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A Single Population Mean Using the Student's t-Distribution

In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this unknown number did not present a problem to statisticians. They used the sample standard deviation s as an estimate for σ and proceeded as before to calculate a confidence interval with close-enough results. However, statisticians ran into problems when the sample size was small. A small sample size caused inaccuracies in the confidence interval.

William S. Gosset (1876–1937) of the Guinness brewery in Dublin, Ireland, ran into this problem. His experiments with hops and barley produced very few samples. Just replacing σ with s did not produce accurate results when he tried to calculate a confidence interval. He realized that he could not use a normal distribution for the calculation; he found that the actual distribution depends on the sample size. This problem led him to discover what is called the Student's t-distribution. The name comes from the fact that Gosset wrote under the pen name Student.

Up until the mid-1970s, some statisticians used the normal distribution approximation for large sample sizes and used the Student's t-distribution only for sample sizes of at most 30. With graphing calculators and computers, the practice now is to use the Student's t-distribution whenever s is used as an estimate for σ.

If you draw a simple random sample of size n from a population that has an approximately normal distribution with mean μ and unknown population standard deviation σ and calculate the t-score t = $\frac{\bar{x} - μ}{(\frac{s}{\sqrt{n}})} \frac{\bar{x} - μ}{(\frac{s}{\sqrt{n}})}$ , then the t-scores follow a Student's t-distribution with n – 1 degrees of freedom. The t-score has the same interpretation as the z-score: It measures how far $\bar{x} \bar{x}$ is from its mean μ. For each sample size n, there is a different Student's t-distribution.

The degrees of freedom (df), n -– 1, are the sample size minus 1.

Properties of the Student's t-distribution

The graph for the Student's t-distribution is similar to the standard normal curve.
The mean for the Student's t-distribution is zero, and the distribution is symmetric about zero.
The Student's t-distribution has more probability in its tails than the standard normal distribution. Figure 8.6 shows the graphs of the student t-distribution for 1, 2 and 5 degrees of freedom: (v), compare to the standard normal distribution (in black).

Figure 8.6
The exact shape of the Student's t-distribution depends on the degrees of freedom. As the degrees of freedom increase, the graph of the Student's t-distribution becomes more like the graph of the standard normal distribution.
The underlying population of individual observations is assumed to be normally distributed with unknown population mean μ and unknown population standard deviation σ. The size of the underlying population is generally not relevant unless it is very small. If it is bell-shaped (normal), then the assumption is met and does not need discussion. Random sampling is assumed, but that is a completely separate assumption from normality.

Calculators and computers can easily calculate any Student's t-probabilities. The TI-83, 83+, and 84+ have a tcdf function to find the probability for given values of t. The grammar for the tcdf command is tcdf(lower bound, upper bound, degrees of freedom). However, for confidence intervals, we need to use inverse probability to find the value of t when we know the probability.

For the TI-84+, you can use the invT command on the DISTRibution menu. The invT command works similarly to the invnorm. The invT command requires two inputs: invT(area to the left, degrees of freedom). The output is the t-score that corresponds to the area we specified.

The TI-83 and 83+ do not have the invT command. (The TI-89 has an inverse T command.)

A probability table for the Student's t-distribution can also be used. The table gives critical t-values that correspond to the confidence level (column) and degrees of freedom (row). (The TI-86 does not have an invT program or command, so if you are using that calculator, you need to use a probability table for the Student's t-distribution.) When using a t-table, note that some tables are formatted to show the confidence level in the column headings, while the column headings in some tables may show only corresponding area in one or both tails.

A Student's t-table (see Appendix H Tables) gives t-scores given the degrees of freedom and the right-tailed probability. The table is very limited. Calculators and computers can easily calculate any Student's t-probabilities.

If the population standard deviation is not known, the error bound for a population mean is

$E B M = (t_{\frac{α}{2}}) (\frac{s}{\sqrt{n}}) E B M = (t_{\frac{α}{2}}) (\frac{s}{\sqrt{n}})$ ,
$t_{\frac{σ}{2}} t_{\frac{σ}{2}}$ is the t-score with area to the right equal to $\frac{α}{2} \frac{α}{2}$ ,
use df = n – 1 degrees of freedom, and
s = sample standard deviation.

The format for the confidence interval is

(\bar{x} - E B M, \bar{x} + E B M) . (\bar{x} - E B M, \bar{x} + E B M) .

8.3

Using the TI-83, 83+, 84, 84+ Calculator

To calculate the confidence interval directly, do the following:
Press STAT.
Arrow over to TESTS.
Arrow down to 8:TInterval and press ENTER (or just press 8).

Example 8.8

Problem

Suppose you do a study of acupuncture to determine how effective it is in relieving pain. You measure sensory rates for 15 subjects with the results given. Use the sample data to construct a 95 percent confidence interval for the mean sensory rate for the population (assumed normal) from which you took the data.
The solution is shown step-by-step and by using the TI-83, 83+, or 84+ calculators.

8.6; 9.4; 7.9; 6.8; 8.3; 7.3; 9.2; 9.6; 8.7; 11.4; 10.3; 5.4; 8.1; 5.5; 6.9

Show solution

Solution

The first solution is step-by-step (Solution A).
The second solution uses the TI-83+ and TI-84 calculators (Solution B).

To find the confidence interval, you need the sample mean, $\bar{x} \bar{x}$ , and the EBM.

$\begin{array}{l} \bar{x} = \frac{8.6 + 9.4 + 7.9 + 6.8 + 8.3 + 7.3 + 9.2 + 9.6 + 8.7 + 11.4 + 10.3 + 5.4 + 8.1 + 5.5 + 6.9}{15} = 8.2267; \\ \begin{array}{l} s = \frac{{(8.6 - \bar{x})}^{2} + {(9.4 - \bar{x})}^{2} + \dots + {(5.5 - \bar{x})}^{2} + {(6.9 - \bar{x})}^{2}}{14} = 1.6722; \end{array} \\ \begin{array}{l} n = 15 \end{array} \end{array} \begin{array}{l} \bar{x} = \frac{8.6 + 9.4 + 7.9 + 6.8 + 8.3 + 7.3 + 9.2 + 9.6 + 8.7 + 11.4 + 10.3 + 5.4 + 8.1 + 5.5 + 6.9}{15} = 8.2267; \\ \begin{array}{l} s = \frac{{(8.6 - \bar{x})}^{2} + {(9.4 - \bar{x})}^{2} + \dots + {(5.5 - \bar{x})}^{2} + {(6.9 - \bar{x})}^{2}}{14} = 1.6722; \end{array} \\ \begin{array}{l} n = 15 \end{array} \end{array}$

df = 15 – 1 = 14 CL, so α = 1 – CL = 1 – 0.95 = 0.05

$\frac{α}{2} \frac{α}{2}$ = 0.025; $t_{\frac{α}{2}} = t_{0.025} t_{\frac{α}{2}} = t_{0.025}$

The area to the right of t_0.025 is 0.025, and the area to the left of t_0.025 is 1 – 0.025 = 0.975.

$t_{\frac{α}{2}} = t_{0.025} = 2.14 t_{\frac{α}{2}} = t_{0.025} = 2.14$ using invT(.975,14) on the TI-84+ calculator.

E B M = (t_{\frac{α}{2}}) (\frac{s}{\sqrt{n}}) E B M = (t_{\frac{α}{2}}) (\frac{s}{\sqrt{n}})

8.4

E B M = (2.14) (\frac{1.6722}{\sqrt{15}}) = 0.924 E B M = (2.14) (\frac{1.6722}{\sqrt{15}}) = 0.924

\bar{x} - E B M = 8 .2267 – 0 .9240 = 7 .3 \bar{x} - E B M = 8 .2267 – 0 .9240 = 7 .3

\bar{x} + E B M = 8 .2267 + 0 .9240 = 9 .15 \bar{x} + E B M = 8 .2267 + 0 .9240 = 9 .15

The 95 percent confidence interval is (7.30, 9.15).

We estimate with 95 percent confidence that the true population mean sensory rate is between 7.30 and 9.15.

Show solution

Solution

Using the TI-83, 83+, 84, 84+ Calculator

Press STAT and arrow over to TESTS.
Arrow down to 8:TInterval and press ENTER (or you can just press 8).
Arrow to Data and press ENTER.
Arrow down to List and enter the list name where you put the data.
There should be a 1 after Freq.
Arrow down to C-level and enter 0.95.
Arrow down to Calculate and press ENTER.
The 95 percent confidence interval is (7.3006, 9.1527).

Note

When calculating the error bound, you can also use a probability table for the Student's t-distribution to find the value of t. The table gives t-scores that correspond to the confidence level (column) and degrees of freedom (row); the t-score is found where the row and column intersect in the table.

Try It 8.8

You do a study of hypnotherapy to determine how effective it is in increasing the number of hours of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results. Construct a 95 percent confidence interval for the mean number of hours slept for the population (assumed normal) from which you took the data.

8.2, 9.1, 7.7, 8.6, 6.9, 11.2, 10.1, 9.9, 8.9, 9.2, 7.5, 10.5

Example 8.9

Problem

A group of researchers is working to understand the scope of industrial pollution in the human body. Industrial chemicals may enter the body through pollution or as ingredients in consumer products. In October 2008, the scientists tested cord-blood samples for 20 newborn infants in the United States. The cord blood of the in utero/newborn group was tested for 430 industrial compounds, pollutants, and other chemicals, including chemicals linked to brain and nervous-system toxicity, immune-system toxicity, reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. Table 8.3 shows how many of the targeted chemicals were found in each infant’s cord blood.

79	145	147	160	116	100	159	151	156	126
137	83	156	94	121	144	123	114	139	99

Table 8.3

Use this sample data to construct a 90 percent confidence interval for the mean number of targeted industrial chemicals to be found in an infant’s blood.

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Solution

Solution A

From the sample data, you can calculate

\begin{array}{l} \begin{array}{l} \bar{x} = \frac{79 + 145 + \dots + 139 + 99}{20} = 127.45 \end{array} \\ s = \frac{{(79 - \bar{x})}^{2} + {(145 - \bar{x})}^{2} + \dots + {(139 - \bar{x})}^{2} + {(99 - \bar{x})}^{2}}{19} = 25.965 . \end{array} \begin{array}{l} \begin{array}{l} \bar{x} = \frac{79 + 145 + \dots + 139 + 99}{20} = 127.45 \end{array} \\ s = \frac{{(79 - \bar{x})}^{2} + {(145 - \bar{x})}^{2} + \dots + {(139 - \bar{x})}^{2} + {(99 - \bar{x})}^{2}}{19} = 25.965 . \end{array}

There are 20 infants in the sample, so n = 20, and df = 20 – 1 = 19.

You are asked to calculate a 90 percent confidence interval: CL = 0.90, so α = 1 – CL = 1 – 0.90 = 0.10. $\frac{α}{2} = 0.05, t_{\frac{α}{2}} = t_{0.05} \frac{α}{2} = 0.05,$

By definition, the area to the right of t_0.05 is 0.05, and so the area to the left of t_0.05 is 1 – 0.05 = 0.95.

Use a table, calculator, or computer to find that t_0.05 = 1.729.

\begin{array}{l} E B M = t_{\frac{α}{2}} (\frac{s}{\sqrt{n}}) = 1.729 (\frac{25.965}{\sqrt{20}}) \approx 10.038 \\ \bar{x} – E B M = 127 .45 – 10 .038 = 117 .412 \\ \bar{x} + E B M = 127 .45 + 10 .038 = 137 .488 \end{array} \begin{array}{l} E B M = t_{\frac{α}{2}} (\frac{s}{\sqrt{n}}) = 1.729 (\frac{25.965}{\sqrt{20}}) \approx 10.038 \\ \bar{x} – E B M = 127 .45 – 10 .038 = 117 .412 \\ \bar{x} + E B M = 127 .45 + 10 .038 = 137 .488 \end{array}

We estimate with 90 percent confidence that the mean number of all targeted industrial chemicals found in cord blood in the United States is between 117.412 and 137.488.

Show solution

Solution

Solution B

Using the TI-83, 83+, 84, 84+ Calculator

Enter the data as a list.
Press STAT and arrow over to TESTS.
Arrow down to 8:TInterval and press ENTER (or you can just press 8). Arrow to Data and press ENTER.
Arrow down to List and enter the list name where you put the data.
Arrow down to Freq and enter 1.
Arrow down to C-level and enter 0.90.
Arrow down to Calculate and press ENTER.
The 90 percent confidence interval is (117.41, 137.49).

Try It 8.9

A random sample of statistics students was asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in Table 8.4. Use the following sample data to construct a 98 percent confidence interval for the mean number of hours statistics students will spend watching television in one week.

0	3	1	20	9
5	10	1	10	4
14	2	4	4	5

Table 8.4