7.2: Confidence Interval for the Mean Using z-values
- Page ID
- 58285
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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 z-values when the population standard deviation is known.
- Ensure the sample is randomly selected to support valid inference.
- Estimate the range where the true population mean is likely to lie at a specified confidence level.
- Determine and meet minimum sample size requirements to ensure accuracy and reliability.
One of the parameters that can be estimated using a confidence interval is the mean \(\mu\). If the population is normally distributed, the population standard deviation \(\sigma\) is known or \(n\ge30\), then the sampling distribution is normal or approximately normal. Thus, the confidence level will be represented as a middle area under a standard normal distribution, and the confidence interval can be derived as follows.
- Construct a standard normal distribution such that the confidence level is the middle area in the distribution.
2. The Z-statistic can fall into the range \(Z_\dfrac{\alpha}{2}\) < \(Z\) <\(Z_\dfrac{\alpha}{2}\). Using the central limit theorem, Z-statistic can be rewritten as \(Z_\dfrac{\alpha}{2}\) < \(\dfrac{\overline{X}-\mu}{\dfrac{\sigma}{\sqrt{n}}}\) <\(Z_\dfrac{\alpha}{2}\). The formula can be solved for \(\mu\) using algebra to derive the confidence interval formula.
\(\overline{X}-Z_\dfrac{\alpha}{2}\left(\dfrac{\sigma}{\sqrt{n}}\right)<\mu<\overline{X}+Z_\dfrac{\alpha}{2}\left(\dfrac{\sigma}{\sqrt{n}}\right)\)
- \(\overline{X}\) is the sample mean.
- \(\sigma\) is the population standard deviation.
- n is the sample size.
- \(Z_\dfrac{\alpha}{2} \) is the z-value that can be found using the Z-Table and the given confidence level.
- \(\mu\) is the population mean that is being estimated.
Examples
A preventative healthcare researcher wishes to estimate the average number of daily steps people take in a Southern California region. The population standard deviation is around 750 steps per day and the distribution is approximately normal. She collects a random sample of 50 people who reside in the region and computes the sample mean, which is 3485 steps per day. Compute a confidence interval for the mean using a confidence level of 98%.
Solution
- List all the given information.
- \(\overline{X}\) = 3485
- \(\sigma\) = 750
- \(n\) = 50
- \(Z_\dfrac{\alpha}{2} \) = 2.33. See Table A.1 below
2. Compute the confidence interval by plugging in given information into the formula and calculating the endpoints using the order of operations.
\(\overline{X}-Z_\dfrac{\alpha}{2} \left(\dfrac{\sigma}{\sqrt{n}}\right)<\mu<\overline{X}+Z_\dfrac{\alpha}{2} \left(\dfrac{\sigma}{\sqrt{n}}\right)\)
\(3485-2.33\cdot\left(\dfrac{750}{\sqrt{50}}\right)<\mu< 3485+2.33\cdot\left(\dfrac{750}{\sqrt{50}}\right)\)
\(3885 - 247.13<\mu<3885 + 247.13\)
\(3637.87<\mu<4132.13\)
An educational researcher wishes to estimate the average age of students who attend school in the evenings at a local college. The ages of students are normally distributed and the population standard deviation is 5.2 years. The researcher collected a random sample of 37 students who attend evening classes at the college and determined the sample mean to be 28.8. Use a 95% confidence level to estimate the average age of evening students who attend the college.
Solution
- List all the given information.
- \(\overline{X}\) = 28.8
- \(\sigma\) = 5.2
- \(n\) = 37
- \(Z_\dfrac{\alpha}{2}\) = 1.96. See Table A.1.
2. Compute the confidence interval by plugging in given information into the formula and calculating the endpoints using the order of operations.
\(\overline{X}-Z_\dfrac{\alpha}{2}\left(\dfrac{\sigma}{\sqrt{n}}\right)<\mu<\overline{X}+Z_\dfrac{\alpha}{2}\left(\dfrac{\sigma}{\sqrt{n}}\right)\)
\(28.8-1.96\cdot\left(\dfrac{5.2}{\sqrt{37}}\right)<\mu< 28.8+1.96\cdot\left(\dfrac{5.2}{\sqrt{37}}\right)\)
\(28.8 - 1.68<\mu< 28.8 + 1.68\)
\(27.12 <\mu< 30.48\)
An arborist who works for a large city in Southern California wants to estimate the mean age of Jacaranda trees in the city. The arborist collected a random sample of 40 Jacaranda trees and determined the average age to be 32.3 years. The ages of the Jacaranda trees are normally distributed with a population standard deviation of 6.5 years. Use the TI 83/84 calculator to estimate the mean age of the Jacaranda trees using a confidence level of 90%.
Solution
- Go into the [STAT] menu and use the right arrows to select [TESTS] then choose [7:ZInterval]. Select [STATS] and type in the information shown in ( \(\PageIndex{4}\) ). Finally, select [CALCULATE] and hit enter.
2. The output displayed in the TI 83/84 calculator screen should mirror the one displayed in the figure below.
3. The confidence interval for the mean age of Jacaranda trees is built using the endpoints displayed between the parenthesis in the figure above.
\(30.61 <\mu< 33.99\)
Minimum Sample Size
The minimum size needed to construct a confidence interval for the mean \(\mu\) can be derived from the confidence interval formula. The margin of error E represents the range of variability around the estimated value. To find the minimum sample size, set \(E = Z_C\left(\dfrac{\sigma}{\sqrt{n}}\right)\) and solve for \(n\). The formula is presented below.
The minimum sample size needed to compute a confidence interval for the mean is defined using the formula below. Always round the final answer upward to the nearest whole number.
\(n=\left(\dfrac{Z_\dfrac{\alpha}{2}\cdot\sigma}{E}\right)^2\)
- \(n\) is the minimum sample size.
- \(\sigma\) is the population standard deviation.
- \(E\) is the margin of error.
- \(Z_\dfrac{\alpha}{2}\) is the z-value that can be found using Table A.1 using the confidence level.
Example of Minimum Sample Size
An entomologist is studying the life span in years of the flam skipper dragonfly (Libellula saturate). It has been determined that the population standard deviation in years is 0.85 years. The entomologist will use a confidence level of 98% and a margin of error of 0.25 years. Compute the minimum sample size needed for \(n\).
Solution
- List the information.
- \(\sigma\) =0.85
- \(E\) =0.25
- \(Z_ \dfrac{\alpha}{2} \) = 2.33 using table A.1.
2. Compute \(n\) by plugging information into the formula.
\(n = \left(\dfrac{Z_\dfrac{\alpha}{2}\cdot\sigma}{E}\right)^2\)
\(n = \left(\dfrac{2.33\cdot0.85}{0.25}\right)^2\)
\(n = 62.758\)
3. Round \(n\) upwards,
\(n \approx 63\)
The minimum sample size is 63.
Author
"7.2: Confidence Interval for the Mean Using z-values" by Alfie Swan is licensed under CC BY 4.0
Exercises
- A consumer research group wants to estimate the average price of a carton of 12 eggs at grocery stores in a certain city. The population standard deviation is known to be $0.40. A random sample of 50 stores shows that the average price of a carton of 12 eggs is $2.85. Construct a 95% confidence interval for the true average price of a carton of 12 eggs in that city. Round to two decimal places.
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- A researcher is studying the yearly tuition costs at private colleges in Southern California. Based on historical data, the population standard deviation is known to be $6,000. A random sample of 64 private colleges shows that the average yearly tuition is $38,500. Construct a 90% confidence interval for the true mean yearly tuition at private colleges in Southern California. Round to two decimal places.
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- A labor research group wants to estimate the average retirement age for employees in California. Based on prior studies, the population standard deviation is known to be 4.5 years. A random sample of 100 recently retired employees in California shows that the average retirement age is 64.2 years. Construct a 98% confidence interval for the true mean retirement age of employees in California.
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- A sports economics researcher wants to estimate the average cost of attending a professional sporting event in Los Angeles (including tickets, parking, and concessions). From previous studies, the population standard deviation is known to be $25. The researcher wants to construct a 95% confidence interval with a margin of error is $5. What is the minimum sample size required? Round to the next whole number.
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- A biologist wants to estimate the average number of days it takes for sparrows to grow from hatching to adulthood. Based on previous research, the population standard deviation is known to be 3.2 days. The biologist wants to construct a 99% confidence interval with a margin of error no greater than 1 day. What is the minimum sample size required?
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- Answers
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