Much of statistical inference concerns the location of the population mean $$\mu$$ for a given parametric distribution. Some of the most common approaches to making inference about $$\mu$$ utilize a test statistic that follows a t distribution.

### One Sample t Test

A one sample t test is used when there is a hypothesized value for the population mean $$\mu$$ of a single quantitative variable.

#### Overview

This test is only appropriate when both of the following are satisfied.

1. The sample is representative of the population. (Having a simple random sample is the best way to do this.)

2. The sampling distribution of the sample mean $$\bar{x}$$ can be assumed to be normal. This is a safe assumption when either (a) the population data can be assumed to be normally distributed or (b) the size of the sample taken from the population is large.

Hypotheses

$$H_0: \mu = \text{some number}$$

$$H_a: \mu \ \left\{\underset{<}{\stackrel{>}{\neq}}\right\} \ \text{some number}$$

Examples: sleepOne

#### R Instructions

Console Help Command: ?t.test()

t.test(object, mu = YourNull, alternative = YourAlternative, conf.level = 0.95)

• object must be a “numeric” vector.
• YourNull is the numeric value from your null hypothesis for $$\mu$$.
• YourAlternative is one of the three options: "two.sided", "greater", "less" and should correspond to your alternative hypothesis.
• The value for conf.level = 0.95 can be changed to any desired confidence level, like 0.90 or 0.99. It should correspond to $$1-\alpha$$.

#### Explanation

In many cases where it is of interest to test a claim about a single population mean $$\mu$$, the one sample t test is used. This is an appropriate decision whenever the sampling distribution of the sample mean can be assumed to be normal and the data represents a simple random sample from the population. (Review of what makes the sampling distribution of the sample mean normal.)

In the figure below, the null hypothesis $$H_0: \mu = \mu_0$$ is represented by the normal distribution (gray) centered at $$\mu_0$$. Note that $$\mu_0$$ is just some specified number. This shows how the null hypothesis represents the assumption about the center of the distribution of the data.

After a hypothesis (null) is established and an alternative hypothesis similarly declared, a simple random sample of data of size $$n$$ is obtained from the population of interest. In the plot above, this is depicted by the points (blue dots) which are centered around their sample mean $$\bar{x}$$.

Above the points (blue dots) is shown a second normal distribution (blue dashed line) which represents the idea that the alternative hypothesis allows for a normal distribution which is potentially more consistent with the data than the one specified under the null hypothesis.

The role of the one sample t test is to measure the probability of a sample mean being as extreme or more extreme from the hypothesized value of $$\mu_0$$ than the one observed assuming the null hypothesis is true. This probability is of course the p-value of the test. This works because the sampling distribution of the sample mean has been assumed to be normal. In this case, the distribution of the test statistic t, $t = \frac{\bar{x}-\mu}{s/\sqrt{n}}$
is known to follow a t distribution with $$n-1$$ degrees of freedom. (The mathematics that provide this result are phenominal! You can consult any advanced statistical textbook for the details.)

The p-value of the one sample t test represents the probability that the test statistic $$t$$ is as extreme or more extreme than the one observed according to a t-distribution with $$n-1$$ degrees of freedom.

If the probability (the p-value) is close enough to zero (smaller than $$\alpha$$) then it is determined that the most plausible hypothesis is the alternative hypothesis, and thus the null is “rejected” in favor of the alternative.

### Paired Samples t Test

The paired samples t test is used when a value is hypothesized for the popluation mean of the differences, $$\mu_d$$, obtained from paired observations.

#### Overview

Paired samples include pre-test post-test scenarios and matched-pairs scenarios, where the only interest is in the difference between the two scores. Such scenarios begin with two sets of measurements for each individual (or each matched-pair). However, in the end, these two measurements are reduced to a single set of “differences”. Thus, paired data is essentially one sample of differences.

Requirements

The test is only appropriate when both of the following are satisfied.

1. The sample of differences is representative of the population differences.

2. The sampling distribution of the sample mean of the differences $$\bar{d}$$ ($$\bar{x}$$ of the differences) can be assumed to be normal. (This second requirement can be assumed to be satisfied when (a) the differences themselves can be assumed to be normal, or (b) when the sample size $$n$$ of the differences is large.)

Hypotheses

$$H_0: \mu_d = \text{some number, but typically 0}$$
$$H_a: \mu_d \ \left\{\underset{<}{\stackrel{>}{\neq}}\right\} \ \text{some number, but typically 0}$$

Examples: sleepPaired

#### R Instructions

Console Help Command: ?t.test()

t.test(object1, object2, paired = TRUE, mu = YourNull, alternative = YourAlternative, conf.level = 0.95)

• object1 must be a “numeric” vector that represents the first sample of data.
• obejct2 must be a “numeric” vector that represents the second sample of data. This vector must be in the same order as the first sample so that the pairing can take place.
• YourNull is the numeric value from your null hypothesis for $$\mu_d$$.
• YourAlternative is one of the three options: "two.sided", "greater", "less" and should correspond to your alternative hypothesis.
• The value for conf.level = 0.95 can be changed to any desired confidence level, like 0.90 or 0.99. It should correspond to $$1-\alpha$$.

#### Explanation

The paired samples t test considers the single mean of all the differences from the paired values. Thus, the paired samples t test essentially becomes a one sample t test on the differences between paired observations. Hence the requirement is that the sampling distribution of the sample mean of the differences, $$\bar{d}$$, can be assumed to be normally distributed. (It is also required that the obtained differences represent a simple random sample of the full population of possible differences.)

The paired samples t test is similar to the independent samples t test scenario, except that there is extra information that allows values from one sample to be paired with a value from the other sample. This pairing of values allows for a more direct analysis of the change or difference individuals experience between the two samples.

The points in the plot below demonstrate how points are paired together, and the only thing of interest are the differences between the paired points.

### Independent Samples t Test

The independent samples t test is used when a value is hypothesized for the difference between two (possibly) different population means, $$\mu_1 - \mu_2$$.

#### Overview

The test is only appropriate when both of the following are satisfied.

1. Both samples are representative of the population. (Simple random samples are the best way to do this.)

2. The sampling distribution of the difference of the sample means $$(\bar{x}_1 - \bar{x}_2)$$ can be assumed to be normal. (This is a safe assumption when the sample size of each group is $$30$$ or greater or when the population data from each group can be assumed to be normal.)

Hypotheses

$$H_0: \mu_1 - \mu_2 = \text{some number, but typically 0}$$ $$H_a: \mu_1 - \mu_2 \ \left\{\underset{<}{\stackrel{>}{\neq}}\right\} \ \text{some number, but typically 0}$$

Examples: sleepInd

#### R Instructions

Console Help Command: ?t.test()

There are two ways to perform the test.

Option 1:

t.test(object1, object2, mu = YourNull, alternative = YourAlternative, conf.level = 0.95)

• object1 must be a “numeric” vector that represents the first sample of data.
• obejct2 must be a “numeric” vector that represents the second sample of data.
• YourNull is the numeric value from your null hypothesis for $$\mu_1-\mu_2$$.
• YourAlternative is one of the three options: "two.sided", "greater", "less" and should correspond to your alternative hypothesis.
• The value for conf.level = 0.95 can be changed to any desired confidence level, like 0.90 or 0.99. It should correspond to $$1-\alpha$$.

Option 2:

t.test(column ~ group, data = YourData, mu = YourNull, alternative = YourAlternative, conf.level = 0.95)

• column must be a “numeric” vector from YourData that represents the data for both samples.
• group must be a “factor” or “character” vector from YourData that represents the group assignment for each observation. There can only be two groups specified in this column of data.
• YourNull is the numeric value from your null hypothesis for $$\mu_1-\mu_2$$.
• YourAlternative is one of the three options: "two.sided", "greater", "less" and should correspond to your alternative hypothesis.
• The value for conf.level = 0.95 can be changed to any desired confidence level, like 0.90 or 0.99. It should correspond to $$1-\alpha$$.

#### Explanation

The first figure below depicts the scenario where the difference in means of two separate normal distributions is non-zero. In other words, the two distributions have different means, $$\mu_1$$ and $$\mu_2$$, respectively. It is worth emphasizing that the values of $$\mu_1$$ and $$\mu_2$$ are unknown to the researcher. The only thing observed are two separate samples of data (blue dots) of sizes $$n_1$$ and $$n_2$$, respectively. For the scenario depicted, the null hypothesis that $$H_0: \mu_1 - \mu_2 = 0$$ (i.e., that $$\mu_1=\mu_2$$) is rejected in favor of the alternative that $$H_a: \mu_1 - \mu_2 \neq 0$$ based on the sample data observed. This dicision would be correct as the true difference in means, $$\mu_1-\mu_2$$ is non-zero in this case.

When the null hypothesis is true, that $$H_0: \mu_1 - \mu_2 = 0$$, then it follows that the test statistic $$t$$ that is obtained by measuring the distance between the two sample means, $$\bar{x}_1-\bar{x}_2$$, and appropriately standardizing the result follows a $$t$$ distribution with degrees of freedom less than or equal to $$n_1+n_2-2$$. Thus, the $$p$$-value of the independent samples $$t$$ test is obtained by using this $$t$$ distribution to calculate the probability of a test statistic $$t$$ being as extreme or more extreme than the one observed assuming the null hypothesis is true. $t = \frac{(\bar{x}_1 - \bar{x}_2) - (\mu_1 - \mu_2)}{\sqrt{s_1/n_1+s_2/n_2 }}$

The plot below demonstrates what data might look like when the null hypothesis is actually true. In other words, when both samples come from the same distribution.