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

Module by: Dr. Barbara Illowsky, Susan Dean

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.

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 1

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 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 variance 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 1.1

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

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 1.2

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 2

  • 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 3

( 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.

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

Solution 1.1

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 1.2

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 = 1.645 z α 2 =1.645

Solution 2

Figure 1
Subfigure 1.1Subfigure 1.2
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.

Glossary

Confidence Level:
The percent expression for the probability that the confidence interval contains the true population parameter. That is, for ex., if CL=90%, then in 90 out of 100 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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