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# 8.4: Student’s t-Distribution

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Use Student’s t\begin{align*}t\end{align*}-distribution to estimate population mean intervals for smaller samples.
• Understand how the shape of Student’s t\begin{align*}t\end{align*}-distribution corresponds to the sample size (which corresponds to a measure called the degrees of freedom).

## Introduction

Back in the early 1900’s, a chemist at a brewery in Ireland discovered that when he was working with very small samples, the distributions of the means differed significantly from the normal distribution. He noticed that as his sample sizes changed, the shape of the distribution changed as well. He published his results under the pseudonym ‘Student’, and this concept and the distributions for small sample sizes are now known as Student’s t\begin{align*}t\end{align*}-distributions.

### Hypothesis Testing with Small Populations and Sample Sizes

Student's t\begin{align*}t\end{align*}-distributions are a family of distributions that, like the normal distribution, are symmetrical, bell-shaped, and centered on a mean. However, the distribution shape changes as the sample size changes. Therefore, there is a specific shape, or distribution, for every sample of a given size (see figure below; each distribution has a different value of k\begin{align*}k\end{align*}, the number of degrees of freedom, which is 1 less than the size of the sample).

We use Student's t\begin{align*}t\end{align*}-distributions in hypothesis testing the same way that we use the normal distribution. Each row in the t\begin{align*}t\end{align*}-distribution table (see link below) represents a different t\begin{align*}t\end{align*}-distribution, and each distribution is associated with a unique number of degrees of freedom (the number of observations minus one). The column headings in the table represent the portion of the area in the tails of the distribution. We use the numbers in the table just as we use z\begin{align*}z\end{align*}-scores.

http://tinyurl.com/ygcc5g9 Follow this link to the Student’s t\begin{align*}t\end{align*}-table.

As the number of observations gets larger, the t\begin{align*}t\end{align*}-distribution approaches the shape of the normal distribution. In general, once the sample size is large enough\begin{align*}-\end{align*}usually about 120\begin{align*}-\end{align*}we would use the normal distribution or a z\begin{align*}z\end{align*}-table instead.

In calculating the t\begin{align*}t\end{align*}-test statistic, we use the following formula:

t=x¯μ0sn\begin{align*}t=\frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}\end{align*}

where:

t\begin{align*}t\end{align*} is the test statistic and has n1\begin{align*}n-1\end{align*} degrees of freedom.

x¯\begin{align*}\bar{x}\end{align*} is the sample mean.

μ0\begin{align*}\mu_0\end{align*} is the population mean under the null hypothesis.

s\begin{align*}s\end{align*} is the sample standard deviation.

n\begin{align*}n\end{align*} is the sample size.

sn\begin{align*}\frac{s}{\sqrt{n}}\end{align*} is the estimated standard error.

Example: A high school athletic director is asked if football players are doing as well academically as the other student athletes at his school. We know from a previous study that the average GPA for the student athletes is 3.10 and that the standard deviation of the sample is 0.54. After an initiative to help improve the GPA of student athletes, the athletic director samples 20 football players and finds that their GPA is 3.18. Is there a significant improvement? Use a 0.05 significance level.

First, we establish our null and alternative hypotheses:

H0:μHa:μ=3.103.10\begin{align*}H_0: \mu &= 3.10\\ H_a: \mu &\neq 3.10\end{align*}

Next, we use our alpha level of 0.05 and the t\begin{align*}t\end{align*}-distribution table to find our critical values. For a two-tailed test with 19 degrees of freedom and a 0.05 level of significance, our critical values are equal to ±2.093\begin{align*}\pm 2.093\end{align*}.

Finally, in calculating the test statistic, we use the formula as shown:

t=x¯μ0sn=3.183.100.54200.66\begin{align*}t=\frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}=\frac{3.18-3.10}{\frac{0.54}{\sqrt{20}}} \approx 0.66\end{align*}

This means that the observed sample mean of the GPA of football players of 3.18 is 0.66 standard errors above the hypothesized value of 3.10. Because the value of the test statistic is less than the critical value of 2.093, we fail to reject the null hypothesis.

Therefore, we can conclude that the difference between the sample mean and the hypothesized value is not sufficient to attribute it to anything other than sampling error. Thus, the athletic director can conclude that the mean academic performance of football players does not differ from the mean performance of other student athletes.

Example: The masses of newly-produced bus tokens are estimated to have a mean of 3.16 grams. A random sample of 11 tokens was removed from the production line, and the mean weight of the tokens was calculated to be 3.21 grams, with a standard deviation of 0.067. What is the value of the test statistic for a test to determine how the mean differs from the estimated mean?

The test statistic for this problem can be calculated as follows:

ttt=x¯μsn=3.213.160.067112.48\begin{align*} t &= \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\\ t &= \frac{3.21-3.16}{\frac{0.067}{\sqrt{11}}}\\ t &\approx 2.48\end{align*}

If the value of t\begin{align*}t\end{align*} from the sample is between the tails of the distribution of t\begin{align*}t\end{align*} constructed by assuming the null hypothesis is true, then the null hypothesis is, in fact, true. On the other hand, if the value of t\begin{align*}t\end{align*} from the sample is way out in a tail of the t\begin{align*}t\end{align*}-distribution, then there is evidence to reject the null hypothesis. When the distribution of t\begin{align*}t\end{align*} is known, if the null hypothesis is true, the location of the value of t\begin{align*}t\end{align*} on the distribution will be between the tails and outside the critical region. The most common method used to determine if this is the case is to find a P\begin{align*}P\end{align*}-value (observed significance level). The P\begin{align*}P\end{align*}-value is a probability that is computed with the assumption that the null hypothesis is true.

The P\begin{align*}P\end{align*}-value for a two-sided test is the area under the t\begin{align*}t\end{align*}-distribution with degrees of freedom of 111=10\begin{align*}11 - 1 = 10\end{align*} that lies above t=2.48\begin{align*}t = 2.48\end{align*} and below t=2.48\begin{align*}t = - 2.48\end{align*}. This P\begin{align*}P\end{align*}-value can be calculated by using technology.

Technology Note: Using the 'tcdf(' Command on the TI-83/84 Calculator to Calculate Probabilities Associated with the t\begin{align*}t\end{align*}-Distribution

Press [2ND][DIST] and use the down arrow to select 'tcdf('. The syntax for this command is 'tcdf(lower bound, upper bound, degrees of freedom)'. This command will return the total area under both tails. To calculate the area under one tail, divide by 2 as shown below:

This means that there is only a 0.016 chance of getting a value of t\begin{align*}t\end{align*} as large as or even larger than the one from this sample. The small P\begin{align*}P\end{align*}-value tells us that the sample is inconsistent with the null hypothesis. Therefore, the population mean differs from the estimated mean of 3.16.

When the P\begin{align*}P\end{align*}-value is close to zero, there is strong evidence against the null hypothesis. On the other hand, when the P\begin{align*}P\end{align*}-value is large, the result from the sample is consistent with the estimated or hypothesized mean, and there is no evidence against the null hypothesis.

A visual picture of the P\begin{align*}P\end{align*}-value can be obtained by using a graphing calculator as follows:

The spread of any t\begin{align*}t\end{align*}-distribution is greater than that of a standard normal distribution. This is due to the fact that in the denominator of the formula, σ\begin{align*}\sigma\end{align*} has been replaced with \begin{align*}s\end{align*}. Since \begin{align*}s\end{align*} is a random quantity changing with various samples, the variability in \begin{align*}t\end{align*} is greater, resulting in a larger spread.

Notice that in the first distribution graph shown above, the spread of the inner curve is small, but in the second graph, both distributions are basically overlapping and are roughly normal. This is due to the increase in the degrees of freedom.

To further illustrate this point, the \begin{align*}t\end{align*}-distributions for 1 and 12 degrees of freedom can be graphed on a graphing calculator. To do so, first press [Y=][2ND][DISTR], choose the 'tpdf(' command, enter 'X' and 1, separated by commas, and close the parentheses. Then go down to Y2 and repeat the process, this time entering 12 instead of 1. Finally, make sure your window is set correctly and press [GRAPH].

The \begin{align*}t\end{align*}-distributions for 1 and 12 degrees of freedom should look similar to the ones shown below (df denotes degrees of freedom):

Notice the difference in the two distributions. The one with 12 degrees of freedom approximates a normal curve.

The \begin{align*}t\end{align*}-distribution can be used with any statistic having a bell-shaped distribution. We already know that the Central Limit Theorem states that the sampling distribution of a statistic will be close to normal with a large enough sample size, but, in fact, the Central Limit Theorem predicts a roughly normal distribution under any of the following conditions:

• The population distribution is normal.
• The sampling distribution is symmetric and the sample size is \begin{align*}\le 15\end{align*}.
• The sampling distribution is moderately skewed and the sample size is \begin{align*}16 \le n \le 30\end{align*}.
• The sample size is greater than 30, without outliers.

In addition to the fact that the \begin{align*}t\end{align*}-distribution can be used with any bell-shaped distribution, it also has some unique properties. These properties are as follows:

• The mean of the distribution equals zero.
• The population standard deviation is unknown.
• The variance is equal to the degrees of freedom divided by the degrees of freedom minus 2. This means that the degrees of freedom must be greater than two to avoid the expression being undefined.
• The variance is always greater than 1, although it approaches 1 as the degrees of freedom increase. This is due to the fact that as the degrees of freedom increase, the distribution is becoming more of a normal distribution.
• Although the \begin{align*}t\end{align*}-distribution is bell-shaped, the smaller sample sizes produce a flatter curve. The distribution is not as mound-shaped as a normal distribution, and the tails are thicker. As the sample size increases and approaches 30, the distribution approaches a normal distribution.
• The population is unimodal and symmetric.

Example: Duracell manufactures batteries that the CEO claims will last 300 hours under normal use. A researcher randomly selected 15 batteries from the production line and tested these batteries. The tested batteries had a mean life span of 290 hours, with a standard deviation of 50 hours. If the CEO’s claim were true, what is the probability that 15 randomly selected batteries would have a life span of no more than 290 hours?

\begin{align*}t &= \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}} \ \text{There are} \ n-1=15-1=14 \ \text{degrees of freedom.}\\ t &= \frac{290-300}{\frac{50}{\sqrt{15}}}\\ t &= \frac{-10}{12.9099}\\ t &= -0.7745967\end{align*}

Using a graphing calculator or a table, the cumulative probability is shown to be 0.226, which means that if the true life span of a battery were 300 hours, there is a 22.6% chance that the life span of the 15 tested batteries would be less than or equal to 290 hours. This is not a high enough level of confidence to reject the null hypothesis and count the discrepancy as significant.

Note: To find this answer using a graphing calculator, press [2ND][DISTR], select the 'tcdf(' command, enter \begin{align*}-0.7745967\end{align*}, 0.7745967, and 14, separated by commas, and press [ENTER]. Subtract the result from 1, and then divide by 2.

Example: You have just taken ownership of a pizza shop. The previous owner told you that you would save money if you bought the mozzarella cheese in a 4.5-pound slab. Each time you purchase a slab of cheese, you weigh it to ensure that you are receiving 72 ounces of cheese. The results of 7 random measurements are 70, 69, 73, 68, 71, 69 and 71 ounces, respectively. Find the test statistic for this scenario.

Begin the problem by determining the mean of the sample and the sample standard deviation. This can be done using a graphing calculator. You should find that \begin{align*}\bar{x}=70.143\end{align*} and \begin{align*}s=1.676\end{align*}. Now calculate the test statistic as follows:

\begin{align*}t &= \frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\\ t &= \frac{70.143-72}{\frac{1.676}{\sqrt{7}}}\\ t &\approx -2.9315\end{align*}

Example: In the last example, the test statistic for testing that the mean weight of the cheese wasn’t 72 ounces was computed. Find and interpret the \begin{align*}P\end{align*}-value.

The test statistic computed in the last example was \begin{align*}-2.9315\end{align*}. Using technology, the \begin{align*}P\end{align*}-value is 0.0262. In other words, the probability that 7 random measurements would give a value of \begin{align*}t\end{align*} greater than 2.9315 or less than \begin{align*}-2.9315\end{align*} is about 0.0262.

Example: In the previous example, the \begin{align*}P\end{align*}-value for testing that the mean weight of cheese wasn’t 72 ounces was determined.

a) State the hypotheses.

b) Would the null hypothesis be rejected at the 10% level? The 5% level? The 1% level?

a)

\begin{align*}H_0: \mu &= 72\\ H_a: \mu &\neq 72\end{align*}

b) Because the \begin{align*}P\end{align*}-value of 0.0262 is less than both 0.10 and 0.05, the null hypothesis would be rejected at these levels. However, the \begin{align*}P\end{align*}-value is greater than 0.01, so the null hypothesis would not be rejected if this level of confidence was required.

## Lesson Summary

A test of significance is done when a claim is made about the value of a population parameter. The test can only be conducted if the random sample taken from the population came from a distribution that is normal or approximately normal. When the sample size is small, you must use \begin{align*}t\end{align*} instead of \begin{align*}z\end{align*} to complete the significance test for a mean.

## Points to Consider

• Is there a way to determine where the \begin{align*}t\end{align*}-statistic lies on a distribution?
• If a way does exist, what is the meaning of its placement?

For an explanation of the \begin{align*}t\end{align*}-distribution and an example using it (7.0)(17.0), see bionicturtledotcom, Student's t distribution (8:32).

## Review Questions

1. You intend to use simulation to construct an approximate \begin{align*}t\end{align*}-distribution with 8 degrees of freedom by taking random samples from a population with bowling scores that are normally distributed with mean, \begin{align*}\mu=110\end{align*}, and standard deviation, \begin{align*}\sigma=20\end{align*}.
1. Explain how you will do one run of this simulation.
2. Produce four values of \begin{align*}t\end{align*} using this simulation.
2. The dean from UCLA is concerned that the students’ grade point averages have changed dramatically in recent years. The graduating seniors’ mean GPA over the last five years is 2.75. The dean randomly samples 30 seniors from the last graduating class and finds that their mean GPA is 2.85, with a sample standard deviation of 0.65. Would a \begin{align*}t\end{align*}-distribution now be the appropriate sampling distribution for the mean? Why or why not?
3. Using the appropriate \begin{align*}t\end{align*}-distribution, test the same null hypothesis with a sample of 30.
4. With a sample size of 30, do you need to have a larger or smaller difference between the hypothesized population mean and the sample mean than with a sample size of 256 to obtain statistical significance? Explain your answer.

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