8.5.1: Practice with Confidence Interval Calculations
 Page ID
 17343
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 \(\sigma\) 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 \(\sigma\) 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 tdistribution. The name comes from the fact that Gosset wrote under the pen name "Student."
Up until the mid1970s, 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 tdistribution whenever \(s\) is used as an estimate for \(\sigma\). If you draw a simple random sample of size \(N\) from a population that has an approximately a normal distribution with mean \(\mu\) and unknown population standard deviation \(\sigma\) and calculate the \(t\)score, then the \(t\)scores follow a Student's tdistribution with \(n – 1\) degrees of freedom. The \(t\)score has the same interpretation as the zscore. It measures how far \(\bar{x}\) is from its mean \(\mu\). For each sample size \(n\), there is a different Student's tdistribution.
The degrees of freedom, \(n – 1\), come from the calculation of the sample standard deviation \(s\). Previously, we used \(n\) deviations (\(x  \bar{x}\) values) to calculate \(s\). 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).
For each sample size \(n\), there is a different Student's tdistribution.
Properties of the Student's \(t\)Distributio
 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 \(\mu\) and unknown population standard deviation \(\sigma\). 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.
A probability table for the Student's \(t\)distribution can be used. The table gives \(t\)scores that correspond to the confidence level (column) and degrees of freedom (row).
A Student's \(t\)table gives \(t\)scores given the degrees of freedom and the righttailed probability. The table is very limited. Calculators and computers can easily calculate any Student's \(t\)probabilities.
Example \(\PageIndex{1}\)
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.
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, \(\bar{x}\), and the \(EBM\).
\(\bar{x} = 8.2267 \)
\(s = 1.6722\)
\(n = 15\)
\(df = 15 – 1 = 14 \)
\(\frac{\alpha}{2} = 0.025\), so \(t_{crit} = t_{0.025}\)
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
\(t_{0.025} = 2.145\) from the Table of Critical tscores.
\[ \begin{align*} EBM &= \left(t\right)\left(\frac{s}{\sqrt{n}}\right) \\[4pt] &= (2.14)\left(\frac{1.6722}{\sqrt{15}}\right) = 0.924 \end{align*}\]
Now it is just a direct application of Equation \ref{confint}:
\[ \begin{align*} \bar{x} – EBM &= 8.2267 – 0.9240 = 7.3 \\[4pt] \bar{x} + EBM &= 8.2267 + 0.9240 = 9.15 \end{align*}\]
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.
You can try one on your own:
Exercise \(\PageIndex{1}\)
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
 Answer

(8.1634, 9.8032)
Example \(\PageIndex{2}\): The Human Toxome Project
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. Table \(\PageIndex{1}\) 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 \(\bar{x} = 127.45\) 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
\[\alpha = 1 – CL = 1 – 0.90 = 0.10 \frac{\alpha}{2} = 0.05, t_{\frac{\alpha}{2}} = t_{0.05}\]
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 \(t_{0.05} = 1.729\).
\[EBM = t \times \left(\frac{s}{\sqrt{n}}\right) = 1.729\times \left(\frac{25.965}{\sqrt{20}}\right) \approx 10.038 \nonumber \]
\[\bar{x} – EBM = 127.45 – 10.038 = 117.412\nonumber \]
\[\bar{x} + EBM = 127.45 + 10.038 = 137.488\nonumber \]
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 \(\PageIndex{3}\)
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 Table \(\PageIndex{2}\). 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
 \(\bar{x} = 6.133\),
 \(s = 5.514\),
 \(n = 15\), and
 \(df = 15 – 1 = 14\).
\(\frac{\alpha}{2} = 0.01 t_{\frac{\alpha}{2}} = t_{0.01} = 2.624\)
\[\bar{x} – EBM = 6.133 – 3.736 = 2.397\nonumber\]
\[\bar{x} + EBM = 6.133 + 3.736 = 9.869\nonumber\]
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.
Reference
 “America’s Best Small Companies.” Forbes, 2013. Available online at http://www.forbes.com/bestsmallcompanies/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 www.fec.gov/data/index.jsp (accessed July 2,2013).
 “Human Toxome Project: Mapping the Pollution in People.” Environmental Working Group. Available online at www.ewg.org/sites/humantoxome...tero%2Fnewborn (accessed July 2, 2013).
 “Metadata Description of Leadership PAC List.” Federal Election Commission. Available online at www.fec.gov/finance/disclosur...pPacList.shtml (accessed July 2, 2013).
Contributors and Attributions
Barbara Illowsky and Susan Dean (De Anza College) with many other contributing authors. Content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at http://cnx.org/contents/30189442699...b91b9de@18.114.
