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Introductory Statistics

8.1 A Single Population Mean using the Normal Distribution

Introductory Statistics8.1 A Single Population Mean using the Normal Distribution
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  1. Preface
  2. 1 Sampling and Data
    1. Introduction
    2. 1.1 Definitions of Statistics, Probability, and Key Terms
    3. 1.2 Data, Sampling, and Variation in Data and Sampling
    4. 1.3 Frequency, Frequency Tables, and Levels of Measurement
    5. 1.4 Experimental Design and Ethics
    6. 1.5 Data Collection Experiment
    7. 1.6 Sampling Experiment
    8. Key Terms
    9. Chapter Review
    10. Practice
    11. Homework
    12. Bringing It Together: Homework
    13. References
    14. Solutions
  3. 2 Descriptive Statistics
    1. Introduction
    2. 2.1 Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs
    3. 2.2 Histograms, Frequency Polygons, and Time Series Graphs
    4. 2.3 Measures of the Location of the Data
    5. 2.4 Box Plots
    6. 2.5 Measures of the Center of the Data
    7. 2.6 Skewness and the Mean, Median, and Mode
    8. 2.7 Measures of the Spread of the Data
    9. 2.8 Descriptive Statistics
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. Bringing It Together: Homework
    16. References
    17. Solutions
  4. 3 Probability Topics
    1. Introduction
    2. 3.1 Terminology
    3. 3.2 Independent and Mutually Exclusive Events
    4. 3.3 Two Basic Rules of Probability
    5. 3.4 Contingency Tables
    6. 3.5 Tree and Venn Diagrams
    7. 3.6 Probability Topics
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Bringing It Together: Practice
    13. Homework
    14. Bringing It Together: Homework
    15. References
    16. Solutions
  5. 4 Discrete Random Variables
    1. Introduction
    2. 4.1 Probability Distribution Function (PDF) for a Discrete Random Variable
    3. 4.2 Mean or Expected Value and Standard Deviation
    4. 4.3 Binomial Distribution
    5. 4.4 Geometric Distribution
    6. 4.5 Hypergeometric Distribution
    7. 4.6 Poisson Distribution
    8. 4.7 Discrete Distribution (Playing Card Experiment)
    9. 4.8 Discrete Distribution (Lucky Dice Experiment)
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. References
    16. Solutions
  6. 5 Continuous Random Variables
    1. Introduction
    2. 5.1 Continuous Probability Functions
    3. 5.2 The Uniform Distribution
    4. 5.3 The Exponential Distribution
    5. 5.4 Continuous Distribution
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  7. 6 The Normal Distribution
    1. Introduction
    2. 6.1 The Standard Normal Distribution
    3. 6.2 Using the Normal Distribution
    4. 6.3 Normal Distribution (Lap Times)
    5. 6.4 Normal Distribution (Pinkie Length)
    6. Key Terms
    7. Chapter Review
    8. Formula Review
    9. Practice
    10. Homework
    11. References
    12. Solutions
  8. 7 The Central Limit Theorem
    1. Introduction
    2. 7.1 The Central Limit Theorem for Sample Means (Averages)
    3. 7.2 The Central Limit Theorem for Sums
    4. 7.3 Using the Central Limit Theorem
    5. 7.4 Central Limit Theorem (Pocket Change)
    6. 7.5 Central Limit Theorem (Cookie Recipes)
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. References
    13. Solutions
  9. 8 Confidence Intervals
    1. Introduction
    2. 8.1 A Single Population Mean using the Normal Distribution
    3. 8.2 A Single Population Mean using the Student t Distribution
    4. 8.3 A Population Proportion
    5. 8.4 Confidence Interval (Home Costs)
    6. 8.5 Confidence Interval (Place of Birth)
    7. 8.6 Confidence Interval (Women's Heights)
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. References
    14. Solutions
  10. 9 Hypothesis Testing with One Sample
    1. Introduction
    2. 9.1 Null and Alternative Hypotheses
    3. 9.2 Outcomes and the Type I and Type II Errors
    4. 9.3 Distribution Needed for Hypothesis Testing
    5. 9.4 Rare Events, the Sample, Decision and Conclusion
    6. 9.5 Additional Information and Full Hypothesis Test Examples
    7. 9.6 Hypothesis Testing of a Single Mean and Single Proportion
    8. Key Terms
    9. Chapter Review
    10. Formula Review
    11. Practice
    12. Homework
    13. References
    14. Solutions
  11. 10 Hypothesis Testing with Two Samples
    1. Introduction
    2. 10.1 Two Population Means with Unknown Standard Deviations
    3. 10.2 Two Population Means with Known Standard Deviations
    4. 10.3 Comparing Two Independent Population Proportions
    5. 10.4 Matched or Paired Samples
    6. 10.5 Hypothesis Testing for Two Means and Two Proportions
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. Bringing It Together: Homework
    13. References
    14. Solutions
  12. 11 The Chi-Square Distribution
    1. Introduction
    2. 11.1 Facts About the Chi-Square Distribution
    3. 11.2 Goodness-of-Fit Test
    4. 11.3 Test of Independence
    5. 11.4 Test for Homogeneity
    6. 11.5 Comparison of the Chi-Square Tests
    7. 11.6 Test of a Single Variance
    8. 11.7 Lab 1: Chi-Square Goodness-of-Fit
    9. 11.8 Lab 2: Chi-Square Test of Independence
    10. Key Terms
    11. Chapter Review
    12. Formula Review
    13. Practice
    14. Homework
    15. Bringing It Together: Homework
    16. References
    17. Solutions
  13. 12 Linear Regression and Correlation
    1. Introduction
    2. 12.1 Linear Equations
    3. 12.2 Scatter Plots
    4. 12.3 The Regression Equation
    5. 12.4 Testing the Significance of the Correlation Coefficient
    6. 12.5 Prediction
    7. 12.6 Outliers
    8. 12.7 Regression (Distance from School)
    9. 12.8 Regression (Textbook Cost)
    10. 12.9 Regression (Fuel Efficiency)
    11. Key Terms
    12. Chapter Review
    13. Formula Review
    14. Practice
    15. Homework
    16. Bringing It Together: Homework
    17. References
    18. Solutions
  14. 13 F Distribution and One-Way ANOVA
    1. Introduction
    2. 13.1 One-Way ANOVA
    3. 13.2 The F Distribution and the F-Ratio
    4. 13.3 Facts About the F Distribution
    5. 13.4 Test of Two Variances
    6. 13.5 Lab: One-Way ANOVA
    7. Key Terms
    8. Chapter Review
    9. Formula Review
    10. Practice
    11. Homework
    12. References
    13. Solutions
  15. A | Review Exercises (Ch 3-13)
  16. B | Practice Tests (1-4) and Final Exams
  17. C | Data Sets
  18. D | Group and Partner Projects
  19. E | Solution Sheets
  20. F | Mathematical Phrases, Symbols, and Formulas
  21. G | Notes for the TI-83, 83+, 84, 84+ Calculators
  22. H | Tables
  23. Index

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 x ¯  = 10 x ¯  = 10 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 x ¯ x ¯ as an estimate for μ and we need the margin of error. Here, the margin of error (EBM) is called the error bound for a population mean (abbreviated EBM). The sample mean x ¯ x ¯ 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,( x ¯ EBM, x ¯ +EBM x ¯ EBM, x ¯ +EBM )

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 8.1

  • 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.

x ¯ x ¯ = 7 and EBM = 2.5

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 8.1

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?

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 x ¯ x ¯ = 10, 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.

This is a normal distribution curve. The peak of the curve coincides with the point 10 on the horizontal axis. The points 5 and 15 are labeled on the axis. Vertical lines are drawn from these points to the curve, and the region between the lines is shaded. The shaded region has area equal to 0.90.
Figure 8.2

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 z-score 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 σ n σ n . The fraction σ n σ n , 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:
  • X ¯ X ¯ is normally distributed, that is, X ¯ X ¯ ~ N ( μ X , σ n ) ( μ X , σ n ) .
  • When the population standard deviation σ is known, we use a normal distribution to calculate the error bound.

Calculating the Confidence Interval

To 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:

  • Calculate the sample mean x ¯ x ¯ from the sample data. Remember, in this section we already know the population standard deviation σ.
  • Find the z-score that corresponds to the confidence level.
  • Calculate the error bound EBM.
  • Construct the confidence interval.
  • Write a sentence that interprets the estimate in the context of the situation in the problem. (Explain what the confidence interval means, in the words of the problem.)

We will first examine each step in more detail, and then illustrate the process with some examples.

Finding the z-score 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 α 2 α 2 .

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

For example, when CL = 0.95, α = 0.05 and α 2 α 2 = 0.025; we write z α 2 z α 2 = z0.025.

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.

z α 2  =  z 0.025  = 1.96 z α 2  =  z 0.025  = 1.96 , using a calculator, computer or a standard normal probability table.

Using the TI-83, 83+, 84, 84+ Calculator

invNorm(0.975, 0, 1) = 1.96

Note

Remember to use the area to the LEFT of z α 2 z α 2 ; 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 = ( z α 2 )( σ n ) ( z α 2 )( σ n )

Constructing the Confidence Interval

  • The confidence interval estimate has the format ( x ¯ EBM, x ¯ +EBM) ( x ¯ EBM, x ¯ +EBM) .

The graph gives a picture of the entire situation.

CL + α 2 α 2 + α 2 α 2 = CL + α = 1.

This is a normal distribution curve. The peak of the curve coincides with the point x-bar on the horizontal axis. The points x-bar - EBM and x-bar + EBM are labeled on the axis. Vertical lines are drawn from these points to the curve, and the region between the lines is shaded. The shaded region has area equal to 1 - a and represents the confidence level. Each unshaded tail has area a/2.
Figure 8.3

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 8.2

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.

Solution 8.2
  • You can use technology to calculate the confidence interval directly.
  • The first solution is shown step-by-step (Solution A).
  • The second solution uses the TI-83, 83+, and 84+ calculators (Solution B).

Solution ATo find the confidence interval, you need the sample mean, x ¯ x ¯ , and the EBM.

  • x ¯ x ¯ = 68
  • EBM = ( z α 2 ) ( z α 2 ) ( σ n ) ( σ n )
  • σ = 3; n = 36; The confidence level is 90% (CL = 0.90)

CL = 0.90 so α = 1 – CL = 1 – 0.90 = 0.10

α 2 α 2 = 0.05 z α 2 = z 0.05 z α 2 = z 0.05

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.

z α 2  =  z 0.05  = 1.645 z α 2  =  z 0.05  = 1.645

using invNorm(0.95, 0, 1) on the TI-83,83+, and 84+ calculators. 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) ( 3 36 ) ( 3 36 ) = 0.8225

x ¯ x ¯ - EBM = 68 - 0.8225 = 67.1775

x ¯ x ¯ + EBM = 68 + 0.8225 = 68.8225

The 90% confidence interval is (67.1775, 68.8225).

Solution 8.2

Solution B

Using the TI-83, 83+, 84, 84+ Calculator

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 x ¯ x ¯ , 36 for n, and .90 for C-level.
Arrow down to Calculate and press ENTER.
The confidence interval is (to three decimal places)(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 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 8.2

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.

Example 8.3

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. Table 8.1 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
Table 8.1

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 8.3

Solution ATo find the confidence interval, start by finding the point estimate: the sample mean.

x ¯ =1.024 x ¯ =1.024

Next, find the EBM. Because you are creating a 98% confidence interval, CL = 0.98.

This is a normal distribution curve. The point z0.01 is labeled at the right edge of the curve and the region to the right of this point is shaded. The area of this shaded region equals 0.01. The unshaded area equals 0.99.
Figure 8.4

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=( z 0.01 ) σ n =(2.326) 0.337 30 =0.1431 EBM=( z 0.01 ) σ n =(2.326) 0.337 30 =0.1431

To find the 98% confidence interval, find   x ¯ ±EBM   x ¯ ±EBM .

x ¯ x ¯ EBM = 1.024 – 0.1431 = 0.8809

x ¯ x ¯ 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 8.3

Solution B

Using the TI-83, 83+, 84, 84+ Calculator

  • 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
    • x ¯ :1.024 x ¯ :1.024
    • n: 30
    • C-level: 0.98
  • Arrow down to Calculate and press ENTER.
  • The confidence interval is (to three decimal places) (0.881, 1.167).

Try It 8.3

Table 8.2 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
Table 8.2

Notice the difference in the confidence intervals calculated in Example 8.3 and the following Try It exercise. 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 8.4

Suppose we change the original problem in Example 8.2 by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score.

Solution 8.4

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

  • x ¯ x ¯ = 68
  • EBM = ( z α 2 )( σ n ) ( z α 2 )( σ n )
  • σ = 3; n = 36; The confidence level is 95% (CL = 0.95).

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

α 2 =0.025 z α 2 = z 0.025 α 2 =0.025 z α 2 = z 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.

z α 2 = z 0.025 =1.96 z α 2 = z 0.025 =1.96

when using invnorm(0.975,0,1) on the TI-83, 83+, or 84+ calculators. (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 = (1.96) ( 3 36 ) ( 3 36 ) = 0.98

x ¯ x ¯ EBM = 68 – 0.98 = 67.02

x ¯ x ¯ + EBM = 68 + 0.98 = 68.98

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

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 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.

Part (a) shows a normal distribution curve. A central region with area equal to 0.90 is shaded. Each unshaded tail of the curve has area equal to 0.05. Part (b) shows a normal distribution curve. A central region with area equal to 0.95 is shaded. Each unshaded tail of the curve has area equal to 0.025.
Figure 8.5
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 8.4

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.

Example 8.5

Suppose we change the original problem in Example 8.2 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?

  • x ¯ x ¯ = 68
  • EBM = ( z α 2 )( σ n ) ( z α 2 )( σ n )
  • σ = 3; The confidence level is 90% (CL=0.90); z α 2 z α 2 = z0.05 = 1.645.
Solution 8.5

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

When n = 100: EBM = ( z α 2 )( σ n ) ( z α 2 )( σ n ) = (1.645) ( 3 100 ) ( 3 100 ) = 0.4935.

Solution 8.5

Solution BIf we decrease the sample size n to 25, we increase the error bound.

When n = 25: EBM = ( z α 2 )( σ n ) ( z α 2 )( σ n ) = (1.645) ( 3 25 ) ( 3 25 ) = 0.987.

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 8.5

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.

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.

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.

Example 8.6

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: EBM = (68.8267.18) 2 (68.8267.18) 2 = 0.82.
Calculate the Sample Mean:
  • If we know the error bound: x ¯ x ¯ = 68.82 – 0.82 = 68
  • If we don't know the error bound: x ¯ x ¯ = (67.18+68.82) 2 (67.18+68.82) 2 = 68.
Try It 8.6

Suppose we know that a confidence interval is (42.12, 47.88). Find the error bound and the sample mean.

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
EBM = ( z α 2 )( σ n ) ( z α 2 )( σ n ) .

The formula for sample size is n = z 2 σ 2 EB M 2 z 2 σ 2 EB M 2 , found by solving the error bound formula for n.

In this formula, z is z α 2 z α 2 , 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 8.7

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 = z0.025 = 1.96, because the confidence level is 95%.
  • n = z 2 σ 2 EB M 2 z 2 σ 2 EB M 2 = ( 1.96 ) 2 ( 15 ) 2 2 2 ( 1.96 ) 2 ( 15 ) 2 2 2 = 216.09 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 8.7

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?

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