Introduction

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

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 $(\alpha )$, 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 $\alpha =1-CL$.

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 $\overline{x}$ = 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.

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 $\frac{\sigma }{\sqrt{n}}$. The fraction $\frac{\sigma }{\sqrt{n}}$ 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.

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 + $\frac{\alpha }{2}$ + $\frac{\alpha }{2}$ = CL + α = 1.

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).

Find a 90 percent 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, $\overline{x}$, and the EBM.

$$\overline{x}\text{ = 68}$$

$$EBM\text{ = }(z_{\frac{\alpha }{2}})(\frac{\sigma }{\sqrt{n}})$$

σ = 3; n = 36;

The confidence level is 90 percent (CL = 0.90).

$$CL\text{ = 0}\text{.90, so }\alpha \text{ = 1 – }CL\text{ = 1 – 0}\text{.90 = 0}\text{.10}\text{.}$$

$$\frac{\alpha }{2}=0.05;z_{\frac{\alpha }{2}}=z_{0.05}$$

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.

$$z_{\frac{\alpha }{2}}\text{ = }{z}_{0.05}\text{ = 1}\text{.645}$$

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)$\left(\frac{3}{\sqrt{36}}\right)$ = 0.8225

$\overline{x}$ – EBM = 6 – 0.8225 = 67.1775

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

The 90 percent 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 3 for σ, 68 for $\overline{x}$, 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 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 minutes.

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

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.

 

Solution 8.3

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

$$\overline{x}=1.024\text{.}$$

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.

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.

$$EBM=(z_{0.01})\frac{\sigma }{\sqrt{n}}=(2.326)\frac{0.337}{\sqrt{30}}=0.1431$$

To find the 98 percent confidence interval, find $\overline{x}\pm EBM$.

$$\overline{x}–EBM=1.024–0.1431=0.8809$$

$$\overline{x}+EBM=\text{1.024<mo>+</mo><mn>0.1431</mn><mo>=</mo><mn>1.1671</mn>}$$

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.

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
  • $\overline{x}:1.024$
  • 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

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.

Solution 8.4

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

$$\overline{x}\text{ = 68}$$

$$EBM\text{ = }(z_{\frac{\alpha }{2}})(\frac{\sigma }{\sqrt{n}})$$

$$\sigma \text{ = 3; }n\text{ = 36}$$

The confidence level is 95 percent (CL = 0.95).

$$CL\text{ = 0}\text{.95, so }\alpha \text{ = 1 – }CL\text{ = 1 – 0}\text{.95 = 0}\text{.05}\text{.}$$

$$\frac{\alpha }{2}=0.025z_{\frac{\alpha }{2}}=z_{0.025}$$

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.

$$z_{\frac{\alpha }{2}}=z_{0.025}=1.96\text{,}$$

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.)

$$EBM\text{ = (1}\text{.96)}\left(\frac{3}{\sqrt{36}}\right)\text{ = 0}\text{.98}$$

$$\overline{x}\text{ – }EBM\text{ = 68 – 0}\text{.98 = 67}\text{.02}$$

$$\overline{x}\text{ + }EBM\text{ = 68 + 0}\text{.98 = 68}\text{.98}$$

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

InterpretationWe 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 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 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.

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.

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.

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.

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 = $\frac{z^2{\sigma }^2}{EB{M}^2}$, 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 $z_{\frac{\alpha }{2}}$, 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.