28.3.1 For single mean

In this section, we will be looking at the Age column from the icu data. So, let’s start by plotting it:

boxplot(icu$Age, ylab = "Age (years)")

As when we started to use the t-distribution, the simplest use for the t-test is to test a single mean. For this, we simply give the function t.test() a numeric vector, such as the Age column from the icu data:

# Simple one sample t-test
t.test(icu$Age)

## 
##  One Sample t-test
## 
## data:  icu$Age
## t = 40.5796, df = 199, p-value < 2.2e-16
## alternative hypothesis: true mean is not equal to 0
## 95 percent confidence interval:
##  54.74861 60.34139
## sample estimates:
## mean of x 
##    57.545

This gives us a 95% confidence interval, which here means that we are 95% confident that the true mean age lies between 54.75 and 60.34 years old and tells us that the mean value is 57.55. This is based on exactly the same methods that we have been working with.

It also gives us a t-statistic, degrees of freedom, and a p-value. However, if you look at the alternative hypothesis you will note that it is testing whether or not the mean is 0. That is probably not a reasonable null hypothesis to have. Instead, we might have predicted that the mean was 60 (e.g., if that was the value at some other hospital). Thus, our H₀ becomes that mean is equal to 60. We can test that mean by setting the argument mu = to our null (recall that mu, the Greek character μ, is the notation for the true population mean).

# Test a different alternate hypothesis
t.test(icu$Age, mu = 60)

## 
##  One Sample t-test
## 
## data:  icu$Age
## t = -1.7312, df = 199, p-value = 0.08496
## alternative hypothesis: true mean is not equal to 60
## 95 percent confidence interval:
##  54.74861 60.34139
## sample estimates:
## mean of x 
##    57.545

Note that our confidence interval did not change, but the t-statistic and p-value (and null hypothesis) did. The t-statistic is calculated the same way that we have been doing it by hand

\[ t = \frac{obs - H_0}{SE} \]

That is the difference between our observed data and the null hypothesis, divided by the calculated standard error (\(SE = \frac{s}{\sqrt{n}}\)). This gives us how different our data are from the null in number of standard errors. The function then calls pt() to determine how likely such a difference (or bigger) was to happen by chance.

Finally, we can also change the size of the confidence interval by setting the conf.level = argument to the size of our desired confidence interval. So, we can calculate a 90% confidence interval like this:

# Test a different alternate hypothesis
t.test(icu$Age, mu = 60, conf.level = 0.90)

## 
##  One Sample t-test
## 
## data:  icu$Age
## t = -1.7312, df = 199, p-value = 0.08496
## alternative hypothesis: true mean is not equal to 60
## 90 percent confidence interval:
##  55.20156 59.88844
## sample estimates:
## mean of x 
##    57.545

Of note, we can pass in any numeric vector, including the value for a subset. So, we could calculate the 90% confidence interval (and test the null hypothesis that the mean is 60 years old) for just women with the following:

# Use this on a subset of data
t.test(icu$Age[icu$Sex == "female"],
       mu = 60, conf.level = 0.90)

## 
##  One Sample t-test
## 
## data:  icu$Age[icu$Sex == "female"]
## t = 0, df = 75, p-value = 1
## alternative hypothesis: true mean is not equal to 60
## 90 percent confidence interval:
##  56.00834 63.99166
## sample estimates:
## mean of x 
##        60

We once again get a confidence interval, though this time our p-value is 1. That is because the observed mean is equal to the null hypothesis. However: that still does not mean that the null hypothesis is (necessarily) true. It only means that we cannot reject it.

28.3.2 For difference in means

The function t.test() also works well for testing for a difference in means, such as between two groups. For this, let’s look at the Age column of icu split by Infection. As always, start by plotting your data.

# Show a box plot of the two groups
plot(Age ~ Infection, data = icu)

There looks to be a small, but not huge, difference between the two groups. We can test it using the t.test() function with the exact same arguments we used to plot the data. This is one of the benefits of the copy-paste approach: once you have plotted the data, you can just copy the line, replace plot with t.test and go.

# Run a t.test
t.test(Age ~ Infection, data = icu)

## 
##  Welch Two Sample t-test
## 
## data:  Age by Infection
## t = -2.2481, df = 193.601, p-value = 0.02569
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -11.6837808  -0.7636741
## sample estimates:
##  mean in group no mean in group yes 
##          54.93103          61.15476

This gives us means of each group, the information for the test, and a confidence interval for the difference between the groups. One important note is the degrees of freedom. When doing these by hand, we were using one less than the smaller sample size. Here, the function instead uses the more complicated formula[*] The formula for the degrees of freedom is technically: \[ \frac{\left( \frac{s_1^2}{n_1} + \frac{s_1^2}{n_1} \right)^2} { \frac{\left( \frac{s_1^2}{n_1} \right)^2}{n_1 - 1} + \frac{\left( \frac{s_2^2}{n_2} \right)^2}{n_2 - 1} } \] Hopefully it is clear why I hid it here. to calculate the degrees of freedom. That difference is why your answers from doing differences in means by hand will differ slightly from the answers given by t.test()

library(gplots)

# ONLY RUN THIS IF YOU NEED TO RE-INSTALL
install.packages("gplots")

# Plot the data with confidence intervals
plotmeans(Age ~ Infection, data = icu)

plotmeans(Age ~ Infection, data = icu,
          n.label = FALSE, connect = FALSE)

# t.test for server's tips
t.test(Age ~ Consciousness,
       data = icu)

## Error in t.test.formula(Age ~ Consciousness, data = icu): grouping factor must have exactly 2 levels

# t.test for Age by Consciousness with out deepStupor
# don't forget the comma to say we are talking about rows
t.test(Age ~ Consciousness,
       data = icu[icu$Consciousness != "deepStupor",])

## 
##  Welch Two Sample t-test
## 
## data:  Age by Consciousness
## t = 1.9803, df = 11.797, p-value = 0.07148
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.8575128 17.6142696
## sample estimates:
##      mean in group coma mean in group conscious 
##                65.40000                57.02162

# t.test for Age by Consciousness with only coma and conscious
# don't forget the comma to say we are talking about rows
t.test(Age ~ Consciousness,
       data = icu[icu$Consciousness == "coma" |
                  icu$Consciousness == "conscious",])

## 
##  Welch Two Sample t-test
## 
## data:  Age by Consciousness
## t = 1.9803, df = 11.797, p-value = 0.07148
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.8575128 17.6142696
## sample estimates:
##      mean in group coma mean in group conscious 
##                65.40000                57.02162

t.test(Age ~ Consciousness == "conscious",
       data = icu)

## 
##  Welch Two Sample t-test
## 
## data:  Age by Consciousness == "conscious"
## t = 2.0466, df = 21.538, p-value = 0.05309
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.1017174 14.0584742
## sample estimates:
## mean in group FALSE  mean in group TRUE 
##            64.00000            57.02162

# Plot the full result
plotmeans(Age ~ Consciousness,
          data = icu,
          n.label = FALSE, connect = FALSE)

# Plot the full result
plotmeans(Age ~ Consciousness == "conscious",
          data = icu,
          n.label = FALSE, connect = FALSE)

# Plot the full result
plotmeans(Age ~ Consciousness == "conscious",
          data = icu,
          legends = c("not conscious","conscious"),
          xlab = "Conscious State",
          n.label = FALSE, connect = FALSE)

28.3.3 Paired data

Our final approach for t.test() will be to assess paired variables. For this, we will compare lab quiz 6 (column labQuiz06) against the lecture quiz 3 (column classQuiz03) from the classGrades data.

This will require some extra manipulation as the lab quizzes are out of 5 points, and the class quizzes are out of 10 points. To handle this, we will represent each as a percentage score and save them in their own variable.

# Store the variables
labQ <- classGrades$labQuiz06 / 5
lecQ <- classGrades$classQuiz03 / 10

We can then plot those scores:

boxplot(labQ, lecQ,
        names = c("lab","lecture"),
        xlab = "Quiz type",
        ylab = "Percent score")

Now, we can use t.test() very similarly to boxplot() and pass in two vectors of data. t.test() will then compare them against each other.

# Run a t.test
t.test(labQ, lecQ)

## 
##  Welch Two Sample t-test
## 
## data:  labQ and lecQ
## t = -1.6932, df = 37.23, p-value = 0.09877
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.22808939  0.02039708
## sample estimates:
## mean of x mean of y 
## 0.7076923 0.8115385

But, if you recall from our earlier chapter, we know more about these data. Specifically we know that they are paired. To handle this with the t.test() function, we add the argument paired = TRUE.

# Run paired t.test
t.test(labQ, lecQ, paired = TRUE)

## 
##  Paired t-test
## 
## data:  labQ and lecQ
## t = -2.0999, df = 25, p-value = 0.04599
## alternative hypothesis: true difference in means is not equal to 0
## 95 percent confidence interval:
##  -0.205695742 -0.001996566
## sample estimates:
## mean of the differences 
##              -0.1038462

Now that is what we had in our previous chapter. This makes it easy to handle paired data, but we still need to remember when it is appropriate. One good rule of thumb is to always try to use the formula approach if possible. If it is not possible, think about why that is the case and you will often (though not always) find that it is because your data are paired measures of different things, rather than a measure of one thing separated by another variable that groups them.

28.3.4 (Not) Working from summary data

Perhaps unfortunately, the t.test() function does not provide a means to work from summary data. This is, perhaps, because the need is rare outside of intro stats courses, or perhaps becuase of the simplicity of the calculations. In either case, to work from summary data, you will have to continue working the problems by hand. Alternatively, you could write a function of your own to handle summary data for t-tests. The choice is yours.

28.4.1 One sample

For a one-sample test, we just give it our successes and trials, along wtih a null hypothesis. So, if we want to test whether or not the proportion of female patients in the ICU is 0.5 (a null that assumes equal numbers of men and women), we would use the following. Note that this treats the first “level” of the table as the “successes”, so we are given an estimate (and confidence interval) for the proportion of female patients. Make suree that you are aware of this when using this tool.

# View data
table(icu$Sex)

## 
## female   male 
##     76    124

# Run one sample test
prop.test(table(icu$Sex), p = 0.5)

## 
##  1-sample proportions test with continuity correction
## 
## data:  table(icu$Sex), null probability 0.5
## X-squared = 11.045, df = 1, p-value = 0.0008893
## alternative hypothesis: true p is not equal to 0.5
## 95 percent confidence interval:
##  0.3132213 0.4514680
## sample estimates:
##    p 
## 0.38

This returns a p-value showing that, with a small value, we can reject the null hypothesis that 50% of patients are male. It also gives us 95% CI. As with t.test() we can change the interval of that CI with conf.level =:

# Get a 90% CI instead
prop.test(table(icu$Sex),
          conf.level = 0.9)

## 
##  1-sample proportions test with continuity correction
## 
## data:  table(icu$Sex), null probability 0.5
## X-squared = 11.045, df = 1, p-value = 0.0008893
## alternative hypothesis: true p is not equal to 0.5
## 90 percent confidence interval:
##  0.3231067 0.4402375
## sample estimates:
##    p 
## 0.38

Note the change in confidence level, showing u the 90% level. In addition note that, by not setting p =, it defaults to 0.5.

28.4.2 Two samples

What if we want to compare two groups, like testing to see if “medical” and “surgery” Service have the same proportion of men and women? First, let’s look at our data.

# Look at the data
table(icu$Service, icu$Sex)

##          
##           female male
##   medical     39   54
##   surgery     37   70

So, our assumptions are met (at least ten in each count). We may want to manually check the proportions too. this is a good idea, especially when using a new test as it gives you a good double-check of the output values.

# Check proportion in each service
prop.table(table(icu$Service, icu$Sex), 1)

##          
##              female      male
##   medical 0.4193548 0.5806452
##   surgery 0.3457944 0.6542056

Now, we can use this table to run our test.

# run the test
prop.test(table(icu$Service, icu$Sex))

## 
##  2-sample test for equality of proportions with continuity
##  correction
## 
## data:  table(icu$Service, icu$Sex)
## X-squared = 0.8518, df = 1, p-value = 0.356
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  -0.07132024  0.21844113
## sample estimates:
##    prop 1    prop 2 
## 0.4193548 0.3457944

Note that the estimates follow the order in the table. Specifcally, the left column is the “successes” (here, being “female” because it is first alphabetically), and the proportions are in the same order as the rows (medical then surgery). If you were ever unsure of this, using the table of proportions above can be especially helpful.

As with the t.test(), this value very nearly matches what we previously calculated by hand. It is using a slightly different statistic, designed to handle more than two groups, which we will explicitly discuss in a chapter very soon.

28.4.3 Entering data from summary

Often, especially when given data for problems in an intro stats course, we are given counts instead of raw data to work with. In this case, we can use prop.test() in a slightly different way. First, let’s consider our tie example from earlier chapters:

President Artman recently started wearing a new tie. You are interested in knowing how students are responding, particularly whether men and women are responding differently. Instead of asking every student on campus, you surveyed a random subset of students (a good, representative sample). In this survey, you asked each student if they “loved” or “hated” the new tie. Your results: 57 men loved it, 41 men hated it; 227 women loved it, 78 women hated it.

Now, for prop.test() we need to submit the two success numbers (number of each sex that loved the tie) as a vector, and the two total trials (total number of each sex) as a separate vector. Therefore, we can enter our data as

# Run two sample test
prop.test( c(57    , 227),
           c(57+41 , 227+78))

## 
##  2-sample test for equality of proportions with continuity
##  correction
## 
## data:  c(57, 227) out of c(57 + 41, 227 + 78)
## X-squared = 8.6615, df = 1, p-value = 0.00325
## alternative hypothesis: two.sided
## 95 percent confidence interval:
##  -0.27862164 -0.04663765
## sample estimates:
##    prop 1    prop 2 
## 0.5816327 0.7442623