Contents

A Single Population Mean Using the Normal Distribution

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 $\bar{x} = 10 \bar{x} = 10$ and we have constructed the 90 percent confidence interval (5, 15), where the margin of error = 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 $\bar{x} \bar{x}$ as an estimate for μ, and we need the margin of error. The margin of error for the population mean is called the error bound for a population mean (EBM). The sample mean, $\bar{x}, \bar{x},$ is the point estimate of the unknown population mean, μ.

The confidence interval (CI) estimate will have the form:

(point estimate – error bound, point estimate + error bound) or, in symbols, ( $\bar{x} - E B M, \bar{x} + E B M \bar{x} - E B M, \bar{x} + E B M$ ).

The margin of error (EBM) depends on the confidence level (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 percentage of confidence intervals that contain the true population parameter when repeated samples are taken. Most often, the person constructing the confidence interval will choose a confidence level of 90 percent or higher, because that person wants to be reasonably certain of his or her conclusions.

Another probability, which is called alpha $(α) (α)$ is related to the confidence level, CL. Alpha is the probability that the confidence interval does not contain the unknown population parameter. Mathematically, alpha can be computed as $α = 1 - C L α = 1 - C L$ .

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.

\bar{x} and E B M = 2 .5 . \bar{x} and E B M = 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 is 95 percent, then we say, "We estimate with 95 percent 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 $\bar{x} \bar{x}$ = 10, and we have constructed the 90 percent confidence interval (5, 15) where EBM = 5.

To get a 90 percent confidence interval, we must include the central 90 percent of the probability of the normal distribution. If we include the central 90 percent, we leave out a total of α = 10 percent in both tails, or 5 percent 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

The critical value 1.645 is the z-score in 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. To capture the central 90 percent, we must go out 1.645 standard deviations on either side of the calculated sample mean. The critical value will change depending on the confidence level of the interval.

It is important that the standard deviation used 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 $\frac{σ}{\sqrt{n}} \frac{σ}{\sqrt{n}}$ . The fraction $\frac{σ}{\sqrt{n}} \frac{σ}{\sqrt{n}}$ 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, the following statements apply:

$\bar{X} \bar{X}$ is normally distributed, that is, $\bar{X} \bar{X}$ ~ N $(μ_{X}, \frac{σ}{\sqrt{n}}) (μ_{X}, \frac{σ}{\sqrt{n}})$ .
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 as follows:

Calculate the sample mean, $\bar{x}, \bar{x},$ 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.
If we denote the critical z-score by $z_{\frac{a}{2}} z_{\frac{a}{2}}$ , and the sample size by n, then the formula for the confidence interval with confidence level $C l = 1 - α C l = 1 - α$ , is given by $(\bar{x} - z_{\frac{a}{2}} \times \frac{σ}{\sqrt{n}}, \bar{x} + z_{\frac{a}{2}} \times \frac{σ}{\sqrt{n}}) . (\bar{x} - z_{\frac{a}{2}} \times \frac{σ}{\sqrt{n}}, \bar{x} + z_{\frac{a}{2}} \times \frac{σ}{\sqrt{n}}) .$
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 $\frac{α}{2} \frac{α}{2}$ .

The z-score that has an area to the right of $\frac{α}{2} \frac{α}{2}$ is denoted by $z_{\frac{α}{2}} z_{\frac{α}{2}}$ .

For example, when CL = 0.95, α = 0.05, and $\frac{α}{2} \frac{α}{2}$ = 0.025, we write $z_{\frac{α}{2}} z_{\frac{α}{2}}$ = $z z$ _0.025.

The area to the right of z_0.025 is 0.025 and the area to the left of z_0.025 is 1 – 0.025 = 0.975.

$z_{\frac{α}{2}} = z_{0. 025} = 1 .96 z_{\frac{α}{2}} = z_{0. 025} = 1 .96$ , using a calculator, computer, or standard normal probability table.

Normal table (see appendices) shows that the probability for 0 to 1.96 is 0.47500, and so the probability to the right tail of the critical value 1.96 is 0.5 – 0.475 = 0.025

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

invNorm(0.975, 0, 1) = 1.96. In this command, the value 0.975 is the total area to the left of the critical value that we are looking to calculate. The parameters 0 and 1 are the mean value and the standard deviation of the standard normal distribution Z.

Note

Remember to use the area to the LEFT of $z_{\frac{α}{2}} z_{\frac{α}{2}}$ . In this chapter, the last two inputs in the invNorm command are 0, 1, because you are using a standard normal distribution Z with mean 0 and standard deviation 1.

Calculating the Margin of Error EBM

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

Margin of error = $(z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}}) . (z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}}) .$

Constructing the Confidence Interval

The confidence interval estimate has the format sample mean plus or minus the margin of error.

The graph gives a picture of the entire situation

CL + $\frac{α}{2} \frac{α}{2}$ + $\frac{α}{2} \frac{α}{2}$ = 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 which population parameter is being estimated (here, a population mean), and state the confidence interval (both endpoints): "We estimate with ___percent 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).

Problem

Find a 90 percent confidence interval for the true (population) mean of statistics exam scores.

Show solution

Solution

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 A

To find the confidence interval, you need the sample mean,

\bar{x} \bar{x}

, and the EBM.

$\bar{x} = 68 \bar{x} = 68$
8.2

$E B M = (z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}}) E B M = (z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}})$
$σ = 3; n = 36; σ = 3;;$
The confidence level is 90 percent (CL = 0.90).

C L = 0 .90, so α = 1 – C L = 1 – 0 .90 = 0 .10 . C L = 0 .90, so α = 1 – C L = 1 – 0 .90 = 0 .10 .

\frac{α}{2} = 0.05, z_{\frac{α}{2}} = z_{0.05} \frac{α}{2} = 0.05, z_{\frac{α}{2}} = z_{0.05}

The area to the right of z_0.05 is 0.05 and the area to the left of z_0.05 is 1 – 0.05 = 0.95.

z_{\frac{α}{2}} = z_{0.05} = 1 .645 z_{\frac{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) $(\frac{3}{\sqrt{36}}) (\frac{3}{\sqrt{36}})$ = 0.8225

$\bar{x} \bar{x}$ – EBM = 68 – 0.8225 = 67.1775

$\bar{x} \bar{x}$ + EBM = 68 + 0.8225 = 68.8225

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

Show solution

Solution

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 3 for σ, 68 for $\bar{x} \bar{x}$ , 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).

Interpretation

We estimate with 90 percent confidence that the true population mean exam score for all statistics students is between 67.18 and 68.82.

Explanation of 90 percent 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 6 minutes. A random sample of 28 pizza delivery restaurants is taken and has a sample mean delivery time of 36 min.

Find a 90 percent 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. For certification from the Federal Communications Commission 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 of a random cell phone company.

Phone Model #	SAR	Phone Model #	SAR	Phone Model #	SAR
800	1.11	1800	1.36	2800	0.74
900	1.48	1900	1.34	2900	0.5
1000	1.43	2000	1.18	3000	0.4
1100	1.3	2100	1.3	3100	0.867
1200	1.09	2200	1.26	3200	0.68
1300	0.455	2300	1.29	3300	0.51
1400	1.41	2400	0.36	3400	1.13
1500	0.82	2500	0.52	3500	0.3
1600	0.78	2600	1.6	3600	1.48
1700	1.25	2700	1.39	3700	1.38

Table 8.1

Problem

Find a 98 percent confidence interval for the true (population) mean of the SARs for cell phones. Assume that the population standard deviation is σ = 0.337.

Show solution

Solution

Solution A

To find the confidence interval, start by finding the point estimate: the sample mean,

\bar{x} = 1.024 . \bar{x} = 1.024 .

This is calculated by adding the specific absorption rate for the 30 cell phones in the sample, and dividing the result by 30.

Next, find the EBM. Because you are creating a 98 percent 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 z_0.01, having the property that the area under the normal density curve to the right of z_0.01 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 z_0.01 = 2.326.

E B M = (z_{0.01}) \frac{σ}{\sqrt{n}} = (2.326) \frac{0.337}{\sqrt{30}} = 0.1431 E B M = (z_{0.01}) \frac{σ}{\sqrt{n}} = (2.326) \frac{0.337}{\sqrt{30}} = 0.1431

To find the 98 percent confidence interval, find $\bar{x} \pm E B M \bar{x} \pm E B M$ .

\bar{x} – E B M = 1 .024 – 0 .1431 = 0 .8809 \bar{x} – E B M = 1 .024 – 0 .1431 = 0 .8809

\bar{x} + E B M = 1 .024 + 0 .1431 = 1 .1671 \bar{x} + E B M = 1 .024 + 0 .1431 = 1 .1671

We estimate with 98 percent 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.

Show solution

Solution

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
- $\bar{x} : 1.024 \bar{x} : 1.024$
- 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 these data to calculate a 93 percent 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
450	1.48	1450	1.53
550	0.8	1550	0.68
650	1.15	1650	1.4
750	1.36	1750	1.24
850	0.77	1850	0.57
950	0.462	1950	0.2
1050	1.36	2050	0.51
1150	1.39	2150	0.3
1250	1.3	2250	0.73
1350	0.7	2350	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 are calculated from different samples, the samples are different sizes, and the intervals are 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

Problem

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

Show solution

Solution

To find the confidence interval, you need the sample mean, $\bar{x} \bar{x}$ , and the EBM.

$\bar{x} = 68 \bar{x} = 68$
$E B M = (z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}}) E B M = (z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}})$
$σ = 3; n = 36 σ = 3;$
The confidence level is 95 percent (CL = 0.95).

C L = 0 .95, so α = 1 – C L = 1 – 0 .95 = 0 .05 . C L = 0 .95, so α = 1 – C L = 1 – 0 .95 = 0 .05 .

\frac{α}{2} = 0.025 z_{\frac{α}{2}} = z_{0.025} \frac{α}{2} = 0.025 z_{\frac{α}{2}} = z_{0.025}

The area to the right of $z z$ _0.025 is 0.025, and the area to the left of $z z$ _0.025 is 1 – 0.025 = 0.975.

z_{\frac{α}{2}} = z_{0.025} = 1.96, z_{\frac{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.)

E B M = (1 .96) (\frac{3}{\sqrt{36}}) = 0 .98 E B M = (1 .96) (\frac{3}{\sqrt{36}}) = 0 .98

\bar{x} – E B M = 68 – 0 .98 = 67 .02 \bar{x} – E B M = 68 – 0 .98 = 67 .02

\bar{x} + E B M = 68 + 0 .98 = 68 .98 \bar{x} + E B M = 68 + 0 .98 = 68 .98

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

Interpretation

We estimate with 95 percent confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98.

Explanation of 95 percent Confidence Level

95 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 percent confidence interval is (67.18, 68.82). The 95 percent confidence interval is (67.02, 68.98). The 95 percent 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 percent confidence interval is wider. For more certainty 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 percent 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.

Problem

Leave everything the same except the sample size. Use the original 90 percent 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?

$\bar{x} \bar{x}$ = 68
EBM = $(z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}}) (z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}})$
σ = 3, the confidence level is 90 percent (CL = 0.90), $z_{\frac{α}{2}} z_{\frac{α}{2}}$ = z_0.05 = 1.645.

Show solution

Solution

Solution A

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

When n = 100, EBM = $(z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}}) (z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}})$ = (1.645) $(\frac{3}{\sqrt{100}}) (\frac{3}{\sqrt{100}})$ = 0.4935.

Show solution

Solution

Solution B

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

When n = 25, EBM = $(z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}}) (z_{\frac{α}{2}}) (\frac{σ}{\sqrt{n}})$ = (1.645) $(\frac{3}{\sqrt{25}}) (\frac{3}{\sqrt{25}})$ = 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 percent confidence interval estimate for the population mean delivery time.

Working Backward 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 backward 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 2.

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 gives the confidence interval and does 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 do not know the sample mean, EBM = $\frac{(68.82 - 67.18)}{2} \frac{(68.82 - 67.18)}{2}$ = 0.82. The margin of error is the quantity that we add and subtract from the sample mean to obtain the confidence interval. Therefore, the margin of error is half of the length of the interval.

Calculate the sample mean:

If we know the error bound, $\bar{x} \bar{x}$ = 68.82 – 0.82 = 68.
If we do not know the error bound, $\bar{x} \bar{x}$ = $\frac{(67.18 + 68.82)}{2} \frac{(67.18 + 68.82)}{2}$ = 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. In this situation, we are given the desired margin of error, EBM, and we need to compute the sample size n.

The formula for sample size is n = $\frac{z^{2} σ^{2}}{E B M^{2}} \frac{z^{2} σ^{2}}{E B M^{2}}$ , found by solving the error bound formula for n. Always round up the value of n to the closest integer.

In this formula, z is the critical value $z_{\frac{α}{2}} z_{\frac{α}{2}}$ , 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 percent 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 = z_0.025 = 1.96, because the confidence level is 95 percent.

n = $\frac{z^{2} σ^{2}}{E B M^{2}} \frac{z^{2} σ^{2}}{E B M^{2}}$ = $\frac{{(1.96)}^{2} {(15)}^{2}}{2^{2}} \frac{{(1.96)}^{2} {(15)}^{2}}{2^{2}}$ = 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 percent 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 percent confident that the sample mean height is within one inch of the true population mean height, how many randomly selected students must be surveyed?