According to the central limit theorem, the sampling distribution of a statistic (like a sample mean) will follow a normal distribution, as long as the sample size is sufficiently large. Therefore, when we know the standard deviation of the population, we can compute a z-score, and use the normal distribution to evaluate probabilities with the sample mean.
But sample sizes are sometimes small, and often we do not know the standard deviation of the population. When either of these problems occur, statisticians rely on the distribution of the t statistic (also known as the t score), whose values are given by:
t = [ x - μ ] / [ s / sqrt( n ) ]
where x is the sample mean, μ is the population mean, s is the standard deviation of the sample, and n is the sample size. The distribution of the t statistic is called the t distribution or the Student t distribution.
Degrees of Freedom
There are actually many different t distributions. The particular form of the t distribution is determined by its degrees of freedom. The degrees of freedom refers to the number of independent observations in a set of data.
When estimating a mean score or a proportion from a single sample, the number of independent observations is equal to the sample size minus one. Hence, the distribution of the t statistic from samples of size 8 would be described by a t distribution having 8 - 1 or 7 degrees of freedom. Similarly, a t distribution having 15 degrees of freedom would be used with a sample of size 16.
For other applications, the degrees of freedom may be calculated differently. We will describe those computations as they come up.
Properties of the t Distribution
The t distribution has the following properties:
- The mean of the distribution is equal to 0 .
- The variance is equal to v / ( v - 2 ), where v is the degrees of freedom (see last section) and v > 2.
- The variance is always greater than 1, although it is close to 1 when there are many degrees of freedom. With infinite degrees of freedom, the t distribution is the same as the standard normal distribution.
When to Use the t Distribution
The t distribution can be used with any statistic having a bell-shaped distribution (i.e., approximately normal). The central limit theorem states that the sampling distribution of a statistic will be normal or nearly normal, if any of the following conditions apply.
- The population distribution is normal.
- The sampling distribution is symmetric, unimodal, without outliers, and the sample size is 15 or less.
- The sampling distribution is moderately skewed, unimodal, without outliers, and the sample size is between 16 and 40.
- The sample size is greater than 40, without outliers.
The t distribution should not be used with small samples from populations that are not approximately normal.
Probability and the Student t Distribution
When a sample of size n is drawn from a population having a normal (or nearly normal) distribution, the sample mean can be transformed into a t score, using the equation presented at the beginning of this lesson. We repeat that equation below:
t = [ x - μ ] / [ s / sqrt( n ) ]
where x is the sample mean, μ is the population mean, s is the standard deviation of the sample, n is the sample size, and degrees of freedom are equal to n - 1.
The t score produced by this transformation can be associated with a unique cumulative probability. This cumulative probability represents the likelihood of finding a sample mean less than or equal to x, given a random sample of size n.
The easiest way to find the probability associated with a particular t score is to use the T Distribution Calculator, a free tool provided by Stat Trek.
Notation and t Scores
Statisticians use tα to represent the t-score that has a cumulative probability of (1 - α). For example, suppose we were interested in the t-score having a cumulative probability of 0.95. In this example, α would be equal to (1 - 0.95) or 0.05. We would refer to the t-score as t0.05
Of course, the value of t0.05 depends on the number of degrees of freedom. For example, with 2 degrees of freedom, that t0.05 is equal to 2.92; but with 20 degrees of freedom, that t0.05 is equal to 1.725.
Note: Because the t distribution is symmetric about a mean of zero, the following is true.
tα = -t1 - alpha And t1 - alpha = -tα
Thus, if t0.05 = 2.92, then t0.95 = -2.92.
Test Your Understanding of This Lesson
Problem 1
Acme Corporation manufactures light bulbs. The CEO claims that an average Acme light bulb lasts 300 days. A researcher randomly selects 15 bulbs for testing. The sampled bulbs last an average of 290 days, with a standard deviation of 50 days. If the CEO's claim were true, what is the probability that 15 randomly selected bulbs would have an average life of no more than 290 days?
Note: There are two ways to solve this problem, using the T Distribution Calculator. Both approaches are presented below. Solution A is the traditional approach. It requires you to compute the t score, based on data presented in the problem description. Then, you use the T Distribution Calculator to find the probability. Solution B is easier. You simply enter the problem data into the T Distribution Calculator. The calculator computes a t score "behind the scenes", and displays the probability. Both approaches come up with exactly the same answer.
Solution A
The first thing we need to do is compute the t score, based on the following equation:
t = [ x - μ ] / [ s / sqrt( n ) ]
t = ( 290 - 300 ) / [ 50 / sqrt( 15) ] = -10 / 12.909945 = - 0.7745966
t = ( 290 - 300 ) / [ 50 / sqrt( 15) ] = -10 / 12.909945 = - 0.7745966
where x is the sample mean, μ is the population mean, s is the standard deviation of the sample, and n is the sample size.
Now, we are ready to use the T Distribution Calculator. Since we know the t score, we select "T score" from the Random Variable drop down box. Then, we enter the following data:
- The degrees of freedom are equal to 15 - 1 = 14.
- The t score is equal to - 0.7745966.
The calculator displays the cumulative probability: 0.226. Hence, if the true bulb life were 300 days, there is a 22.6% chance that the average bulb life for 15 randomly selected bulbs would be less than or equal to 290 days.
Solution B:
This time, we will work directly with the raw data from the problem. We will not compute the t score; the T Distribution Calculator will do that work for us. Since we will work with the raw data, we select "Sample mean" from the Random Variable drop down box. Then, we enter the following data:
- The degrees of freedom are equal to 15 - 1 = 14.
- Assuming the CEO's claim is true, the population mean equals 300.
- The sample mean equals 290.
- The standard deviation of the sample is 50.
The calculator displays the cumulative probability: 0.226. Hence, there is a 22.6% chance that the average sampled light bulb will burn out within 290 days.
Problem 2
Suppose scores on an IQ test are normally distributed, with a mean of 100. Suppose 20 people are randomly selected and tested. The standard deviation in the sample group is 15. What is the probability that the average test score in the sample group will be at most 110?
Solution:
To solve this problem, we will work directly with the raw data from the problem. We will not compute the t score; the T Distribution Calculator will do that work for us. Since we will work with the raw data, we select "Sample mean" from the Random Variable drop down box. Then, we enter the following data:
- The degrees of freedom are equal to 20 - 1 = 19.
- The population mean equals 100.
- The sample mean equals 110.
- The standard deviation of the sample is 15.
We enter these values into the T Distribution Calculator. The calculator displays the cumulative probability: 0.996. Hence, there is a 99.6% chance that the sample average will be no greater than 110.
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