4.5: Confidence Intervals

A confidence interval provides a range of values that will likely include the actual mean. For example, if we wanted to know the average age of people who love bananas in Colorado, we might not be able to talk to every person in Colorado. But we could talk to a representative sample and come up with an educated guess or calculation to give us a window of ages. This window is the confidence interval and when certain procedures are followed we can be confident to a certain level that the actual answer is in that window.

There are a few different ways to calculate confidence intervals, based on the information you have. Here is a video on one way. Continue reading for other approaches.

This video references the Inverse Normal Distribution App

and Three Column Notes

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. (EBM is the margin of error, see below)

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]\overline{{x}}[/latex] is the point estimate of the unknown population mean μ. A point estimate can be thought of as an educated guess of the population mean. One way a point estimate is obtained is by collecting a sample. For the scope of this course, you will be provided with the point estimate.

The confidence interval estimate will have the form:

(point estimate – error bound, point estimate + error bound) or, in symbols,
[latex]\displaystyle{(\overline{{x}}-{E}{B}{M},\overline{{x}}+{E}{B}{M})}[/latex]

This means, if you know the mean and the margin of error, you can compute the error bound (sometimes referred to as EMB). For example:

Example 1: Suppose BrandA Batteries has a mean life of 10 hours with a 90% margin of error of 1.5 hours.  Find the confidence interval.

Solution: The point estimate is our 10 hours. Our error bound it 1.5 hours.

[latex]\displaystyle{(\overline{{x}}-{E}{B}{M},\overline{{x}}+{E}{B}{M})}[/latex] =

(10 -1.5, 10+1.5) =

(8.5, 11.5)

This means we are 90% confident that the battery life of BrandA will be between 8.5 and 11.5 hours.

Example 2: Suppose that we are 95% confident that the average life of a ski lift is between 17 and 27 years. What is the margin of error?

Solution: First, find the average of 17 and 27.

[latex]{\frac{17+27}{2}}[/latex]

=22

Next, find the positive difference between one of the end bounds and the averate:

27-22 = 7

Thus, the margin of error with a 95% confidence interval is 7 years.

More about the Margin of Error

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.

Mathematically, α + CL = 1.

Example

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 (or margin of error) for the mean is 2.5.

[latex]\displaystyle\overline{{x}}={7}{\quad\text{and}\quad}{E}{B}{M}={2.5}[/latex]

The confidence interval is (7 – 2.5, 7 + 2.5), and calculating the values gives (4.5, 9.5).

If the confidence level (CL) is 95%, then we say that, “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?

(11.8, 18.2)

 

 

Calculating the Error Bound (EBM)/Margin of Error

The error bound formula for an unknown population mean μ when the population standard deviation σ is known is

  • EBM = ([latex]\displaystyle{z}_{\frac{{\alpha}}{{2}}})(\frac{{\sigma}}{{\sqrt{n}}})[/latex]

Constructing the Confidence Interval

  • The confidence interval estimate has the format

    ([latex]\displaystyle\overline{x}[/latex]–EBM,[latex]\displaystyle\overline{x}[/latex]+EBM).

The graph gives a picture of the entire situation.

CL + [latex]\displaystyle\frac{{\alpha}}{{2}}+\frac{{\alpha}}{{2}}={\text{CL}}+{\alpha}=1[/latex]

Graph of how to construct a confidence interval for CL = 1-alpha

Finding the z-score for the Stated Confidence Level

In this section you will want to use the following FREE apps to find the area above or below a z-score and to find the z-score. With both apps, you will use a mean of 0 and a SD (standard deviation) of 1, unless otherwise indicated.

FREE Inverse Normal Distribution APP (use when you know the area under the curve and want to find the z-score)

FREE app: Normal Distribution APP (Use when you know the z-score and want to find the area above or below a z-score)

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 α/2.

The z-score that has an area to the right of α/2 is denoted by zα2.

For example, when CL = 0.95, α = 0.05 and α/2 = 0.025; we write [latex]\displaystyle{z}_{\frac{{\alpha}}{{2}}})[/latex]=0.025.

The area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 – 0.025 = 0.975.

zα/2=z0.025=1.96, using the the FREE Inverse Normal Distribution APP

Put in a mean of 0, SD (standard deviation) of 1, and area of 0.025 above z to get 1.96

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).”

Watch: Interpreting Confidence Level Example  to help you better understand how to interpret a confidence interval.

Example

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.

Solution:

To find the confidence interval, you need the sample mean, and the EBM.

[latex]\overline{x}={68, }{EBM}=({z}_{\frac{{\alpha}}{{2}}})(\frac{{\sigma}}{{\sqrt{n}}})[/latex]

[latex]\displaystyle\sigma=3[/latex]

n = 6

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 z0.05 is 0.05 and the area to the left of z0.05 is 1 – 0.05 = 0.95

To find [latex]\displaystyle\frac{{{z}_{\alpha}}}{{2}}={z}_{0.05}=1.645[/latex]

You will use:

FREE Inverse Normal Distribution APP to find the z-score

We want to find the z-score when the are to the right of z is 0.05. Mean of 0, SD of 1 (this will always be the case unless otherwise stated), and then select 0.05 as the area above. This will give you a z-score of 1.645. Plug this into the equation:

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).

Interpretation: We 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 Level

Ninety 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.

Answer: (34.1347, 37.8653)

Example

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 SCH-R750 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 SGH-A227 1.13
Huawei M835 Ideos 0.82 Motorola Razr2 V9 0.52 SGH-a107 GoPhone 0.3
Kyocera DuraPlus 0.78 Motorola V195s 1.6 Sony W350a 1.48
Kyocera K127 Marbl 1.25 Nokia 1680 1.39 T-Mobile 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 :

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.

Graph of area under the curve to the right of z 0.01 is 0.01.

You need to find z0.01 having the property that the area under the normal density curve to the right of z0.01 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 z0.01 = 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.

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

[latex]\displaystyle\overline{x}[/latex] = 0.940

[latex]\displaystyle\frac{{\alpha}}{{2}}=\frac{{1-CL}}{{2}}=\frac{{1-0.93}}{{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 = 1.0765

We 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

Suppose we change the original problem in Example 2 by using a 95% confidence level. 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

EBM =([latex]\displaystyle{z}_{\frac{{\alpha}}{{2}}})(\frac{{\sigma}}{{\sqrt{n}}}[/latex])

[latex]\displaystyle{\sigma}={3}[/latex]

n = 36

CL = 0.95 so α = 1 – CL = 1 – 0.95 = 0.05

The area to the right of z0.025 is 0.025 and the area to the left of z0.025 is 1 – 0.025 = 0.975.

Use the free inverse normal app previously listed.

Notice that the EBM is larger for a 95% confidence level in the original problem.

Interpretation

We 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 Level

Ninety-five percent of all confidence intervals constructed in this way contain the true value of the population mean statistics exam score.

Comparing the Results

The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). 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. To be more confident that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider.

Graphs of changing the confidence interval from 09% to 95%

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

Refer back to the pizza-delivery Try It exercise. The population standard deviation is six minutes and the sample mean deliver time is 36 minutes. Use a sample size of 20. Find a 95% confidence interval estimate for the true mean pizza delivery time.

(33.37, 38.63)

Example

Suppose we change the original problem in Example 2 to see what happens to the error bound if the sample size is changed.

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?

  • = 68
  • EBM =
  • σ = 3; The confidence level is 90% (CL=0.90); .

Solution A:

If we increase the sample size n to 100, we decrease the error bound.

Solution B:

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

Refer back to the pizza-delivery Try It exercise. The mean delivery time is 36 minutes and the population standard deviation is six minutes. Assume the sample size is changed to 50 restaurants with the same sample mean. Find a 90% confidence interval estimate for the population mean delivery time.

(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

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.

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

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 = z0.025 = 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?

35 students

 

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 z-score with the property that the area to the right of the z-score is  ∝2 this is the z-score 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)

References

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“Headcount Enrollment Trends by Student Demographics Ten-Year 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. Grummer-Strawn, 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/2000growthchart-us.pdf (accessed July 2, 2013).

La, Lynn, Kent German. “Cell Phone Radiation Levels.” c|net part of CBX Interactive Inc. Available online at http://reviews.cnet.com/cell-phone-radiation-levels/ (accessed July 2, 2013).

“Mean Income in the Past 12 Months (in 2011 Inflaction-Adjusted Dollars): 2011 American Community Survey 1-Year 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).

Online Statistics Education: A Multimedia Course of Study (http://onlinestatbook.com/). Project Leader: David M. Lane, Rice University.