37 8.1 A Single Population Mean using the Normal Distribution
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of [latex]\displaystyle\overline{{x}}={10}[/latex] and we have constructed the 90% confidence interval (5, 15) where EBM = 5.
Calculating the Confidence Interval
To construct a confidence interval for a single unknown population mean μ, where the population standard deviation is known, we need [latex]\displaystyle\overline{{x}}[/latex] is the point estimate of the unknown population mean μ.
The confidence interval estimate will have the form:
(point estimate – error bound, point estimate + error bound) or,
in symbols, [latex]\displaystyle{(\overline{{x}}  {EBM},\overline{{x}} + {EBM})}[/latex]
The margin of error (EBM) depends on the confidence level (abbreviated CL). The confidence level is often considered the probability that the calculated confidence interval estimate will contain the true population parameter. However, it is more accurate to state that the confidence level is the percent of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, it is the choice of the person constructing the confidence interval to choose a confidence level of 90% or higher because that person wants to be reasonably certain of his or her conclusions.
There is another probability called alpha (α). α is related to the confidence level, CL. α is the probability that the interval does not contain the unknown population parameter.
Given that CL is the probability that the calculated confidence interval estimate will contain the true population parameter,

Example 1
Suppose we have collected data from a sample. We know the sample mean but we do not know the mean for the entire population.
The sample mean is seven, and the error bound for the mean is 2.5.
[latex]\displaystyle\overline{{x}}={7}[/latex]. At 95% confidence level, EBM = 2.5.
General form to find confidence interval: [latex]\displaystyle{(\overline{{x}}{EBM},\overline{{x}}+{EBM})}[/latex] 
The confidence interval is (7 – 2.5, 7 + 2.5), and calculating the values gives (4.5, 9.5).
Since we calculate the interval at 95% confidence level, we estimate with 95% confidence that the true value of the population mean is between 4.5 and 9.5.”
Try It
Suppose we have data from a sample. The sample mean is 15, and the error bound for the mean is 3.2.
What is the confidence interval estimate for the population mean?
[practicearea rows=”1″][/practicearea]
Show Answer
(11.8, 18.2)
A confidence interval for a population mean with a known standard deviation is based on the fact that the sample means follow an approximately normal distribution. Suppose that our sample has a mean of [latex]\displaystyle\overline{{x}}={10}[/latex], and we have constructed the 90% confidence interval (5, 15) where EBM = 5.
To get a 90% confidence interval, we must include the central 90% of the probability of the normal distribution. If we include the central 90%, we leave out a total of α = 10% in both tails, or 5% in each tail, of the normal distribution.
To capture the central 90%, we must go out 1.645 “standard deviations” on either side of the calculated sample mean. The value 1.645 is the zscore from a standard normal probability distribution that puts an area of 0.90 in the center, an area of 0.05 in the far left tail, and an area of 0.05 in the far right tail.
It is important that the “standard deviation” used must be appropriate for the parameter we are estimating, so in this section we need to use the standard deviation that applies to sample means, which is[latex]\displaystyle\frac{{\sigma}}{{\sqrt{n}}}[/latex]. The fraction [latex]\displaystyle\frac{{\sigma}}{{\sqrt{n}}}[/latex], is commonly called the “standard error of the mean” in order to distinguish clearly the standard deviation for a mean from the population standard deviation σ.
In summary, as a result of the central limit theorem:
 [latex]\displaystyle\overline{X}[/latex], that is, [latex]\displaystyle\overline{X}{\sim}{N}\left({\mu}_{x}, \frac{{\sigma}}{{\sqrt{n}}}\right)[/latex]
 When the population standard deviation σ is known, we use a normal distribution to calculate the error bound.
Calculating the Confidence IntervalTo construct a confidence interval estimate for an unknown population mean, we need data from a random sample. The steps to construct and interpret the confidence interval are:

We will first examine each step in more detail, and then illustrate the process with some examples.
Finding the zscore for the Stated Confidence Level
When we know the population standard deviation σ, we use a standard normal distribution to calculate the error bound EBM and construct the confidence interval. We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution Z ~ N(0, 1).
The confidence level, CL, is the area in the middle of the standard normal distribution.
CL = 1 – α, so α is the area that is split equally between the two tails.
Each of the tails contains an area equal to [latex]\frac{\alpha}{2}[/latex].
The zscore that has an area to the right of [latex]\frac{\alpha}{2}[/latex] is denoted by [latex]{Z}_{\frac{\alpha}{2}}[/latex].
For example, when CL = 0.95, α = 0.05,
then [latex]\frac{\alpha}{2}[/latex] = 0.025. We can write [latex]{Z}_{\frac{\alpha}{2}}[/latex]_{ }= z_{0.025}.
The area to the right of z_{0.025} is 0.025 and the area to the left of z_{0.025} is 1 – 0.025 = 0.975.
[latex]{Z}_{\frac{\alpha}{2}}[/latex] = z_{0.025 }= 1.96, using a calculator, computer or a standard normal probability table.
To find out z_{0.025}, we can use TICalculator.
TICalculator: invNorm(0.975, 0, 1)
We will find z_{0.025 }= 1.96
NoteRemember to use the area to the LEFT of ; in this chapter the last two inputs in the invNorm command are 0, 1, because you are using a standard normal distribution Z ~ N(0, 1). 
Calculating the Error Bound (EBM)
The error bound formula for an unknown population mean μ when the population standard deviation σ is known is
 EBM = ([latex]{Z}_{\frac{{\alpha}}{{2}}})(\frac{{\sigma}}{{\sqrt{n}}})[/latex]
The confidence interval estimate has the format ([latex]\overline{x}[/latex]–EBM,[latex]\overline{x}[/latex]+EBM).
The graph gives a picture of the entire situation.
Blueshaded area + 2 white areas on each side
= CL + [latex]\displaystyle\frac{{\alpha}}{{2}}+\frac{{\alpha}}{{2}}[/latex]
= [latex]{\text{CL}}+{\alpha}[/latex]
= 1
Writing the Interpretation
The interpretation should clearly state the confidence level (CL), explain what population parameter is being estimated (here, a population mean), and state the confidence interval (both endpoints). “We estimate with ___% confidence that the true population mean (include the context of the problem) is between ___ and ___ (include appropriate units).”
Example 2
Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams).
Find a 90% confidence interval for the true (population) mean of statistics exam scores.
You can use technology to calculate the confidence interval directly. The first solution is shown stepbystep (Solution A). The second solution uses the TI83, 83+, and 84+ calculators (Solution B). 
Solution A:
To find the confidence interval, you need the sample mean, and the EBM.
[latex]\overline{x}={68}[/latex], [latex]\displaystyle\sigma=3[/latex], n = 6
[latex]{EBM}=({Z}_{\frac{{\alpha}}{{2}}})(\frac{{\sigma}}{{\sqrt{n}}})[/latex]
The confidence level is 90% (CL = 0.90)
CL = 0.90 so α = 1 – CL = 1 – 0.90 = 0.10
[latex]\displaystyle\frac{{\alpha}}{{2}}=0.05[/latex], [latex]\displaystyle{z}_{\frac{{\alpha}}{{2}}}={z}_{0.05}[/latex]
The area to the right of z_{0.05} is 0.05 and the area to the left of z_{0.05 }is 1 – 0.05 = 0.95.
[latex]\displaystyle\frac{{{z}_{\alpha}}}{{2}}={z}_{0.05}=1.645[/latex]
TI83/84: invNorm(0.95, 0, 1)
This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.
EBM = (1.645)([latex]\displaystyle\frac{{3}}{{\sqrt{36}}}[/latex])= 0.8225
[latex]\displaystyle\overline{x}[/latex] – EBM = 68 – 0.8225 = 67.1775
[latex]\displaystyle\overline{x}[/latex] + EBM = 68 + 0.8225 = 68.8225
The 90% confidence interval is (67.1775, 68.8225).
Solution B:
Press STAT
and arrow over to TESTS
.
Arrow down to 7: ZInterval
.
Press ENTER
.
Arrow to Stats
and press ENTER
.
Arrow down and enter three for σ, 68 for[latex]\displaystyle\overline{X}[/latex], 36 for n, and .90 for Clevel
.
Arrow down to Calculate
and press ENTER
.
The confidence interval is (67.178, 68.822).
InterpretationWe estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82. Explanation of 90% Confidence LevelNinety percent of all confidence intervals constructed in this way contain the true mean statistics exam score. For example, if we constructed 100 of these confidence intervals, we would expect 90 of them to contain the true population mean exam score. 
Try It
Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 minutes.
Find a 90% confidence interval estimate for the population mean delivery time.
Show Answer
(34.1347, 37.8653)
Example 3
The Specific Absorption Rate (SAR) for a cell phone measures the amount of radio frequency (RF) energy absorbed by the user’s body when using the handset. Every cell phone emits RF energy. Different phone models have different SAR measures. To receive certification from the Federal Communications Commission (FCC) for sale in the United States, the SAR level for a cell phone must be no more than 1.6 watts per kilogram. This table shows the highest SAR level for a random selection of cell phone models as measured by the FCC.
Phone Model  SAR  Phone Model  SAR  Phone Model  SAR 

Apple iPhone 4S  1.11  LG Ally  1.36  Pantech Laser  0.74 
BlackBerry Pearl 8120  1.48  LG AX275  1.34  Samsung Character  0.5 
BlackBerry Tour 9630  1.43  LG Cosmos  1.18  Samsung Epic 4G Touch  0.4 
Cricket TXTM8  1.3  LG CU515  1.3  Samsung M240  0.867 
HP/Palm Centro  1.09  LG Trax CU575  1.26  Samsung Messager III SCHR750  0.68 
HTC One V  0.455  Motorola Q9h  1.29  Samsung Nexus S  0.51 
HTC Touch Pro 2  1.41  Motorola Razr2 V8  0.36  Samsung SGHA227  1.13 
Huawei M835 Ideos  0.82  Motorola Razr2 V9  0.52  SGHa107 GoPhone  0.3 
Kyocera DuraPlus  0.78  Motorola V195s  1.6  Sony W350a  1.48 
Kyocera K127 Marbl  1.25  Nokia 1680  1.39  TMobile Concord  1.38 
Find a 98% confidence interval for the true (population) mean of the Specific Absorption Rates (SARs) for cell phones.
Assume that the population standard deviation is σ = 0.337.
Solution A:
To find the confidence interval, start by finding the point estimate: the sample mean, [latex]\displaystyle\overline{x}[/latex] = 1.024
Next, find the EBM. Because you are creating a 98% confidence interval, CL = 0.98.
You need to find z_{0.01} having the property that the area under the normal density curve to the right of z_{0.01} is 0.01 and the area to the left is 0.99.
Use your calculator, a computer, or a probability table for the standard normal distribution to find z_{0.01 }= 2.326.
EBM = ([latex]\displaystyle{Z}_{0.01})(\frac{{\sigma}}{{\sqrt{n}}})=(2.236)\frac{{0.337}}{{\sqrt{30}}}=0.1431[/latex]
To find the 98% confidence interval, find[latex]\displaystyle\overline{x}\pm{EBM}[/latex]
[latex]\displaystyle\overline{x}[/latex] – EBM = 1.024 – 0.1431 = 0.8809
[latex]\displaystyle\overline{x}[/latex] + EBM = 1.024 + 0.1431 = 1.1671
We estimate with 98% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8809 and 1.1671 watts per kilogram.
Solution B:
 Press STAT and arrow over to TESTS.
 Arrow down to 7: ZInterval.
 Press ENTER.
 Arrow to Stats and press ENTER.
 Arrow down and enter the following values:
 σ: 0.337
 [latex]\displaystyle\overline{x}[/latex]: 1.024
 n: 30
 Clevel: 0.98
 Arrow down to Calculate and press ENTER.
 The 98% confidence interval is (0.881, 1.167).(to three decimal places)
Try It
This table shows a different random sampling of 20 cell phone models. Use this data to calculate a 93% confidence interval for the true mean SAR for cell phones certified for use in the United States. As previously, assume that the population standard deviation is σ = 0.337.
Phone Model SAR Phone Model SAR Blackberry Pearl 8120 1.48 Nokia E71x 1.53 HTC Evo Design 4G 0.8 Nokia N75 0.68 HTC Freestyle 1.15 Nokia N79 1.4 LG Ally 1.36 Sagem Puma 1.24 LG Fathom 0.77 Samsung Fascinate 0.57 LG Optimus Vu 0.462 Samsung Infuse 4G 0.2 Motorola Cliq XT 1.36 Samsung Nexus S 0.51 Motorola Droid Pro 1.39 Samsung Replenish 0.3 Motorola Droid Razr M 1.3 Sony W518a Walkman 0.73 Nokia 7705 Twist 0.7 ZTE C79 0.869 [practicearea rows=”2″][/practicearea]
Show Answer
[latex]\displaystyle\overline{x}[/latex] = 0.940
[latex]\displaystyle\frac{{\alpha}}{{2}}=\frac{{1CL}}{{2}}=\frac{{10.93}}{{2}}[/latex] = 0.035
[latex]\displaystyle{z}_{0.05}[/latex] = 1.812
EBM=[latex]\displaystyle({z}_{0.05})(\frac{{\sigma}}{{\sqrt{n}}})=(1.182)(\frac{{0.337}}{{\sqrt{20}}}[/latex] = 0.1365 [latex]\displaystyle\overline{x}[/latex]– EBM = 0.940 – 0.1365 = 0.8035
[latex]\displaystyle\overline{x}[/latex]+EBM = 0.940 + 0.1365 = 1.0765We estimate with 93% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8035 and 1.0765 watts per kilogram.
Notice the difference in the confidence intervals calculated in Example 3 and the Try It just completed. These intervals are different for several reasons: they were calculated from different samples, the samples were different sizes, and the intervals were calculated for different levels of confidence. Even though the intervals are different, they do not yield conflicting information. The effects of these kinds of changes are the subject of the next section in this chapter.
Changing the Confidence Level or Sample Size
Example 4
Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of three points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68. Find a confidence interval estimate for the population mean exam score (the mean score on all exams).
Find a 95% confidence interval for the true (population) mean statistics exam score.
Solution:
To find the confidence interval, you need the sample mean,[latex]\displaystyle\overline{x}[/latex], and the EBM.
[latex]\displaystyle\overline{x}[/latex] = 68, [latex]\displaystyle{\sigma}={3}[/latex], n = 36
EBM =([latex]\displaystyle{z}_{\frac{{\alpha}}{{2}}})(\frac{{\sigma}}{{\sqrt{n}}}[/latex])
CL = 0.95 so α = 1 – CL = 1 – 0.95 = 0.05
The area to the right of z_{0.025} is 0.025 and the area to the left of z_{0.025} is 1 – 0.025 = 0.975.
TI83/84: invnorm(0.975,0,1)
This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.
InterpretationWe estimate with 95% confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98. Explanation of 95% Confidence Level95% of all confidence intervals constructed in this way contain the true value of the population mean statistics exam score. 
Comparing the ResultsIn Example 2, the 90% confidence interval is (67.18, 68.82). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider. 
Summary: Effect of Changing the Confidence Level
 Increasing the confidence level increases the error bound, making the confidence interval wider.
 Decreasing the confidence level decreases the error bound, making the confidence interval narrower.
Try It
Suppose average pizza delivery times are normally distributed with an unknown population mean and a population standard deviation of six minutes. The sample mean deliver time of 20 deliverers is 36 minutes.
Find a 95% confidence interval estimate for the true mean pizza delivery time.
[practicearea rows=”1″][/practicearea]
Show Answer
(33.37, 38.63)
Example 5
What happens to the error bound(EBM) if the sample size is changed?
In example 2, we suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of 3 points. A random sample of 36 scores is taken and gives a sample mean (sample mean score) of 68.
We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82. 
Leave everything the same except the sample size. Use the original 90% confidence level.
What happens to the error bound and the confidence interval if we increase the sample size and use n = 100 instead of n = 36?
What happens if we decrease the sample size to n = 25 instead of n = 36?
 [latex]\mu[/latex] = 68
 σ = 3
 n = 100
 The confidence level is 90% (CL=0.90); .
Solution:
EBM = ([latex]{Z}_{\frac{{\alpha}}{{2}}})(\frac{{\sigma}}{{\sqrt{n}}})[/latex]
If we increase the sample size n to 100, we decrease the error bound.
If we decrease the sample size n to 25, we increase the error bound.
Summary: Effect of Changing the Sample Size
 Increasing the sample size causes the error bound to decrease, making the confidence interval narrower.
 Decreasing the sample size causes the error bound to increase, making the confidence interval wider.
Try It
The mean pizza delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is 50 restaurants. Find a 90% confidence interval estimate for the population mean delivery time.
[practicearea rows=”1″][/practicearea]
Show Answer
(34.6041, 37.3958)
Working Backwards to Find the Error Bound or Sample Mean
When we calculate a confidence interval, we find the sample mean, calculate the error bound, and use them to calculate the confidence interval. However, sometimes when we read statistical studies, the study may state the confidence interval only. If we know the confidence interval, we can work backwards to find both the error bound and the sample mean.
Finding the Error Bound
 From the upper value for the interval, subtract the sample mean,
 OR, from the upper value for the interval, subtract the lower value. Then divide the difference by two.
Finding the Sample Mean
 Subtract the error bound from the upper value of the confidence interval,
 OR, average the upper and lower endpoints of the confidence interval.
Example 6
Notice that there are two methods to perform each calculation. You can choose the method that is easier to use with the information you know.
Suppose we know that a confidence interval is (67.18, 68.82) and we want to find the error bound. We may know that the sample mean is 68, or perhaps our source only gave the confidence interval and did not tell us the value of the sample mean.
Calculate the Error Bound:
 If we know that the sample mean is 68: EBM = 68.82 – 68 = 0.82.
 If we don’t know the sample mean: .
Calculate the Sample Mean:
 If we know the error bound: = 68.82 – 0.82 = 68
 If we don’t know the error bound: .
Try It
Suppose we know that a confidence interval is (42.12, 47.88). Find the error bound and the sample mean.
[practicearea rows=”1″][/practicearea]
Show Answer
Sample mean is 45, error bound is 2.88
Calculating the Sample Size n
If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size.
The error bound formula for a population mean when the population standard deviation is known is
The formula for sample size is , found by solving the error bound formula for n.
In this formula, z is , corresponding to the desired confidence level. A researcher planning a study who wants a specified confidence level and error bound can use this formula to calculate the size of the sample needed for the study.
Example 7
The population standard deviation for the age of Foothill College students is 15 years. If we want to be 95% confident that the sample mean age is within two years of the true population mean age of Foothill College students, how many randomly selected Foothill College students must be surveyed?
 From the problem, we know that σ = 15 and EBM = 2.
 z = z_{0.025} = 1.96, because the confidence level is 95%.
 using the sample size equation.
 Use n = 217: Always round the answer UP to the next higher integer to ensure that the sample size is large enough.
Therefore, 217 Foothill College students should be surveyed in order to be 95% confident that we are within two years of the true population mean age of Foothill College students.
Try It
The population standard deviation for the height of high school basketball players is three inches. If we want to be 95% confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed?
[practicearea rows=”1″][/practicearea]
Show Answer
35 students
References
“American Fact Finder.” U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/nav/jsf/pages/searchresults.xhtml?refresh=t (accessed July 2, 2013).
“Disclosure Data Catalog: Candidate Summary Report 2012.” U.S. Federal Election Commission. Available online at http://www.fec.gov/data/index.jsp (accessed July 2, 2013).
“Headcount Enrollment Trends by Student Demographics TenYear Fall Trends to Most Recently Completed Fall.” Foothill De Anza Community College District. Available online at http://research.fhda.edu/factbook/FH_Demo_Trends/FoothillDemographicTrends.htm (accessed September 30,2013).
Kuczmarski, Robert J., Cynthia L. Ogden, Shumei S. Guo, Laurence M. GrummerStrawn, Katherine M. Flegal, Zuguo Mei, Rong Wei, Lester R. Curtin, Alex F. Roche, Clifford L. Johnson. “2000 CDC Growth Charts for the United States: Methods and Development.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/growthcharts/2000growthchartus.pdf (accessed July 2, 2013).
La, Lynn, Kent German. “Cell Phone Radiation Levels.” cnet part of CBX Interactive Inc. Available online at http://reviews.cnet.com/cellphoneradiationlevels/ (accessed July 2, 2013).
“Mean Income in the Past 12 Months (in 2011 InflactionAdjusted Dollars): 2011 American Community Survey 1Year Estimates.” American Fact Finder, U.S. Census Bureau. Available online at http://factfinder2.census.gov/faces/tableservices/jsf/pages/productview.xhtml?pid=ACS_11_1YR_S1902&prodType=table (accessed July 2, 2013).
“Metadata Description of Candidate Summary File.” U.S. Federal Election Commission. Available online at http://www.fec.gov/finance/disclosure/metadata/metadataforcandidatesummary.shtml (accessed July 2, 2013).
“National Health and Nutrition Examination Survey.” Centers for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed July 2, 2013).
Concept Review
In this module, we learned how to calculate the confidence interval for a single population mean where the population standard deviation is known. When estimating a population mean, the margin of error is called the error bound for a population mean ( EBM). A confidence interval has the general form:
(lower bound, upper bound) = (point estimate – EBM, point estimate + EBM)
The calculation of EBM depends on the size of the sample and the level of confidence desired. The confidence level is the percent of all possible samples that can be expected to include the true population parameter. As the confidence level increases, the corresponding EBM increases as well. As the sample size increases, the EBM decreases. By the central limit theorem,
Given a confidence interval, you can work backwards to find the error bound ( EBM) or the sample mean. To find the error bound, find the difference of the upper bound of the interval and the mean. If you do not know the sample mean, you can find the error bound by calculating half the difference of the upper and lower bounds. To find the sample mean given a confidence interval, find the difference of the upper bound and the error bound. If the error bound is unknown, then average the upper and lower bounds of the confidence interval to find the sample mean.
Sometimes researchers know in advance that they want to estimate a population mean within a specific margin of error for a given level of confidence. In that case, solve the EBM formula for n to discover the size of the sample that is needed to achieve this goal:
Formula Review
[latex]\displaystyle\overline{X}{\sim}{N}\left({\mu}_{x}, \frac{{\sigma}}{{\sqrt{n}}}\right)[/latex]. The distribution of sample means is normally distributed with mean equal to the population mean and standard deviation given by the population standard deviation divided by the square root of the sample size.
The general form for a confidence interval for a single population mean, known standard deviation, normal distribution is given by
(lower bound, upper bound) = (point estimate – EBM, point estimate + EBM)
=([latex]\displaystyle\overline{x}[/latex] – EBM, [latex]\displaystyle\overline{x}[/latex]+EBM)
=([latex]\displaystyle\overline{x}{z}_{\frac{{\alpha}}{{\sqrt{n}}}}, \overline{x}+{z}_{\frac{{\alpha}}{{\sqrt{n}}}}[/latex])
EBM = [latex]\displaystyle{z}_{\frac{{\alpha}}{{\sqrt{n}}}}[/latex]= the error bound for the mean, or the margin of error for a single population mean; this formula is used when the population standard deviation is known.
CL = confidence level, or the proportion of confidence intervals created that are expected to contain the true population parameter
α = 1 – CL = the proportion of confidence intervals that will not contain the population
[latex]\displaystyle{z}_{\frac{{\alpha}}{{\sqrt{n}}}}[/latex]= the zscore with the property that the area to the right of the zscore is ∝2 this is the zscore used in the calculation of “EBM where α = 1 – CL.
n = [latex]\displaystyle\frac{{{z}^{2}{\sigma}^{2}}}{{{EBM}^{2}}}[/latex] = the formula used to determine the sample size (n) needed to achieve a desired margin of error at a given level of confidence
General form of a confidence interval
(lower value, upper value) = (point estimate−error bound, point estimate + error bound)
To find the error bound when you know the confidence interval
error bound = upper value−point estimate OR error bound =[latex]\displaystyle\frac{{\text{upper}{value}{lower}{value}}}{{2}}[/latex]
Single Population Mean, Known Standard Deviation, Normal Distribution
Use the Normal Distribution for Means, Population Standard Deviation is Known EBM=[latex]\displaystyle{z}_{\frac{{\alpha}}{{2}}\cdot\frac{{\sigma}}{{\sqrt{n}}}}[/latex]
The confidence interval has the format EBM = ([latex]\displaystyle\overline{x}[/latex] – EBM, [latex]\displaystyle\overline{x}[/latex]+EBM)