A confidence interval for a population mean, when the population standard deviation is known, is based on the conclusion of the Central Limit Theorem that the sampling distribution of 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.
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 ¯ – E B M , x ¯ + E B M x ¯ – E B M , x ¯ + E B M )
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 their 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.
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."
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.
μ = X ¯ ± Z α σ n = 2 ± 1 . 96 ( 0 . 1 ) = 2 ± 0 . 196 1 . 804 ≤ μ ≤ 2 . 196 μ = X ¯ ± Z α σ n = 2 ± 1 . 96 ( 0 . 1 ) = 2 ± 0 . 196 1 . 804 ≤ μ ≤ 2 . 196
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: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:
We will first examine each step in more detail, and then illustrate the process with some examples.
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.
invNorm (0.975, 0, 1) = 1.96
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).
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)."
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.
The solution is shown step by step:
The formula for a confidence interval for an unknown population mean assuming we know the population standard deviation is:
X ¯ - Z α σ n ≤ μ ≤ X ¯ + Z α ( σ n ) X ¯ - Z α σ n ≤ μ ≤ X ¯ + Z α ( σ n )For a 90% confidence interval, visualize an area of 0.90 centered under the normal curve (See Figure 8.3). The remaining area for the two tails of the normal distribution is then 0.10, which indicates that the area in the left tail is one-half of 0.10, which is 0.05. The corresponding z-score that cuts off an area of 0.05 in the left tail is 1.645.
In this example we are given that the population standard deviation σ = 3 σ = 3 .
We are also given that the sample size n = 36 and the sample mean X ¯ = 68 X ¯ = 68 .
Substituting these values in the confidence interval formula results in the following:
X ¯ - Z α σ n ≤ μ ≤ X ¯ + Z α ( σ n ) X ¯ - Z α σ n ≤ μ ≤ X ¯ + Z α ( σ n ) 68 - 1 . 645 3 36 ≤ μ ≤ 68 + 1 . 645 3 36 68 - 1 . 645 3 36 ≤ μ ≤ 68 + 1 . 645 3 36 68 - 0 . 8225 ≤ μ ≤ 68 + 0 . 8225 68 - 0 . 8225 ≤ μ ≤ 68 + 0 . 8225 67 . 1775 ≤ μ ≤ 68 . 8225 67 . 1775 ≤ μ ≤ 68 . 8225We estimate with 90% confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82.
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).
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.
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.
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.
| SAR Data for a sample of 30 cell phones | ||
|---|---|---|
| 1.11 | 1.36 | 0.74 |
| 1.48 | 1.34 | 0.5 |
| 1.43 | 1.18 | 0.4 |
| 1.3 | 1.3 | 0.867 |
| 1.09 | 1.26 | 0.68 |
| 0.455 | 1.29 | 0.51 |
| 1.41 | 0.36 | 1.13 |
| 0.82 | 0.52 | 0.3 |
| 0.78 | 1.6 | 1.48 |
| 1.25 | 1.39 | 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.
To 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.
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.
E B M = ( z 0.01 ) σ n = ( 2.326 ) 0.337 30 = 0.1431 E B M = ( z 0.01 ) σ n = ( 2.326 ) 0.337 30 = 0.1431
To find the 98% confidence interval, find x ¯ ± E B M x ¯ ± E B M .
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.
