenced-based statements about causes of disease or other conditions of concern to the researcher, then how the subjects are selected for study directly impacts our ability to make generalizable conclusions. The most important concept to learn about inference in statistical science is that your sample of subjects upon which all measurements and treatments are conducted, ideally should be a random selection of individuals from a well-defined reference population.

The primary benefit of random sampling is that it strengthens our confidence in the links between cause and effect. Often after an intervention trial is complete, differences among the treatment groups will be observed. Groups of subjects who participated in sixteen weeks of “vigorous” aerobic exercise training show reduced systolic blood pressure compared to those subjects who engaged in light exercise for the same period of time (Cox et al 1996). But how do we know that exercise training caused the difference in blood pressure between the two treatment groups? Couldn’t the differences be explained by chance differences in the subjects? Age, body mass index (BMI), overall health, family history, etc?

How can we account for these additional differences among the subjects? If you are thinking like an experimental biologist, then the word “control” is likely coming to the foreground. Why not design a study in which all 60 subjects are the same age, the same BMI, the same general health, the same family … history…? Hmm. That does not work. Even if you decide to control age, BMI, and general health categories, you can imagine the increased effort and cost to the project in trying to recruit subjects based on such narrow criteria. So, control per se is not the general answer.

If done properly, random sampling makes these alternative explanations less likely. Random sampling implies that other factors that may causally contribute to differences in the measured outcome, but themselves are not measured or included as a focus of the research study, should be the same, on average, among our different treatment groups. The practical benefits of proper random sampling is that recruiting subjects gets easier — fewer subjects will be needed because you are not trying to control dozens of factors that may (or may not!) contribute to differences in your outcome variable. The downside to random sampling is that the variability of the outcomes within your treatment groups will tends to increase. As we will see when we get to statistical inference, large variability within groups will make it less likely that any statistical difference between the treatment groups will be observed.

Demonstrate the benefits of random sampling as a method to control for extraneous factors.

The study reported by Cox et al included 60 obese men between the ages of 20 and 50. A reasonable experimental design decision would suggest that the 60 subjects be split into the two treatment groups such that both groups had 30 subjects for a balanced design. Subjects who met all of the research criteria and who had signed the informed consent agreement are to be placed into the treatment groups and there are many ways that group assignment could be accomplished. One possibility, the researchers could assign the first 30 people that came into the lab to the Vigorous exercise group and the remaining 30 then would be assigned to the Light exercise group. Intuitively I think we would all agree that this is a suspect way to design an experiment, but more importantly, why shouldn’t you use this convenient method?

Just for argument’s sake, imagine that their subjects came in one at a time, and, coincidentally, they did so by age. The first person was age 21, the second was 22, and so on up to the 30th person who was 50. Then, the next group came in, again, coincidentally in order of ascending age. If you calculate the simple average age for each group you will find that they are identical (35.5 years). On the surface, this looks like we have controlled for age: both treatment groups have subjects that are the same age. A second option is to sort the subjects into the two treatment groups so that a 21 year old is in Group A, and the other 21 year old is in Group B, and so on. Again, the average age of Group A subjects and of Group B subjects would be the same and therefore controlled with respect to any covariation between age and change in blood pressure. However, there are other variables that may covary with blood pressure, and by controlling one, we would need to control the others. Randomization provides a better way.

I will demonstrate how randomization tends to distribute the values in such a way that the groups will not differ appreciably for the nuisance variables like age and BMI differences and, by extension, any other covariable. The R work is attached following the Reading list. The take-home message: After randomly selecting subjects for assignment to the treatment groups, the apparent differences between Group A and Group B for both age and BMI are substantially diminished. No attempt to match by age and by BMI is necessary. The numbers are shown in the table and then in two graphics (Fig 1, Fig 2) derived from the table.

Table 1. Mean age and BMI for subjects in two treatment groups A and B where subjects were assigned randomly or by convenience to treatment groups.

Group	Random assignment of subjects to treatment groups	Convenience assignment of subjects to treatment groups
A	35.2	28
B	35.8	43
A	32.49	28.99
B	32.87	37.37

Just for emphasis, the means from Table 1 are presented in the next two figures (Fig 1 and Fig 2).

Figure 1. Age of subjects by groups (A = blue, B = red) with and without randomized assignment of subjects to treatment groups.

Figure 2. BMI of subjects by groups (A = blue, B = red) with and without randomized assignment of subjects to treatment groups.

Note that the apparent difference between A and B for BMI disappear once proper randomization of subjects was accomplished. In conclusion, a random sample is an approach to experimental design that helps to reduce the influence other factors may have on the outcome variable (eg, change in blood pressure after 16 weeks of exercise). In principle, randomization should protect a project because, on average, these influences will be represented randomly for the two groups of individuals. This reasoning extends to unmeasured and unknown causal factors as well.

This discussion was illustrated by random assignment of subjects to treatment groups. The same logic applies to how to select subjects from a population. If the sampling is large enough, then a random sample of subjects will tend to be representative of the variability of the outcome variable for the population and representative also of the additional and unmeasured cofactors that may contribute to the variability of the outcome variable.

What about observational studies? How does randomization work?

However, if you do cannot obtain a random sample, then conclusions reached may be sample-specific, biased. …perhaps the group of individuals that likes to exercise on treadmills just happens to have a higher cardiac output because they are larger than the individuals that like to exercise on bicycles. This nonrandom sample, an example of confounding where an extraneous or nuisance variable, body size, is associated with both the independent variable (exercise preference) and the dependent variable (cardiac output), will bias your results and can lead to incorrect interpretation of results.

Alternatively, perhaps a researcher elects to study the cardiac output of treadmill users versus bicycle users. The researcher sets up their testing station inside a gym’s powerlifting zone, where the largest individuals in the gym hang out. Inevitably because of this choice, the researcher over recruits larger individuals into the treadmill group, leading to a falsely elevated average cardiac output for treadmill users. This is an example of ascertainment bias — the sample is not representative of all treadmill users at the gym.

Random sampling is CRUCIAL in epidemiology, opinion survey work, most aspects of health, drug studies, medical work with human subjects. It’s difficult and very costly to do… so most surveys you hear about, especially polls reported from Internet sites, are NOT conducted using random sampling (included in the catch-all term “probability sampling“)!! As an aside, most opinion survey work involves complex sample designs involving some form of geographic clustering (eg, all phone numbers in a city, random sample among neighborhoods).

Random sampling is the ideal if generalizations are to be made about data, but strictly random sampling is not appropriate for all kinds of studies. Consider the question of whether or not EMF exposure is a risk factor for developing cancer (Pool 1990). These kinds of studies are observational: at least in principle, we wouldn’t expect that housing and therefore exposure to EMF is manipulated (cf. discussion Walker 2009). Thus, epidemiologists will look for patterns: if EMF exposure is linked to cancer, then more cases of cancer should occur near EMF sources compared to areas distant from EMF sources. Thus, the hypothesis is that an association between EMF exposure and cancer occurs non-randomly, whereas cancers occurring in people not exposed to EMF are random. Unfortunately, clusters can occur even if the process that generates the data is random.

Compare Graph A and Graph B (Fig 3). One of the graphs resulted from a random process and the other was generated by a non-random process. Note that the claim can be rephrased about the probability that each grid has a point, eg, it’s like Heads/Tails of 16 tosses of a coin. We can see clusters of points in Graph B; Graph A lacks obvious clusters of points — there is a point in each of the 16 cells of the grid. Although both patterns could be random, the correct answer in this case is Graph B.

Figure 3. An example of clustering resulting from a random sampling process (Graph B). In contrast, Graph A was generated so that a point was located within each grid.

Hard to tell — and truth be told, random process could generate either one. To convince you, Figure 4 reveals the grouping that was done for Group A, in contrast with Group B which was drawn with the truly random process.

Figure 4. Same graphs as Figure 3, but with ellipses around the grouped data (hard to tell, but the centroids are the larger points). 

Implications for sampling

The graphic below shows the transmission grid in the continental United States (Fig 5). How would one design a random sampling scheme overlaid against the obviously heterogeneous distribution of the grid itself? If a random sample was drawn, chances are good that no population would be near a grid in many of the western states, but in contrast, the likelihood would increase in the eastern portion of the United States where the population and therefore transmission grid is more densely placed.

Figure 5. Map of electrical transmission grid for continental United States of America. Image source https://openinframap.org/#3/24.61/-101.16

For example, you want to test whether or not EMF affects human health, and your particular interest is in whether or not there exists a relationship between living close to high voltage towers or transfer stations and brain cancer. How does one design a study, keeping in mind the importance of randomization for our ability to generalize and assign causation?  This is a part of epidemiology which strives to detect whether clusters of disease are related to some environmental source. It is an extremely difficult challenge. For the record, no clear link to EMF and cancer has been found, but reports do appear from time to time (eg, report on a cluster of breast cancer in men working in office adjacent to high EMF, Milham 2004).

 

Questions

1. Use the sample() with and without replacement on the object (see help with R below)

fruit <- c("apple", "banana", "grape", "kiwi", "pear", "pineapple", "tomato")

a) set of 3

b) set of 4

2. Confirm the claim by calculating the probability of Graph A result vs Graph B result (see R script below).

Quiz Chapter 5.5

Importance of randomization in experimental design

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---

R code!

Code you type is shown in red; responses or output from R are shown in blue. Recall that statements preceded by the hash # are comments and are not read by R (ie, no need for you to type them).

First, create some variables. Vectors aa and bb contain my two age sequences.

 aa = seq(21,50)
 bb = seq(21,50)

Second, append vector bb to the end of vector aa

 age = append(aa,bb)
 age           #submit the vector (variable) name to print the records for verification - looks good! 
  [1] 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37
 [48] 38 39 40 41 42 43 44 45 46 47 48 49 50

Third, get the average age for the first group (the aa sequence)  and for the second group (the bb sequence). Lots of ways to do this, I made a two subsets from the combined age variable; could have just as easily taken the mean of aa and the mean of bb (same thing!).

 A = age[1:30]
 mean(A)
 [1] 35.5
 B = age[31:60]
 mean(B)
 [1] 35.5

Fourth, start building a data frame, then sort it by age. Will be adding additional variables to this data frame

 ex.random = data.frame(age)
 AO.all = sort(ex.random$age)
 AO.age           #submit the vector (variable) name to print the records for verification - looks good! 
  [1] 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44
 [48] 44 45 45 46 46 47 47 48 48 49 49 50 50

Fifth, divide the variable again into two subsets of 30 and get the averages

 AO = AO.age[1:30]
 AO
  [1] 21 21 22 22 23 23 24 24 25 25 26 26 27 27 28 28 29 29 30 30 31 31 32 32 33 33 34 34 35 35
 mean(AO)
 [1] 28 
 BO = AO.age[31:60]
 BO
  [1] 36 36 37 37 38 38 39 39 40 40 41 41 42 42 43 43 44 44 45 45 46 46 47 47 48 48 49 49 50 50
 mean(BO)
 [1] 43

Sixth, create an index variable, random order without replacement

 rand.index = sample(1:60,60,replace=F)

Add the new variable to our existing data frame, then print it to check that all is well

 ex.random$rand = rand.index
 ex.random
    age rand
 1   21   43
 2   22   15
 3   23   17
 4   24   35
 5   25   19
 6   26   18
 7   27   22
 8   28   31
 9   29   12
 10  30   44
 11  31   24
 12  32    5
 13  33    2
 14  34   50
 15  35   23
 16  36   20
 17  37   41
 18  38   56
 19  39   36
 20  40    8
 21  41   45
 22  42   38
 23  43   42
 24  44   46
 25  45   16
 26  46   21
 27  47   28
 28  48   10
 29  49   32
 30  50   54
 31  21   57
 32  22   51
 33  23   27
 34  24   40
 35  25   14
 36  26   48
 37  27   26
 38  28   58
 39  29    9
 40  30   11
 41  31    4
 42  32   52
 43  33   37
 44  34   53
 45  35    6
 46  36   34
 47  37   39
 48  38    7
 49  39    1
 50  40   47
 51  41   33
 52  42   60
 53  43   49
 54  44   30
 55  45   29
 56  46   55
 57  47   13
 58  48    3
 59  49   25
 60  50   59

Seventh, select for our first treatment group the first 30 subjects from the randomized index. There are again other ways to do this, but sorting on the index variable means that the subject order will be change too.

 AR.age = ex.random[order(ex.random$rand),]     #created a new data frame to distinguish it from the presorted version "ex.random"

Print the new data frame to confirm that the sorting worked. It did. we can see that the rows have been sorted by ascending order based on the index variable.

 AR.age
    age rand
 49  39    1
 13  33    2
 58  48    3
 41  31    4
 12  32    5
 45  35    6
 48  38    7
 20  40    8
 39  29    9
 28  48   10
 40  30   11
 9   29   12
 57  47   13
 35  25   14
 2   22   15
 25  45   16
 3   23   17
 6   26   18
 5   25   19
 16  36   20
 26  46   21
 7   27   22
 15  35   23
 11  31   24
 59  49   25
 37  27   26
 33  23   27
 27  47   28
 55  45   29
 54  44   30
 8   28   31
 29  49   32
 51  41   33
 46  36   34
 4   24   35
 19  39   36
 43  33   37
 22  42   38
 47  37   39
 34  24   40
 17  37   41
 23  43   42
 1   21   43
 10  30   44
 21  41   45
 24  44   46
 50  40   47
 36  26   48
 53  43   49
 14  34   50
 32  22   51
 42  32   52
 44  34   53
 30  50   54
 56  46   55
 18  38   56
 31  21   57
 38  28   58
 60  50   59
 52  42   60

Eighth, create our new treatment groups, again of n = 30 each, then get the means ages for each group.

 AR = AR.age$age[1:30]
 mean(AR)
 [1] 35.16667
 AR2 = AR.all$all[31:60]
 mean(AR2)
 [1] 35.83333

Get the minimum and maximum values for the groups

 min(AR)
 [1] 22
 min(AR2)
 [1] 21
 max(AR)
 [1] 49
 max(AR2)
 [1] 50

Ninth, create a BMI variable drawn from a normal distribution with coefficient of variation equal to 20%. The first group with we will call cc

 cc = rnorm(n=30,m=27.5, sd=5.5)          #mean was 27.5 for this group with standard deviation of 5.5

The second group called dd

 dd = rnorm(n=30,m=37.5, sd=7.5)          #mean was 37.5 for this group with standard deviation of 7.5

Create a new variable called BMI by joining cc and dd

 BMI=append(cc,dd)
 BMI            #print out BMI to confirm. Looks good!
  [1] 27.87528 27.83250 31.88703 34.99041 24.06751 23.50952 22.57779 31.48394 31.04321 25.60258 25.41081 22.34619 34.62213 36.41348 41.17740
 [16] 20.56529 27.25238 21.85205 32.11690 32.37168 23.11314 33.29110 34.99106 38.22016 18.72105 26.22030 25.13412 27.50475 34.79361 32.81267
 [31] 47.57872 27.58428 40.17211 38.22195 26.91893 37.02784 53.72671 34.94727 30.35245 38.32571 40.52111 36.15627 30.36592 36.20397 47.63142
 [46] 40.30846 36.47643 50.86804 43.63741 37.84994 42.82665 41.71008 28.44976 24.57906 42.37762 38.38512 35.22879 31.34063 34.02996 27.28038

Add the BMI variable to our data frame.

 ex.random$BMI = BMI
 ex.random              #Print out the revised data frame. Looks good. We now have three variables: age, the index "rand," and now BMI
    age rand      BMI
 1   21   43 27.87528
 2   22   15 27.83250
 3   23   17 31.88703
 4   24   35 34.99041
 5   25   19 24.06751
 6   26   18 23.50952
 7   27   22 22.57779
 8   28   31 31.48394
 9   29   12 31.04321
 10  30   44 25.60258
 11  31   24 25.41081
 12  32    5 22.34619
 13  33    2 34.62213
 14  34   50 36.41348
 15  35   23 41.17740
 16  36   20 20.56529
 17  37   41 27.25238
 18  38   56 21.85205
 19  39   36 32.11690
 20  40    8 32.37168
 21  41   45 23.11314
 22  42   38 33.29110
 23  43   42 34.99106
 24  44   46 38.22016
 25  45   16 18.72105
 26  46   21 26.22030
 27  47   28 25.13412
 28  48   10 27.50475
 29  49   32 34.79361
 30  50   54 32.81267
 31  21   57 47.57872
 32  22   51 27.58428
 33  23   27 40.17211
 34  24   40 38.22195
 35  25   14 26.91893
 36  26   48 37.02784
 37  27   26 53.72671
 38  28   58 34.94727
 39  29    9 30.35245
 40  30   11 38.32571
 41  31    4 40.52111
 42  32   52 36.15627
 43  33   37 30.36592
 44  34   53 36.20397
 45  35    6 47.63142
 46  36   34 40.30846
 47  37   39 36.47643
 48  38    7 50.86804
 49  39    1 43.63741
 50  40   47 37.84994
 51  41   33 42.82665
 52  42   60 41.71008
 53  43   49 28.44976
 54  44   30 24.57906
 55  45   29 42.37762
 56  46   55 38.38512
 57  47   13 35.22879
 58  48    3 31.34063
 59  49   25 34.02996
 60  50   59 27.28038

Tenth, repeat our protocol from before: Set up two groups each with 30 subjects, calculate the means for the variables and then sort by the random index and get the new group means.

 AO = ex.random$BMI[1:30]
 mean(AO)
 [1] 28.99333
 BO = ex.random$BMI[31:60]
 mean(BO)
 [1] 37.36943

All we did was confirm that the unsorted groups had mean BMI of around 27.5 and 37.5 respectively. Now, proceed to sort by the random index variable. Go ahead and create a new data frame

 AR.age = ex.random[order(ex.random$rand),]
 AR.age                 #Print out the new data frame to confirm. Looks good.
    age rand      BMI
 49  39    1 43.63741
 13  33    2 34.62213
 58  48    3 31.34063
 41  31    4 40.52111
 12  32    5 22.34619
 45  35    6 47.63142
 48  38    7 50.86804
 20  40    8 32.37168
 39  29    9 30.35245
 28  48   10 27.50475
 40  30   11 38.32571
 9   29   12 31.04321
 57  47   13 35.22879
 35  25   14 26.91893
 2   22   15 27.83250
 25  45   16 18.72105
 3   23   17 31.88703
 6   26   18 23.50952
 5   25   19 24.06751
 16  36   20 20.56529
 26  46   21 26.22030
 7   27   22 22.57779
 15  35   23 41.17740
 11  31   24 25.41081
 59  49   25 34.02996
 37  27   26 53.72671
 33  23   27 40.17211
 27  47   28 25.13412
 55  45   29 42.37762
 54  44   30 24.57906
 8   28   31 31.48394
 29  49   32 34.79361
 51  41   33 42.82665
 46  36   34 40.30846
 4   24   35 34.99041
 19  39   36 32.11690
 43  33   37 30.36592
 22  42   38 33.29110
 47  37   39 36.47643
 34  24   40 38.22195
 17  37   41 27.25238
 23  43   42 34.99106
 1   21   43 27.87528
 10  30   44 25.60258
 21  41   45 23.11314
 24  44   46 38.22016
 50  40   47 37.84994
 36  26   48 37.02784
 53  43   49 28.44976
 14  34   50 36.41348
 32  22   51 27.58428
 42  32   52 36.15627
 44  34   53 36.20397
 30  50   54 32.81267
 56  46   55 38.38512
 18  38   56 21.85205
 31  21   57 47.57872
 38  28   58 34.94727
 60  50   59 27.28038
 52  42   60 41.71008

Get the means of the new groups

 AR = AR.age$BMI[1:30]
 mean(AR)
 [1] 32.49004
 min(AR)
 [1] 18.72105
 max(AR)
 [1] 53.72671
 AR2 = AR.all$BMI[31:60]
 mean(AR2)
 [1] 33.87273
 min(AR2)
 [1] 21.85205
 max(AR2)
 [1] 47.57872

That’s all of the work!

Data set

nonrandom

X	Y
0.779	0.747
0.098	1.248
0.697	2.811
0.59	3.553
1.295	0.075
1.858	1.207
1.172	2.388
1.597	3.051
2.051	0.242
2.081	1.546
2.016	2.738
2.587	3.088
3.833	0.088
3.776	1.264
3.017	2.254
3.657	3.712

random

X1	Y1
1.157	3.203
2.407	0.62
1.033	1.946
3.676	2.866
3.57	2.703
3.056	0.685
1.568	3.355
1.041	1.753
0.643	2.295
2.884	0.687
3.53	2.591
2.838	2.431
3.945	1.518
0.34	1.64
2.45	0.386
3.921	2.976

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Chapter 5 contents

• The basics explained
• Experiments
• Experimental and Sampling units
• Replication, Bias, and Nuisance Variables
• Clinical trials
• Importance of randomization in experimental design
• Sampling from Populations
• References and suggested readings