### 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, ($\overline{x}\u2013EBM,\overline{x}+EBM$).

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 $\alpha =1-CL$.

### 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.

$$\overline{x}\text{and}\mathrm{EBM}\text{=2}\text{.5}\text{.}$$

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 $\overline{x}$ = 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.

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*.

**In summary, as a result of the central limit theorem, the following statements apply:**

- $\overline{X}$ is normally distributed, that is, $\overline{X}$ ~
*N*$\left({\mu}_{X},\frac{\sigma}{\sqrt{n}}\right)$.
**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, $\overline{x}\text{,}$ 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 ${z}_{\frac{a}{2}}$, and the sample size by *n*, then the formula for the confidence interval with confidence level $Cl=1-\alpha $, is given by $(\overline{x}-{z}_{\frac{a}{2}}\times \frac{\sigma}{\sqrt{n}},\overline{x}+{z}_{\frac{a}{2}}\times \frac{\sigma}{\sqrt{n}}).$
- 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 $\frac{\alpha}{2}$.

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

For example, when *CL* = 0.95, *α* = 0.05, and $\frac{\alpha}{2}$ = 0.025, we write ${z}_{\frac{\alpha}{2}}$ = *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.

${z}_{\frac{\alpha}{2}}\text{=}{z}_{0.\text{025}}\text{=1}\text{.96}$, 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 ${z}_{\frac{\alpha}{2}}$; 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 = $\left({z}_{\frac{\alpha}{2}}\right)\left(\frac{\sigma}{\sqrt{n}}\right)\text{.}$

#### 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.

#### 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).

$$CL\text{=0}\text{.90, so}\alpha \text{=1\u2013}CL\text{=1\u20130}\text{.90 =0}\text{.10}\text{.}$$

$$\frac{\alpha}{2}=\text{}0.05;{z}_{\frac{\alpha}{2}}={z}_{0.05}$$

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.

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 $\overline{x}\pm EBM$.

$$\overline{x}\u2013EBM=1.024\u20130.1431=0.8809$$

$$\overline{x}+EBM=\text{1.024+0.1431=1.1671}$$

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.

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).

$$CL\text{=0}\text{.95, so}\alpha \text{=1\u2013}CL\text{=1\u20130}\text{.95=0}\text{.05}\text{.}$$

$$\frac{\alpha}{2}=0.025\text{\hspace{1em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{z}_{\frac{\alpha}{2}}={z}_{0.025}$$

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.)

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

$$\overline{x}\text{\u2013}EBM\text{=68\u20130}\text{.98=67}\text{.02}$$

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

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.

**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.

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?

- $\overline{x}$ = 68
*EBM* = $\left({z}_{\frac{\alpha}{2}}\right)\left(\frac{\sigma}{\sqrt{n}}\right)$
*σ* = 3; the confidence level is 90 percent (*CL *= 0.90); ${z}_{\frac{\alpha}{2}}$ = *z*_{0.05} = 1.645.

Solution 8.5

Solution AIf we *increase* the sample size *n* to 100, we *decrease* the margin of error.

When *n* = 100, *EBM* = $\left({z}_{\frac{\alpha}{2}}\right)\left(\frac{\sigma}{\sqrt{n}}\right)$ = (1.645)$\left(\frac{3}{\sqrt{100}}\right)$ = 0.4935.

Solution 8.5

Solution BIf we *decrease* the sample size *n* to 25, we *increase* the error bound.

When *n* = 25, *EBM* = $\left({z}_{\frac{\alpha}{2}}\right)\left(\frac{\sigma}{\sqrt{n}}\right)$ = (1.645)$\left(\frac{3}{\sqrt{25}}\right)$ = 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.