Exercise 1.1.1.1

1.1. Logic of T-tests

Even when sample means come from the same population, it is very unlikely that any two sample means will ever be exactly the same. Imagine obtaining a mean depression score of exactly 30 for CBT and 30 for psychodynamic therapy for a sample of 6 in each group, it seems very unlikely. This is the result of sampling error, the natural error we expect to see in our sample estimates.

However, the bigger the size of the difference in sample means, the more likely there is to be a true difference in population means, rather than being merely the result of sampling error.

The t-test assesses the size of the difference between sample means or mean difference, by testing the likelihood that sample means come from the same population. In the case of the CBT and psychodynamic example we could

1. fail to reject \(H_{0}\). We conclude that post-treatment depression is equal across CBT and psychodynamic therapies (i.e. suggesting that neither therapy is superior)
   • Any difference in sample means is due to sampling error
2. reject \(H_{0}\) for \(H_{1}\). We conclude that post-treatment depression is lower for CBT relative to psychodynamic therapy (i.e. CBT is more effective than psychodynamic therapy in treating depression)
   • Difference in sample means reflects a combination of both sampling error and the true difference in population means

How does the t-test accomplish this? Roughly speaking,

1. based on the Central Limit Theorem (CLT), we know that sample means will differ from each other, on average, by the standard error
2. t-test calculates the standard error and compares the 2 sample means in terms of the standard error
3. t-test sets up a signal to noise ratio, where signal is the difference in means, and noise is the size of the standard error estimate.
4. t-test then assesses whether the signal is large enough relative to noise to conclude that the sample means come from different populations

If the idea of ‘signal to noise ratio’ is new to you, or you don’t immediately grasp it, don’t worry, in the next section we will be looking more at this idea.

1.2. Sample estimates from the same population: An illustration of “Noise”

Light and Relaxation therapy for insomnia
Let’s say we know that treated insomnia patients typically get 5.5 hours of sleep on average, with a standard deviation of 2. We draw a random sample of 20 insomnia patients who received light therapy, and obtain a sample mean (\(\bar{X}\)) of hours slept of 6.5, and standard deviation of 2. We draw another random sample of 20 insomnia patients who received relaxation therapy, and who obtained a sample mean (\(\bar{X}\)) of hours of sleep of 6 with a standard deviation of 2.5.

Let’s now assume that there is no difference in the effect of light and relaxation therapy on hours of sleep acquired.

light.therapy =rnorm(10, mean = 6.5, sd = 2)
relaxation.therapy =rnorm(10, mean = 6, sd = 2)

c(mean(light.therapy), mean(relaxation.therapy))

1.3. Sample estimates from different populations

Light and Relaxation therapy for insomnia
Now let’s say that the average hours slept does in fact differ between insomnia patients treated with light therapy versus those treated with relaxation therapy. The population mean (\({\mu}\)) for light therapy is 7 hours, with a standard deviation of 1.5 (\({\sigma}\)), whilst the population mean for relaxation therapy is 5 hours (\({\mu}\)), with a standard deviation of 1.5 (\({\sigma}\))

See a visual depiction of different population distributions for light and relaxation therapy below. Notice that the distributions appear to have different central tendency (means or medians) but there is still quite a lot of overlap.

2.1 General form of the t-test

There are three types of t-tests, but they all follow the same form:

\[ \begin{aligned} t =\frac{difference\ between\ the\ means}{standard\ error} \end{aligned} \]

or in other words…

\[ \begin{aligned} t =\frac{signal}{noise} \end{aligned} \]

As mentioned previously, the t-test is much like the z-test except that the population standard deviation is unknown (known) for the t-test (z-test), and thus sample information is used to estimate the population standard deviation. See this difference illustrated in the formula below

\[ \begin{aligned} z =\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \ vs \ t =\frac{{\bar{X}_{1}} - {\bar{X}_{2}}}{\frac{s}{\sqrt{n}}} \end{aligned} \] \(\bar{X}_{1}\) = mean of sample 1 (e.g. Light therapy)
\(\bar{X}_{2}\) = mean of sample 2 (e.g. Relaxation therapy)

2.2 Degrees of freedom (df)

Degrees of freedom (df) refer to the number of values that are free to vary, a rule that is commonly applied in inferential tests. Degrees of freedom relate to the standard deviation, and deviations from the mean must always add up to 0. This is because the mean is the mid-point between minimum and maximum values, where the sum of deviations for values greater than the mean (positive sum) will equal the sum of deviations for values less than the mean (negative sum).

Degrees of freedom formula for t-tests:

\[ \begin{aligned} {single\ group\ or\ matched\ pairs}, df = n-1 \end{aligned} \]

df = degrees of freedom
n = sample size

\[ \begin{aligned} {two\ independent\ groups}, df = n_{1}+ n_{2}-2 \end{aligned} \]

df = degrees of freedom
\(n_{1}\) = group 1 sample size
\(n_{2}\) = group 2 sample size

These degrees of freedom formulas suggest that a degree of freedom is lost when a t-test is computed.

One easy way to think about degrees of freedom:

Let’s say we have a sample of 5 scores including: (1,3,5,7,9), whose mean is 5 and the sum of the deviations from the mean is therefore (1-5)+(3-5)+(5-5)+(7-5)+(9-5) = 0

If we were given all the values except for one, we could work out the last value. Thus, this value is “fixed” and therefore is not free to vary. In other words, the freedom to be varied has been lost for the last value.

2.3 Hypothesis testing and t-tests

Hypothesis testing with respect to the t-test is very much like the z-test, where the null hypothesis is being tested (i.e. the null hypothesis is assumed to be true)

In the case of a t-test:

1. The null hypothesis (\(H_{0}\)) states that there is no difference between means
2. The alternate hypothesis (\(H_{1}\)) states that either
   • there is a difference between means (non-directional hypothesis) or
   • one mean is greater than/ smaller than another mean (directional hypothesis)

In symbolic form:

1. \(H_{0}\): \({\mu_{1}}\) = \({\mu_{2}}\)
2. \(H_{1}\): \(\mu_{1} \neq \mu_{2}\) OR \(\mu_{1} > \mu_{2}\) OR \(\mu_{1} < \mu_{2}\)

An example (Treatments for insomnia):

1. \(H_{0}\): mean hours of sleep is equal for insomnia patients receiving light or relaxation therapy

• \(H_{0}\): \({\mu_{Light therapy}}\) = \({\mu_{Relaxation therapy}}\)

2. \(H_{1}\): mean hours of sleep is higher for insomnia patients receiving light therapy versus those receiving relaxation therapy (directional hypothesis)

• \(H_{1}\): \({\mu_{Light therapy}}\) > \({\mu_{Relaxation therapy}}\)

3.1 Normality assumption

T-tests assume your data is normally distributed. Put more formally, that your samples come from populations that are normally distributed.

In order to check for normality, one can draw histograms of the data for each group.

Caffeine and muscle metabolism
Of 26 participants, 13 were randomly allocated to either a placebo or caffeine condition (where one either received a pure caffeine pill or a sugar pill), after which participants engaged in an exercise task where their muscle metabolism was measured. Muscle metabolism was measured using the participant’s respiratory exchange ratio (RER), where RER is the ratio of carbon dioxide produced to oxygen consumed (%)

... = c(...)

... = c(...)

placebo = c(92, 107, 95,95,90, 98,98, 96, 99, 102, 100, 103, 99)

caffeine= c(94, 95, 93,90,92, 97,98, 94, 96, 94, 91, 94, 95)

par(mfrow = c(...,...))



hist(..., xlab ="...", main= "...")

hist(..., xlab ="...", main= "...")

par(mfrow = c(1,2))

hist(Placebo, xlab= "Respiratory Exchange Ratio (%)", main ="Distribution of muscle metabolism\n under Placebo condition")

hist(Caffeine, xlab = "Respiratory Exchange Ratio (%)", main ="Distribution of muscle metabolism\n under Caffeine condition")

3.2 Homogeneity of variance assumption

Homogeneity of variance refers to the similarity or uniformity of groups. More specifically, it means that the two groups need to have similar/equal variances

#You many want to calculate variance for each group first, in order to find which group has the larger variance

var(...)/var(...)

var(Placebo)/var(Caffeine)

3.3 Independence assumption

Groups need to be independent of one another in order for a t-test to be run. In other words, there should be different, unrelated people in each group. In the study of caffeine on muscle metabolism, this would mean that participants who received the caffeine pill need to differ from those who received the sugar pill.

The independence assumption ensures that the scores in each group have no relationship with each other.

Note that the independence assumption does not apply to paired-sample t-tests

Exercise 4.0.0.1

4.1 Single sample t-test

In this section, we will be discussing single sample t-tests

What is required in order to run a single sample t-test? 1. Single sample (i.e. group of participants) 2. Known population mean 3. Unknown population variance/standard deviation

A single sample t-test is much like a z-test, except that the population variance is unknown. If the population variance were known, then a single sample z-test could be run.

See single sample t-test formula below:

\[ \begin{aligned} t =\frac{\overline{X} - \mu}{\frac{s}{\sqrt{n}}} \end{aligned} \] \(\overline{X}\) = sample average
\({\mu}\) = population average
\({s}\) = sample standard deviation
\({n}\) = sample size

Smoking and systolic blood pressure study 1
In a series of studies investigating the relationship between smoking and systolic blood pressure, systolic blood pressure was measured amongst 10 smokers. Data is stored in the .csv file named “smoking”. The data set contains the following variables

smoking = read.csv("...")

smoking = read.csv("smoking.csv")

tstat1=(mean(...$...) - 102)/(sd(...$...)/sqrt(...))
tstat1

tstat1=(mean(smoking$sbp) - 102)/(sd(smoking$sbp)/sqrt(10))
tstat1

1-pt(tstat1, df = ...)

1-pt(tstat1, df = 9)

?t.test

t.test(...$..., alternative ="...", mu=...) # mu = population mean

t.test(smoking$sbp, alternative ="two.sided", mu=102) # mu = population mean

Exercise 5.1.1.1

var(...$...)/var(...$...)

var(smoking$sbp)/var(nonsmoking$sbp)

Exercise 5.1.1.2

var.estimate=((10-1)*...(...$...)+ (10-1)*...(...$...))/(10+10-2)
var.estimate

var.estimate=((10-1)*var(smoking$sbp)+ (10-1)*var(nonsmoking$sbp))/(10+10-2)
var.estimate

Exercise 5.1.1.3

ITtest.tstat =(mean(...$...)- mean(...$...))/sqrt(var.estimate*((1/...)+ (1/...)))
ITtest.tstat

var.estimate = (9*var(smoking$sbp) + 9*var(nonsmoking$sbp))/(18)
ITtest.tstat = (mean(smoking$sbp) - mean(nonsmoking$sbp))/sqrt(var.estimate*(2/10))
ITtest.tstat

Exercise 5.1.1.4

t.test(...$..., ...$..., alternative ="...")

t.test(smoking$sbp, nonsmoking$sbp, alternative ="greater")

Exercise 6.1.1.1

.../(.../sqrt(...))

1.7/(2/sqrt(10))

Exercise 6.1.1.2

t.test(...$..., ...$..., alternative ="...", paired =...)

t.test(smoking.treat$pre.sbp, smoking.treat$post.sbp, alternative ="two.sided", paired =TRUE)

Exercise 7.1.1.1

...=read.csv("...")

...(...)

fulldata =read.csv("smoking.fulldata.csv")

head(fulldata,6)

Exercise 7.1.1.2

t.test(...~..., data = ...)

t.test(sbp~smoker, data = fulldata)

Move on to and complete Statistics Tutorial Assignment 6 on Amathuba (Activities | Assignments)