Learning Outcomes
- Interpret the Student’s t probability distribution as the sample size changes
- Discriminate between problems applying the normal and the Student’s t distributions
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.
Excel has a T.DIST function that finds probabilities for left-tailed regions of the Student’s t-distribution. The function syntax is T.DIST(x, deg_freedom, cumulative) where [latex]x[/latex] is the numeric value at which to evaluate the distribution, deg-freedom is the degrees of freedom, and cumulative is a logical value that determines the form of the function and takes either TRUE or FALSE (or similarly 1 or 0). As with NORM.DIST in previous sections, we will only use cumulative = TRUE or 1).
The T.INV function in Excel is used to find the value of [latex]t[/latex] if we know the probability. The syntax is T.INV(probability, deg_freedom) where probability is the area to the left of the desired [latex]t[/latex] and deg_freedom is the degrees of freedom.
The TI-83,83+, 84 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 t-scores that correspond to the confidence level (column) and degrees of freedom (row). 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{{s}}{{\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],
- use df = n – 1 degrees of freedom, and
- s = sample standard deviation.
The format for the confidence interval is:
([latex]\displaystyle\overline{x}[/latex] – EBM, [latex]\displaystyle\overline{x}[/latex]+EBM)
or
[latex]\overline{x}[/latex]– EBM ≤ μ ≤ [latex]\overline{x}[/latex] + EBM
Calculate the Confidence Interval Directly
Using Excel:
- Use function CONFIDENCE.T to calculate the EBM. The function has syntax CONFIDENCE.T(alpha, standard_dev, size) where alpha is the significance level (1 – CL), the standard deviation is the population standard deviation (or estimated by the sample standard deviation, s), and size is the sample size n.
- Construct confidence interval using EBM.
Using TI83/84:
- Press STAT.
- Arrow over to TESTS. Arrow down to 8:TInterval and press ENTER (or just press 8).
- Arrow to
Data
and pressENTER
. - Arrow down and enter the name of the list where the data is stored.
- Enter
Freq
: 1EnterC-Level
- Arrow down to
Calculate
and pressEnter
Summary of Requirements:
- The sample is a simple random sample.
- Either the sample size is at least 30 (n > 30) or the sample is from a normally distributed population.
Example
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.
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:
To find the 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 so 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
[latex]\displaystyle{t}_{\frac{{\alpha}}{{2}}}={t}_{0.025}={2.14}[/latex] using invT(.975, 14) on the TI-84+ calculator or T.INV(0.975, 14) on Excel
EBM = [latex]\displaystyle({t}_{\frac{{\alpha}}{{2}}})(\frac{{s}}{{\sqrt{n}}})[/latex]
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.
Note: The EBM can also be found in Excel with CONFIDENC.T(0.05, 1.6722, 15)
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 hours 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. Use Excel to find the sample mean and the sample standard deviation.
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
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:
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.
Example
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:
[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}}})={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 2.4 and 9.9 hours.
Alternate Solution (if using Excel):
First find significance level by taking 1-confidence level, so for this problem 1-0.98=0.02.
Type =CONFIDENCE.T(0.02, 5.514, 15) = 3.736 = EBM
[latex]\displaystyle\overline{x}[/latex] – EBM = 6.133 – 3.736 = 2.397
[latex]\displaystyle\overline{x}[/latex]+EBM= 16.133 -+3.736= 9.869
Alternate Solution (if using TI83/84):
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] is 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{t}_\frac{{\alpha}}{{2}}[/latex] for a given confidence level.
Formula Review
s = the standard deviation of sample values.
t = [latex]\displaystyle\frac{{\overline{x}-\mu}}{{\frac{{s}}{{\sqrt{n}}}}}[/latex] is 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
EBM: [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.
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} -{t}_{\frac{{\alpha}}{{2}}}(\frac{{s}}{{\sqrt{n}}}),\overline{x} +{t}_{\frac{{\alpha}}{{2}}}(\frac{{s}}{{\sqrt{n}}})[/latex]
Candela Citations
- A Single Population Mean using the Student t Distribution. Provided by: OpenStax. Located at: http://cnx.org/contents/30189442-6998-4686-ac05-ed152b91b9de@17.44:52/Introductory_Statistics. License: CC BY: Attribution
- Confidence Intervals for One Mean: Sigma Not Known (t Method). Authored by: jbstatistics. Located at: https://youtu.be/bFefxSE5bmo. License: All Rights Reserved. License Terms: Standard YouTube License
- Introductory Statistics . Authored by: Barbara Illowski, Susan Dean. Provided by: Open Stax. Located at: http://cnx.org/contents/30189442-6998-4686-ac05-ed152b91b9de@17.44. License: CC BY: Attribution. License Terms: Download for free at http://cnx.org/contents/30189442-6998-4686-ac05-ed152b91b9de@17.44