38 8.2 A Single Population Mean using the Student t Distribution
In practice, we rarely know the population standard deviation. In the past, when the sample size was large, this 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. Goset (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 only 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 a normal distribution with mean μ and unknown population standard deviation σ and calculate the t-score: [latex]\displaystyle{t}=\frac{{\overline{{x}}-\mu}}{{\frac{{s}}{\sqrt{{n}}}}}[/latex] is from its mean μ. For each sample size n, there is a different Student’s t-distribution.
The degrees of freedom, n – 1, come from the calculation of the sample standard deviation s. Because the sum of the deviations is zero, we can find the last deviation once we know the other n – 1 deviations. The other n – 1 deviations can change or vary freely. We call the number n – 1 the degrees of freedom (df).
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 because the spread of the t-distribution is greater than the spread of the standard normal. So the graph of the Student’s t-distribution will be thicker in the tails and shorter in the center than the graph of the standard normal distribution.
- The exact shape of the Student’s t-distribution depends on the degrees of freedom. As the degrees of freedom increases, the graph of 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 doesn’t 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 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 t-scores 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 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.
The notation for the Student’s t-distribution (using T as the random variable) is:
T ~ tdf where df = n – 1.
For example, if we have a sample of size n = 20 items, then we calculate the degrees of freedom as df = n – 1 = 20 – 1 = 19 and we write the distribution as T ~ t19.
If the population standard deviation is not known, the error bound for a population mean is:
EBM = [latex]\displaystyle({t}_{\frac{{\alpha}}{{2}}})(\frac{{\sigma}}{{\sqrt{n}}})[/latex]
- [latex]\displaystyle({t}_{\frac{{\alpha}}{{2}}}[/latex]is the t-score with area to the right equal to[latex]\displaystyle\frac{{\alpha}}{{2}}[/latex],
- degree of freedom, df = n – 1, and
- s = sample standard deviation.
The format for the confidence interval is: ([latex]\displaystyle\overline{x}[/latex] – EBM, [latex]\displaystyle\overline{x}[/latex] + EBM)
In other words, the formula for the confidence interval is
([latex]\displaystyle\overline{x}[/latex] – [latex]\displaystyle({t}_{\frac{{\alpha}}{{2}}})(\frac{{\sigma}}{{\sqrt{n}}})[/latex], [latex]\displaystyle\overline{x}[/latex] + [latex]\displaystyle({t}_{\frac{{\alpha}}{{2}}})(\frac{{\sigma}}{{\sqrt{n}}})[/latex])
Calculate the Confidence Interval bu using TI-Calculator:Press STAT. Arrow over to TESTS.Arrow down to 8:TInterval and press ENTER (or just press 8). |
Example 1
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% 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
|
Solution A:
To find the 95% confidence interval, you need the sample mean, , and the EBM.
[latex]\displaystyle\overline{X}[/latex] = 8.2267, s = 1.6722, n = 15, df = 15 – 1 = 14
CL = 0.95.
Then α = 1 – CL = 1 – 0.95 = 0.05
[latex]\displaystyle\frac{{\alpha}}{{2}}[/latex] = 0.025
[latex]\displaystyle{t}_{\frac{{\alpha}}{{2}}} = {t}_{0.025}[/latex]
The area to the right of [latex]\displaystyle{t}_{0.025}[/latex] is 0.025, and the area to the left of[latex]\displaystyle{t}_{o.o25}[/latex] is 1 – 0.025 = 0.975
TI-84+ Calculator : invT(.975,14) or use t-table to find [latex]\displaystyle{t}_{\frac{{\alpha}}{{2}}}[/latex] when df = 14.
[latex]\displaystyle{t}_{\frac{{\alpha}}{{2}}}={t}_{0.025}={2.14}[/latex]
EBM = [latex]\displaystyle({t}_{\frac{{\alpha}}{{2}}})(\frac{{s}}{{\sqrt{n}}})[/latex]
Hence, EBM = [latex]\displaystyle(2.14)(\frac{{1.6722}}{{\sqrt{15}}})=(0.924)[/latex]
[latex]\displaystyle\overline{x}[/latex] – EBM = 8.2267 – 0.9240 = 7.3
[latex]\displaystyle\overline{x}[/latex] + EBM =8.2267 + 0.9240 = 9.15
The 95% confidence interval is (7.30, 9.15).
We estimate with 95% confidence that the true population mean sensory rate is between 7.30 and 9.15.
Solution B
Press STAT
and arrow over toT ESTS
.
Arrow down to 8:TInterval
and press ENTER
(or you can just press8
).
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% confidence interval is (7.3006, 9.1527)
| Note: When calculating the error bound, a probability table for the Student’s t-distribution can also be used 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
You do a study of hypnotherapy to determine how effective it is in increasing the number of hourse of sleep subjects get each night. You measure hours of sleep for 12 subjects with the following results.
Construct a 95% 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
[practice-area rows=”1″][/practice-area]
Show Answer
(8.1634, 9.8032)
Example 2
The Human Toxome Project (HTP) 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 at HTP 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, and reproductive toxicity, and fertility problems. There are health concerns about the effects of some chemicals on the brain and nervous system. This table 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 |
Use this sample data to construct a 90% confidence interval for the mean number of targeted industrial chemicals to be found in an in infant’s blood.
Solution A:
From the sample, you can calculate [latex]\displaystyle\overline{x}[/latex]=127.45
and s = 25.965. There are 20 infants in the sample, so n = 20, and df = 20 – 1 = 19.
You are asked to calculate a 90% confidence interval: CL = 0.90, so [latex]\displaystyle\alpha[/latex]= 1-CL = 1-0.90 = 0.10
[latex]\displaystyle\frac{{\alpha}}{{2}}[/latex]= 0.05
[latex]\displaystyle({t}_{\frac{{\alpha}}{{2}}}={t}_{0.05}[/latex]
By definition, the area to the right of t0.05 is 0.05 and so the area to the left of t0.05 is 1 – 0.05 = 0.95.
Use a table, calculator, or computer to find that t0.05 = 1.729.
EBM = [latex]\displaystyle{t}_{\frac{{\alpha}}{{2}}}(\frac{{s}}{{\sqrt{n}}})[/latex]={1.729}([latex]\displaystyle\frac{{25.965}}{{\sqrt{20}}}={10.038}[/latex]
[latex]\displaystyle\overline{x}[/latex] – EBM = 127.45 – 10.038 = 117.412
[latex]\displaystyle\overline{x}[/latex]+EBM= 127.45 + 10.038= 137.488
We estimate with 90% 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.
Solution B:
Enter the data as a list.
Press STAT
and arrow over toTESTS
.
Arrow down to 8:TInterval
and press ENTER
(or you can just press8
).
Arrow to Data and pressENTER
.
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% confidence interval is (117.41, 137.49).
Example 3
A random sample of statistics students were asked to estimate the total number of hours they spend watching television in an average week. The responses are recorded in This table. Use this sample data to construct a 98% 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 |
Solution A:
[latex]\displaystyle\overline{x}[/latex]= 6.133, s = 5.514, n= 15, and df = 15-1=14
CL = 0.98, so [latex]\displaystyle\alpha[/latex] = 1- CL = 1.0.98 = 0.02
[latex]\displaystyle({t}_{\frac{{\alpha}}{{2}}}={t}_{0.01}[/latex]
[latex]\displaystyle({t}_{\frac{{\alpha}}{{2}}}={t}_{0.01}={2.624}[/latex]
EBM = [latex]\displaystyle{t}_{\frac{{\alpha}}{{2}}}(\frac{{s}}{{\sqrt{n}}})[/latex]={2.624}(\frac{{5.514}}{{\sqrt{15}}}={3.736}[/latex]
[latex]\displaystyle\overline{x}[/latex] – EBM = 6.133 – 3.736 = 2.397
[latex]\displaystyle\overline{x}[/latex]+EBM= 16.133 -+3.736= 9.869
We estimate with 98% confidence that the mean number of all hours that statistics students spend watching television in one week is between 2.397 and 9.869.
Solution B:
Enter the data as a list.
PressSTAT
and arrow over to TESTS
.
Arrow down to8:TInterval
.
PressENTER
.
Arrow toData
and pressENTER
.
Arrow down and enter the name of the list where the data is stored.
EnterFreq
: 1Enter C-Level
: 0.98
Arrow down toCalculate
and pressEnter
.
The 98% confidence interval is (2.3965, 9,8702).
References
“America’s Best Small Companies.” Forbes, 2013. Available online at http://www.forbes.com/best-small-companies/list/ (accessed July 2, 2013).
Data from Microsoft Bookshelf.
Data from http://www.businessweek.com/.
Data from http://www.forbes.com/.
“Disclosure Data Catalog: Leadership PAC and Sponsors Report, 2012.” Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2,2013).
“Human Toxome Project: Mapping the Pollution in People.” Environmental Working Group. Available online at http://www.ewg.org/sites/humantoxome/participants/participant-group.php?group=in+utero%2Fnewborn (accessed July 2, 2013).
“Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataLeadershipPacList.shtml (accessed July 2, 2013).
Concept Review
In many cases, the researcher does not know the population standard deviation, σ, of the measure being studied. In these cases, it is common to use the sample standard deviation, s, as an estimate of σ. The normal distribution creates accurate confidence intervals when σ is known, but it is not as accurate when s is used as an estimate. In this case, the Student’s t-distribution is much better. Define a t-score using the following formula:
[latex]\displaystyle\frac{{\overline{x}-\mu}}{{\frac{{s}}{{\sqrt{n}}}}}[/latex]
The t-score follows the Student’s t-distribution with n – 1 degrees of freedom. The confidence interval under this distribution is calculated with EBM = [latex]\displaystyle{t}_{\frac{{\alpha}}{{2}}}(\frac{{s}}{{\sqrt{n}}})[/latex] where[latex]\displaystyle{t}_{\frac{{\alpha}}{{2}}}[/latex]s the t-score with area to the right equal to [latex]\displaystyle\frac{{\alpha}}{{2}}[/latex] s is the sample standard deviation, and n is the sample size. Use a table, calculator, or computer to find [latex]\displaystyle\frac{{\alpha}}{{2}}[/latex] for a given a.
Formula Review
s = the standard deviation of sample values.
t = [latex]\displaystyle\frac{{\overline{x}-\mu}}{{\frac{{s}}{{\sqrt{n}}}}}[/latex]
s the formula for the t-score which measures how far away a measure is from the population mean in the Student’s t-distribution
df = n – 1; the degrees of freedom for a Student’s t-distribution where n represents the size of the sample
T~tdf the random variable, T, has a Student’s t-distribution with df degrees of freedom
s the formula for the t-score which measures how far away a measure is from the population mean in the Student’s t-distribution
df = n – 1; the degrees of freedom for a Student’s t-distribution where n represents the size of the sample
T~tdf the random variable, T, has a Student’s t-distribution with df degrees of freedom
[latex]\displaystyle{t}_{\frac{{\alpha}}{{2}}}(\frac{{s}}{{\sqrt{n}}})[/latex] = the error bound for the population mean when the population standard deviation is unknown
is the t-score in the Student’s t-distribution with area to the right equal to
The general form for a confidence interval for a single mean, population standard deviation unknown, Student’s t is given by (lower bound, upper bound)
= (point estimate – EBM, point estimate + EBM) = [latex]\displaystyle(\overline{x} -\frac{{ts}}{{\sqrt{n}}},\overline{x} +\frac{{ts}}{{\sqrt{n}}})[/latex]