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Confidence Intervals: Confidence Interval, Single Population Mean, Population Standard Deviation Known, Normal

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

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To construct a confidence interval for a single unknown population mean μμ where the population standard deviation is known, we need x¯ x as an estimate for μμ and a margin of error. Here, the margin of error is called the error bound for a population mean (abbreviated EBM). The margin of error depends on the confidence level (abbreviated CL). The confidence level is the probability that the confidence interval produced contains the true population parameter. Most often, it is the choice of the person constructing the confidence interval to choose a confidence level of 90% or higher because he wants to be reasonably certain of his conclusions.

There is another probability called alpha (αα). αα is the probability that the sample produced a point estimate that is not within the appropriate margin of error of the unknown population parameter.

Example 1

Suppose the sample mean is 7 and the error bound for the mean is 2.5.

Problem 1

x¯ = x = _______ and EBM = EBM= _______.

Solution

x¯ = x = 7 and EBM = EBM= 2.5.

The confidence interval is ( 7 - 2.5 , 7 + 2.5 ) (7-2.5,7+2.5).

If the confidence level (CL) is 95%, then we say we are 95% confident that the true population mean is between 4.5 and 9.5.

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 we have constructed the 90% confidence interval (5, 15) where x¯ = 10 x =10 and EBM = 5 EBM=5. To get a 90% confidence interval, we must include the central 90% of the sample means. If we include the central 90%, we leave out a total of 10 % or 5% in each tail of the normal distribution. To capture the central 90% of the sample means, we must go out 1.645 standard deviations on either side of the calculated sample mean. The 1.645 is the z-score from a standard normal table that has area to the right equal to 0.05 (5% area in the right tail). The graph shows the general situation.

Normal distribution curve with values of 5 and 15 on the x-axis. Vertical upward lines from points 5 and 15 extend to the curve. The confidence interval area between these two points is equal to 0.90.

To summarize, resulting from the Central Limit Theorem,

X¯ X is normally distributed, that is, X¯ X ~ N ( μ X , σ X n ) N( μ X , σ X n )

Since the population standard deviation, σ, is known, we use a normal curve.

The confidence level, CLCL, is CL = 1 - α CL=1-α. Each of the tails contains an area equal to α 2 α 2 .

The z-score that has area to the right of α2 α2 is denoted by z α 2 z α 2 .

For example, if α 2 = 0.025 α 2=0.025, then area to the right = 0.025 =0.025 and area to the left = 1 - 0.025 = 0.975 =1-0.025=0.975 and z α 2 = z 0.025 = 1.96 z α 2 = z 0.025 =1.96 using a calculator, computer or table. Using the TI83+ or 84 calculator function, invNorm, you can verify this result. invNorm(.975,0,1)=1.96(.975,0,1)=1.96.

The error bound formula for a single population mean when the population standard deviation is known is

EBM = z α 2 σ n EBM= z α 2 σ n

The confidence interval 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 = CL + α = 1 CL+ α 2 + α 2 =CL+α=1.

Normal distribution curve displaying the confidence interval formulas and corresponding area formulas.

Example 2

Problem 1

Suppose scores on exams in statistics are normally distributed with an unknown population mean and a population standard deviation of 3 points. A sample of 36 scores is taken and gives a sample mean (sample average score) of 68. Find a 90% confidence interval for the true (population) mean of statistics exam scores.

  • The first solution is step-by-step.
  • The second solution uses the TI-83+ and TI-84 calculators.

Solution A

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

  • a. x¯ = 68 x =68
  • b. EBM = z α 2 ( σ n ) EBM= z α 2 ( σ n )
  • c. σ = 3 σ=3
  • d. n = 36 n=36

CL = 0.90 CL = 0.90 so a = 1 - CL = 1 - 0.90 = 0.10 a=1-CL=1-0.90=0.10

Since α 2 = 0.05 α 2 =0.05, then z α 2 = z .05 = 1.645 z α 2 = z .05 =1.645

from a calculator, computer or standard normal table. For the table, see the Table of Contents 15. Tables.

Therefore, EBM = 1.645 ( 3 36 ) = 0.8225 EBM=1.645( 3 36 )=0.8225

This gives x¯ - EBM = 68 - 0.8225 = 67.18 x -EBM=68-0.8225=67.18

and x¯ + EBM = 68 + 0.8225 = 68.82 x +EBM=68+0.8225=68.82

The 90% confidence interval is (67.18, 68.82).

Solution B

The TI-83+ and TI-84 caculators simplify this whole procedure. Press STAT and arrow over to TESTS. Arrow down to 7:ZInterval. Press ENTER. Arrow to Stats and press ENTER. Arrow down and enter 3 for σσ, 68 for x¯ x , 36 for nn, and .90 for C-level. Arrow down to Calculate and press ENTER. The confidence interval is (to 3 decimal places) (67.178, 68.822).

We can find the error bound from the confidence interval. From the upper value, subtract the sample mean or subtract the lower value from the upper value and divide by two. The result is the error bound for the mean (EBM).

EBM = 68.822 - 68 = 0.822 EBM=68.822-68=0.822 or EBM = ( 68.822 67.178 ) 2 = 0.822 EBM= ( 68.822 67.178 ) 2 =0.822

We can interpret the confidence interval in two ways:

  1. We are 90% confident that the true population mean for statistics exam scores is between 67.178 and 68.822.
  2. Ninety percent of all confidence intervals constructed in this way contain the true average 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.

Now for the same problem, find a 95% confidence interval for the true (population) mean of scores. Draw the graph. The sample mean, standard deviation, and sample size are:

Problem 2

  • a. x¯ = x =
  • b. σ = σ=
  • c. n = n=

Solution

  • a. x¯  =  68 x  =  68
  • b. σ  =  3 σ = 3
  • c. n  =  36 n = 36

The confidence level is CL = 0.95CL= 0.95. Graph:

The confidence interval is (use technology)

Problem 3

( x¯ - EBM , x¯ + EBM ) = (_______, _______) ( x -EBM, x +EBM)=(_______, _______). The error bound EBM =EBM = _______.

Solution

( x¯ - EBM , x¯ + EBM ) = (67.02 , 68.98) ( x -EBM, x +EBM)=(67.02 , 68.98). The error bound EBM =EBM = 0.98.

We can say that we are 95 % confident that the true population mean for statistics exam scores is between 67.02 and 68.98 and that 95% of all confidence intervals constructed in this way contain the true average statistics exam score.

Example 3

Suppose we change the previous problem.

Problem 1

Leave everything the same except the sample size. For this problem, we can examine the impact of changing nn to 100100 or changing nn to 2525.

  • a. x¯ = 68 x =68
  • b. σ = 3 σ=3
  • c. z α 2 = 1.645 z α 2 =1.645

Solution A

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

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

Solution B

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

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

Problem 2

Leave everything the same except for the confidence level. We increase the confidence level from 0.90 to 0.95.

  • a. x¯ = 68 x =68
  • b. σ = 3 σ=3
  • c. z α 2 z α 2 changes from 1.645 1.645 to 1.961.96.

Solution

Figure 1
(a) (b)
Normal distribution curve with 0.90 confidence interval area blocked off and corresponding residual areas.Normal distribution curve with 0.95 confidence interval area blocked off and corresponding residual areas.

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.

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 EBM= z α 2 σ n
  • Solving for nn gives you an equation for the sample size.
  • n= z α 2 2 σ 2 EBM 2 n= z α 2 2 σ 2 EBM 2

Example 4

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 2 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=15σ=15
  • EBM=2EBM=2
  • z α 2 z α 2 = 1.96 =1.96 because the confidence level is 95%95%.

Using the equation for the sample size, we have

  • n= z α 2 2 σ 2 EBM 2 n= z α 2 2 σ 2 EBM 2
  • n= 1.96 2 15 2 2 2 n= 1.96 2 15 2 2 2
  • n=216.09n=216.09
  • Round the answer to the next higher value to ensure that the sample size is as large as it should be. Therefore, 217217 Foothill College students should be surveyed for us to be 95%95% confident that we are within 22 years of the true population age of Foothill College students.

Note:

In reality, we usually do not know the population standard deviation so we estimate it with the sample standard deviation or use some other way of estimating it (for example, some statisticians use the results of some other earlier study as the estimate).

Glossary

Confidence Level (CL):
The percent expression for the probability that the confidence interval contains the true population parameter. For example, if the CL=90%CL=90%, then in 9090 out of 100100 samples the interval estimate will enclose the true population parameter.
Error Bound for a Population Mean (EBM):
The margin of error. Depends on the confidence level, sample size, and known or estimated population standard deviation.

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