7.3: Confidence Interval for the Mean Using t-values
- Page ID
- 58286
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Compute confidence intervals for the mean using t-values when the population standard deviation is unknown.
- Use the sample standard deviation to estimate variability.
- Understand that the t-distribution accounts for additional uncertainty, especially with small sample sizes.
- Recognize that the t-distribution approaches the normal distribution as sample size increases.
t-distribution:
For most confidence intervals, the population standard deviation will not be known. Thus, the distribution used to estimate the population mean will most likely not be normal. The t-distribution will be used instead of the normal distribution when 𝜎 is unknown.
The t-distribution was discovered by William S. Gosset in 1908. He was a statistician employed by Guinness Brewery who was not allowed to use his name when publishing his discovery. Thus, he used the pseudonym “student”, and his distribution was known as Student’s t- distribution. A graph of the t-distribution is listed below
- It is symmetric.
- The mean is 0.
- The mean, median, and mode are all equal and located at the center of the distribution.
- The graph never touches the horizontal axis.
- The variance is greater than 1.
- The graph is a family of curves based on the concept of degrees of freedom (d.f. = n - 1).
- As the sample size n gets larger, the shape of the t-distribution becomes normal.
Confidence Interval for One Population Mean (t-Interval) Formula:
The confidence interval formula can be found by using a confidence level as the middle area of a t-distribution and then using algebra to solve for \(\mu\) using the corresponding t-values associated with the confidence level. A graph of the confidence level in the t-distribution along with corresponding t-values is presented in the graph below.
The range of values on the horizontal axis leads to the following inequality \(-t_ \dfrac{\alpha}{2} <t<t_ \dfrac{\alpha}{2} \). The value t can be replaced with the following expression \(-t_ \dfrac{\alpha}{2} <\dfrac{\overline{X}-\mu}{\dfrac{s}{\sqrt{n}}}<t_ \dfrac{\alpha}{2} \). Finally solving for \(\mu\) in the center will lead to the confidence interval formula.
\(\overline{X}-t_ \dfrac{\alpha}{2} \left(\dfrac{s}{\sqrt{n}}\right)<\mu<\overline{X}+t_ \dfrac{\alpha}{2} \left(\dfrac{s}{\sqrt{n}}\right)\)
- \(\overline{X}\) is the sample mean.
- s is the sample standard deviation.
- n is the sample size.
- \(T_ \dfrac{\alpha}{2} \) is the t-value found in Table A2 using the confidence level and d.f. = n - 1.
- \(\mu\) is the population mean that is being estimated.
Examples:
Suppose a researcher wishes to estimate the age of a student at a California community college who passes an elementary statistics class. The researcher collects a sample of 11 random students. The sample mean age is 22.6 and the sample standard deviation is 1.88. The researcher decides to use a confidence level of 95%.
Solution:
Step 1) Label the given statistics.
- \(\overline{X}\) = 22.6 (sample mean)
- s = 1.88 (sample standard deviation)
- n = 11 (sample size)
- Confidence level = 95%
Step 2) Look up the \(t_\dfrac{\alpha}{2}\) value in Table A.2 using the confidence level and d.f. = n - 1 = 11 - 1 = 10. In Table A.2, go down the first column to 10 degrees of freedom. Then go over to the column headed with 95%. Thus \(t_{\dfrac{\alpha}{2}}=2.228\). (See Example \(\PageIndex{1}\).)
- \(t_\dfrac{\alpha}{2}\) = 2.228
Step 3) Plug the information into the formula and use the order of operations to calculate the two endpoints.
\(\overline{X}-t_\dfrac{\alpha}{2}\left(\dfrac{s}{\sqrt{n}}\right)<\mu<\overline{X}+t_\dfrac{\alpha}{2}\left(\dfrac{s}{\sqrt{n}}\right)\)
\(22.6-2.228\left(\dfrac{1.88}{\sqrt{11}}\right)<\mu<22.6+2.228\left(\dfrac{1.88}{\sqrt{11}}\right)\)
\(22.6-1.26<\mu<22.6+1.26\)
\(21.34<\mu<23.86\)
A random sample of 20 IQ scores of famous people was taken information from the website of IQ of Famous People ("IQ of famous," 2013) and then using a random number generator to pick 20 of them. The data are in Example \(\PageIndex{2}\) (this is the same data set that was used in Example \(\PageIndex{2}\)). Find a 98% confidence interval for the average IQ of a famous person.
| 158 | 180 | 150 | 137 | 109 |
|---|---|---|---|---|
| 225 | 122 | 138 | 145 | 180 |
| 118 | 118 | 126 | 140 | 165 |
| 150 | 170 | 105 | 154 | 118 |
- State the random variable and the parameter in words.
- State and check the assumptions for a confidence interval.
- Find the sample statistic and confidence interval.
- Statistical Interpretation
- Real World Interpretation
Solution
1. x = IQ score of a famous person
\(\mu\) = mean IQ score of a famous person
2.
- A random sample of 20 IQ scores was taken. This was stated in the problem.
- The population of IQ scores is normally distributed. This was shown in Example \(\PageIndex{2}\).
3. Sample Statistic:
\(\overline{x} = 145.4\)
\(s = 29.27\)
Calculate the degrees of freedom, df = n - 1 = 20 - 1 = 19, and use the confidence level, which is 98%, to look up the t-value. In table A.2, go down the first column to 19 degrees of freedom. Then go over to the column headed with 98%. Thus \(t_{\dfrac{\alpha}{2}}=2.539\). (See Example \(\PageIndex{2}\).)
.png?revision=1)
Figure \(\PageIndex{5}\): Table of Critical Values for the t-distribution at Degrees of Freedom of 19
\(\overline{X}-t_\dfrac{\alpha}{2}\left(\dfrac{s}{\sqrt{n}}\right)<\mu<\overline{X}+t_\dfrac{\alpha}{2}\left(\dfrac{s}{\sqrt{n}}\right)\)
\(145.4-2.539\left(\dfrac{29.27}{\sqrt{20}}\right)<\mu<145.4+2.539\left(\dfrac{29.27}{\sqrt{20}}\right)\)
\(145.4-16.6<\mu<145.4+16.6\)
\(128.8<\mu<162\)
4. There is a 98% chance that \(128.8<\mu<162\) contains the mean IQ score of a famous person.
5. The mean IQ score of a famous person is between 128.8 and 162.
For this example, the TI-83/84 is used to compute a confidence interval for the mean. In this problem, a random sample of 10 final exam scores for a statistics class is collected from multiple sections over the past 20 years. The scores are provided in the data set below. Construct a confidence interval for the mean with a confidence level of 99%.
| 85 | 98 | 74 | 56 | 62 |
|---|---|---|---|---|
| 95 | 87 | 77 | 60 | 84 |
Solution
1. On the TI-83/84: Go into the [STAT] menu. Select [EDIT] and make sure [1:EDIT] is highlighted. Hit enter, and then type the data into \(L_1\).
2. Go into the [STAT] menu and use the right arrows to select [TESTS] then choose [8:TInterval]. Select [DATA] and type in the information shown in ( \(\PageIndex{4}\) ). Finally, select [CALCULATE] and hit enter.
.png?revision=1)
3. The output should look as follows.
4. There is a 99% chance that 62.7 points < \(\mu\) < 92.9 points contains the mean score of the final exam.
5. The mean final exam score for the statistics class is between 62.7 and 92.9 years.
Authors
"7.3: Confidence Interval for the Mean Using t-values" by Toros Berberyan, Tracy Nguyen, and Alfie Swan is licensed under CC BY-SA 4.0
Attributions
"8.3: One-Sample Interval for the Mean" by Kathryn Kozak is licensed under CC BY-SA 4.0
Exercises
- A film analyst wants to estimate the average length of movies released in the past year. A random sample of 12 movies was selected, and the sample had an average length of 118.4 minutes with a sample standard deviation of 15.2 minutes. Construct a 90% confidence interval to estimate the true average movie length.
Scan the QR code or click on it to open the MyOpenMath version of the above question with step-by-step guidance.
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- An education researcher wants to estimate the average age at which students graduate with a bachelor's degree. A random sample of 10 recent graduates was selected. The sample had an average age of 25.7 years with a sample standard deviation of 2.4 years. Construct a 95% confidence interval to estimate the true average graduation age for students earning a bachelor’s degree.
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- A veterinarian wants to estimate the average age of pet dogs in her clinic's care. She randomly selects 15 dogs and records their ages (in years). Construct a 95% confidence interval for the true average age of pet dogs based on this sample. Round the answers to two decimal places.
The sample data is shown below: Data (ages in years):
4.5, 6.2, 3.8, 5.1, 7, 4.9, 6.3, 5.5, 3.7, 4.8, 6.1, 5.4, 4.6, 5.9, 6.4
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- A transportation researcher wants to estimate the average commute time from home to work for residents in the Los Angeles region. She randomly selects 18 commuters and records their one-way commute times (in minutes). Construct a 99% confidence interval to estimate the true average commute time for LA commuters.
The data is shown below: Commute times (in minutes):
36, 42, 55, 47, 38, 50, 60, 33, 40, 45, 58, 53, 41, 39, 44, 49, 52, 46
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- Answers
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If you are an instructor and want the solutions to all the exercise questions for each section, please email Toros Berberyan.