The t distribution

The t-distribution

The t-distribution is used to make inferences about a population’s mean when working with small samples or when the population standard deviation is unknown. It comes up constantly in real research because we usually don’t know the true population standard deviation — we have to estimate it from the sample.

Definitions

t-distribution

A probability distribution used to make inferences about a population means when the sample size is small, the population standard deviation is unknown, or both. It looks similar to the normal distribution, but is flatter with heavier tails. Those heavier tails reflect the extra uncertainty that comes from using the sample standard deviation in place of the population standard deviation.

When to use the t-distribution:

1. The sample size is small (rule of thumb: $n<30$ is considered small).
2. The population standard deviation is unknown (when $\sigma$ is unknown, we use the sample standard deviation $s$ instead).
3. If $n$ is small, the population should be approximately normally distributed. For larger samples ($n>30$), the population does not necessarily need to be normally distributed because the central limit theorem tells us that the t-distribution can still be used even if the population itself isn’t normally distributed.

Procedure for a one-sample t-test

Step 1: Verify conditions

• The sample is random and independent.
• The population standard deviation $\sigma$ is unknown, or the sample size is small.

Step 2: Determine the degrees of freedom

Definitions

Degrees of freedom ($df$)

A value that identifies which specific t-distribution to use, based on sample size.

$$\begin{array}{rl}df& =n-1\text{(single sample)}\\ df& =n_1+n_2-2\text{(two independent samples)}\end{array}$$

Step 3: Find the significance level

The significance level $\alpha$ depends on the confidence level. Recall that the significance level is $\alpha$ and the confidence level is $1-\alpha$:

$$\begin{array}{rl}\text{Confidence level }95\%,\alpha & =0.05\\ \text{Confidence level }99\%,\alpha & =0.01\\ \text{Confidence level }90\%,\alpha & =0.10\end{array}$$

Step 4: Perform the t-test

For a one-sample t-test, the null hypothesis is:

$$\begin{array}{rl}{H}_0& :\mu ={\mu }_0\end{array}$$

with one of the following alternatives:

$$\begin{array}{rl}{H}_a& :\mu \ne {\mu }_0\text{(two-tailed)}\\ H_a& :\mu >{\mu }_0\text{(right-tailed)}\\ H_a& :\mu <{\mu }_0\text{(left-tailed)}\end{array}$$

Compute the t-test statistic:

$$\begin{array}{rl}t& =\frac{\bar{x}-{\mu }_0}{s/\sqrt{n}}\end{array}$$

Make a decision:

• If $|t|>t_{\text{critical}}$, reject $H_0$.
• Otherwise, fail to reject $H_0$.

Practice problem

Example:

A registered dietician claims that the average sodium content of a certain brand of soup is no more than $600$ milligrams per serving. A consumer group suspects that the true average sodium content is higher than that. They randomly select $12$ cans of this soup and find the sodium content (in mg) is as follows:

$645,610,605,630,615,625,640,650,635,620,600,655$

Assume that the sodium content in the population is approximately normally distributed.

Part a.

Why should a t-distribution be used here?

Solution:

Part b.

State the hypotheses.

Let $\mu$ be the true mean sodium content per serving.

Solution:

Part c.

Calculate the test statistic.

There are two ways to do this: by hand using a scientific calculator, or with a TI-84.

Method 1: Using a scientific calculator

Sample size:

$$\begin{array}{rl}n& =12\end{array}$$

Sample mean:

$$\begin{array}{rl}\bar{x}& =\frac{645+610+\cdots +655}{12}\\ & =\frac{7530}{12}\\ & =627.5\text{ mg}\end{array}$$

Sample standard deviation, using:

$$\begin{array}{rl}s& =\sqrt{\frac{1}{n-1}\sum _{i=1}^n(x_i-\bar{x}{)}^2}\end{array}$$

Compute each squared deviation:

$$\begin{array}{rl}(645-627.5{)}^2& =306.25\\ (610-627.5{)}^2& =306.25\\ (605-627.5{)}^2& =506.25\\ (630-627.5{)}^2& =6.25\\ (615-627.5{)}^2& =156.25\\ (625-627.5{)}^2& =6.25\\ (640-627.5{)}^2& =156.25\\ (650-627.5{)}^2& =506.25\\ (635-627.5{)}^2& =56.25\\ (620-627.5{)}^2& =56.25\\ (600-627.5{)}^2& =756.25\\ (655-627.5{)}^2& =756.25\end{array}$$

Sum the squared deviations:

$$\begin{array}{rl}\sum (x_i-\bar{x}{)}^2& =3362.5\end{array}$$

Divide by $n-1=11$ and take the square root:

$$\begin{array}{rl}s& =\sqrt{\frac{3362.5}{11}}\\ & =\sqrt{305.68}\\ & \approx 17.49\end{array}$$

Standard error:

$$\begin{array}{rl}\text{SE}& =\frac{s}{\sqrt{n}}\\ & =\frac{17.49}{\sqrt{12}}\\ & =\frac{17.49}{3.464}\\ & \approx 5.05\end{array}$$

Test statistic:

$$\begin{array}{rl}t& =\frac{\bar{x}-{\mu }_0}{\text{SE}}\\ & =\frac{627.5-600}{5.05}\\ & =\frac{27.5}{5.05}\end{array}$$

Method 2: Using a TI-84 calculator

• Step 1: Enter your data. Go to STAT, select EDIT, then enter the values $645,610,\dots ,655$ into L1.
• Step 2: Find the mean and standard deviation. Press STAT, move to the CALC menu, then choose 1: 1-Var Stats. Press ENTER and make sure L1 is the selected list. The results display $\bar{x}$ (sample mean), $s_x$ (sample standard deviation), and $n$ (sample size):

$$\begin{array}{rl}\bar{x}& \approx 627.5\\ s& \approx 17.49\\ n& =12\end{array}$$

• Step 3: Find the test statistic. Press STAT, move to TESTS, then select 2: T-Test. Choose Inpt: Data, ${\mu }_0:600$, List: L1, Freq: 1, and $\mu :>{\mu }_0$. Scroll to Calculate and press ENTER. The calculator will display the test statistic, p-value, $\bar{x}$, $s$, and $n$.

Solution:

Part d.

Find the p-value.

There are two ways to find (or approximate) the p-value: using a t-table or using a TI-84.

Method 1: Using the t-table

Go to the t-table for the right-tailed test (since $H_a:\mu >{\mu }_0$). Find the row with $11$ degrees of freedom:

$$\begin{array}{rl}df& =n-1\\ & =12-1\\ & =11\end{array}$$

Look across the row to see where a t-value of $5.45$ would fall. The closest listed value is $4.437$, which corresponds to a p-value of $0.001$. Since $5.45$ is even farther into the right tail, the p-value is less than $0.001$. This isn’t the exact p-value, but it’s enough to decide common significance levels such as $0.05$ or $0.01$.

Method 2: Using the TI-84 calculator (recommended)

Press STAT, scroll to TESTS, and choose 2: T-Test. Choose Inpt: Stats, then enter:

$$\begin{array}{rl}{\mu }_0& =600\\ \bar{x}& =627.5\\ s& \approx 17.49\\ n& =12\end{array}$$

Choose the correct alternative hypothesis (in this case, $\mu >{\mu }_0$ for right-tailed). Scroll to Calculate and press ENTER.

Solution:

Part e.

Conclude and explain it in context. Use the $0.05$ significance level.

Solution: