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Q1- In Chapter 8 (attached) of your text, read the interpretation of confidence intervals. Find a journal article from the library that incorporates repeated sampling. This may be stated or implied. Use the explanation in the last paragraph of the section on interpretation of confidence intervals to describe your data. Cite your sources.
Q2- Find an article that incorporates research of confidence intervals to market a new product. Summarize the information gained from your research. Based on the research, is it wise to go ahead and market the product? Support your ideas with cited sources.
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8.3 CONFIDENCE INTERVAL FOR A POPULATION
MEAN
For point estimation, a single number lies in the forefront even though a standard error is attached. Instead,
it is often more desirable to produce an interval of values that is likely to contain the true value of the
parameter.
Ideally, we would like to be able to collect a sample and then use it to calculate an interval that would
definitely contain the true value of the parameter. This goal, however, is not achievable because of sample-
to-sample variation. Instead, we insist that before sampling the proposed interval will contain the true value
with a specified high probability. This probability, called the level of confidence, is typically taken as .90,
.95, or .99.
To develop this concept, we first confine our attention to the construction of a confidence interval for a
population mean μ, assuming that the population is normal and the standard deviation σ is known. This
restriction helps to simplify the initial presentation of the concept of a confidence interval. Later on, we will
treat the more realistic case where σ is also unknown.
A probability statement about based on the normal distribution provides the cornerstone for the
development of a confidence interval. From Chapter 7, recall that when the population is normal, the
distribution of is also normal. It has mean μ and standard deviation . Here μ is unknown, but
is a known number because the sample size n is known and we have assumed that σ is known.
The normal table shows that the probability is .95 that a normal random variable will lie within 1.96
standard deviations from its mean. For , we then have
as shown in Figure 3. Now, the relation
and
as we can see by transposing from one side of an inequality to the other. Therefore, the event
is equivalent to
In essence, both events state that the difference lies between -1.96 and . Thus, the
probability statement
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can also be expressed as
This second form tells us that, before we sample, the random interval from to
will include the unknown parameter μ with a probability of .95. Because σ is assumed to
be known, both the upper and lower endpoints can be computed as soon as the sample data are available.
Guided by the above reasonings, we say that the
interval
or its realization is a 95% confidence interval for μ when the
population is normal and σ known.
Example 5
Calculating a Confidence Interval—Normal Population σ Known
The daily carbon monoxide (CO) emission from a large production plant will be measured on 25
randomly selected weekdays. The production process is always being modified and the current mean
value of daily CO emissions μ is unknown. Data collected over several years confirm that, for each
year, the distribution of CO emission is normal with a standard deviation of .8 ton.
Suppose the sample mean is found to be tons. Construct a 95% confidence interval for the
current daily mean emission μ.
SOLUTION
The population is normal, and the observed value
is a 95% confidence interval for μ. Since μ is unknown, we cannot determine whether or not μ lies in
this interval.
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Figure 3 Normal distribution of .
Referring to the confidence interval obtained in Example 5, we must not speak of the probability of the
fixed interval (2.39, 3.01) covering the true mean μ. The particular interval (2.39, 3.01) either does or does
not cover μ, and we will never know which is the case.
We need not always tie our discussion of confidence intervals to the choice of a 95% level of confidence.
An investigator may wish to specify a different high probability. We denote this probability by 1 − α and
speak of a 100(1 − α)% confidence interval. The only change is to replace 1.96 with , where
denotes the upper α/2 point of the standard normal distribution (i.e., the area to the right of is α/2, as
shown in Figure 2).
In summary, when the population is normal and σ is known, a 100(1 − α)% confidence interval for μ is
given by
INTERPRETATION OF CONFIDENCE INTERVALS
To better understand the meaning of a confidence statement, we use the computer to perform repeated
samplings from a normal distribution with μ = 100 and σ = 10. Ten samples of size 7 are selected, and a
95% confidence interval is computed from each. For the first sample, and the
interval is 104.3 ± 7.4, or 96.9 to 111.7. This and the other intervals are illustrated in Figure 4, where each
vertical line segment represents one confidence interval. The midpoint of a line is the observed value of
for that particular sample. Also note that all the intervals are of the same length 2 × 1.96 . Of
the 10 intervals shown, 9 cover the true value of μ. This is not surprising, because the specified
probability .95 represents the long-run relative frequency of these intervals covering the true μ = 100.
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Figure 4 Interpretation of the confidence interval for μ.
Because confidence interval statements are the most useful way to communicate information obtained from
a sample, certain aspects of their formulation merit special emphasis. Stated in terms of a 95% confidence
interval for μ, these are:
1. Before we sample, a confidence interval is a random interval
that attempts to cover the true value of the parameter μ.
2. The probability
interpreted as the long-run relative frequency over many repetitions of sampling asserts that about
95% of the intervals will cover μ.
3. Once is calculated from an observed sample, the interval
which is a realization of the random interval, is presented as a 95% confidence interval for μ. A
numerical interval having been determined, it is no longer sensible to speak about the probability of
its covering a fixed quantity μ.
4. In any application we never know if the 95% confidence interval covers the unknown mean μ.
Relying on the long-run relative frequency of coverage in property 2, we adopt the terminology
confidence once the interval is calculated.
At this point, one might protest, “I have only one sample and I am not really interested in repeated
sampling.” But if the confidence estimation techniques presented in this text are mastered and followed each
time a problem of interval estimation arises, then over a lifetime approximately 95% of the intervals will
cover the true parameter. Of course, this is contingent on the validity of the assumptions underlying the
techniques—independent normal observations here.
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LARGE SAMPLE CONFIDENCE INTERVALS FOR μ
Having established the basic concepts underlying confidence interval statements, we now turn to the more
realistic situation for which the population standard deviation σ is unknown. We require the sample size n to
be large in order to dispense with the assumption of a normal population. The central limit theorem then
tells us that is nearly normal whatever the form of the population. Referring to the normal distribution of
in Figure 5 and the discussion accompanying Figure 2, we again have the probability statement
(Strictly speaking, this probability is approximately 1 − α for a nonnormal population.) Even though the
interval
will include μ with the p robability 1 − α, it does not serve as a confidence interval because it involves the
unknown quantity σ. However, because n is large, replacing with its estimator does not
appreciably affect the probability statement. Summarizing, we find that the large sample confidence interval
for μ has the form
Large Sample Confidence Interval for μ
When n is large, a 100(1 − α)% confidence interval for μ is given by
where S is the sample standard deviation.
Example 6
A Confidence Interval for the Mean Time of Community Service
Refer to the data in Example 1, consisting of 40 measurements of the time spent on community
service during the past month. The summary statistics are
Compute (a) 90% and (b) 80% confidence intervals for the mean number of hours per month.
SOLUTION
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The sample size n = 40 is large, so a normal approximation for the distribution of the sample mean
is appropriate. From the sample data, we know that
(a) With 1 − α = .90, we have α/2 = .05, and ,
The 90% confidence interval for the population mean of number of hours worked μ becomes
or approximately (3.2, 5.9) hours per month. This means that we can be 90% confident that
the mean hours per month μ is in the interval 3.2 to 5.9 hours. We have this confidence
because about 90% of the random samples of 40 students would produce intervals
that contain μ.
(b) With 1 − α = .80, we have α/2 = .10, and , so
The 80% confidence interval for μ becomes (4.55 − 1.05, 4.55 + 1.05) or (3.5, 5.6) hours per
month. Comparing the two confidence intervals, we note that the 80% confidence interval is
shorter than the 90% interval. A shorter interval seems to give a more precise location for μ
but suffers from a lower long-run frequency of being correct.
Figure 5 Normal distribution of .
CONFIDENCE INTERVAL FOR A PARAMETER
The concept of a confidence interval applies to any parameter, not just the mean. It requires that a lower
limit L and an upper limit U be computed from the sample data. Then the random interval from L to U must
have the specified probability of covering the true value of the parameter. The large sample 100(1 − α)%
confidence interval for μ has
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Definition of a Confidence Interval for a Parameter
An interval (L, U) is a 100(1 − α)% confidence interval for a parameter if
and the endpoints L and U are computable from the sample.
Example 7
A Confidence Interval for the Mean Time to Complete a Test
Madison recruits for the fire department need to complete a timed test that simulates working
conditions. It includes placing a ladder against a building, pulling out a section of fire hose, dragging
a weighted object, and crawling in a simulated attic environment. The times, in seconds, for recruits
to complete the test for Madison firefighter are
425 389 380 421 438 331 368 417 403 416 385 315
427 417 386 386 378 300 321 286 269 225 268 317
287 256 334 342 269 226 291 280 221 283 302 308
296 266 238 286 317 276 254 278 247 336 296 259
270 302 281 228 317 312 327 288 395 240 264 246
294 254 222 285 254 264 277 266 228 347 322 232
365 356 261 293 354 236 285 303 275 403 268 250
279 400 370 399 438 287 363 350 278 278 234 266
319 276 291 352 313 262 289 273 317 328 292 279
289 312 334 294 297 304 240 303 255 305 252 286
297 353 350 276 333 285 317 296 276 247 339 328
267 305 291 269 386 264 299 261 284 302 342 304
336 291 294 323 320 289 339 292 373 410 257 406
374 268
Obtain a 95% confidence interval for the mean time of recruits who complete the test.
SOLUTION
A computer calculation gives
SAMPLE SIZE 158
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MEAN 307.77
STD DEV 51.852
Since 1 − α = .95, α/2 = .025, and , the large sample 95% confidence interval for μ
becomes
or
When the sample size is large, the sample also contains information on the shape of the distribution
that can be elicited by graphical displays. Figure 6 gives the stem-and-leaf display, with the data
rounded to two places, accompanied by the fsmpplot. The confidence interval pertains to the mean of
a population with a long right-hand tail.
Figure 6 A stem-and-leaf display and fsmpplot give more information about the form of the
population.
Exercises
8.15 Refer to Example 2, Chapter 2, where Table 2 records the number of items returned by 30
persons to a large discount department store in late December. The summary statistics are
n = 30 s = 1.303.
Obtain a 98% confidence interval for μ, the population mean number of items returned.
Answer:
(2.03, 3.14)
8.16
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A company wants to check the consistency of electronic copies of signatures for consumer
credit purchases. A sample of 49 electronic signatures are available from the same customer.
One measure of consistency in signing is the total length that the script is outside the signature
fsmp. The sample of 49 signatures yielded a mean length of .21 with a standard deviation
of .19 centimeter. Obtain a 99% confidence interval for this customer’s population mean
length outside the fsmp.
8.17 Each day of the year, a large sample of cellular phone calls is selected and a 95% confidence
interval is calculated for the mean length of all cellular phone calls on that day. Of these 365
confidence intervals, one for each day of the year, approximately how many will cover their
respective population means? Explain your reasoning.
8.18 A forester measures 100 needles off a pine tree and finds centimeters and s = 0.7
centimeter. She reports that a 95% confidence interval for the mean needle length is
(a) Is the statement correct?
(b) Does the interval (2.96, 3.24) cover the true mean? Explain.
8.19 In a study on the nutritional qualities of fast foods, the amount of fat was measured for a
random sample of 35 hamburgers of a particular restaurant chain. The sample mean and
standard deviation were found to be 30.2 and 3.8 grams, respectively. Use these data to
construct a 95% confidence interval for the mean fat content in hamburgers served in these
restaurants.
Answer:
(28.94, 31.46) grams
8.20 In the same study described in Exercise 8.19, the sodium content was also measured for the
sampled hamburgers, and the sample mean and standard deviation were 658 and 47
milligrams, respectively. Determine a 98% confidence interval for the true mean sodium
content.
8.21 An entomologist sprayed 120 adult Melon flies with a specific low concentration of malathion
and observed their survival times. The mean and standard deviation were found to be 18.3 and
5.2 days, respectively. Use these data to construct a 99% confidence interval for the true mean
survival time.
Answer:
(17.1, 19.5) days
8.22 Students are asked about the number of songs they downloaded from a pay-for-songs Web
site the last month. From a random sample of 39 students, the sample mean was 4.7 with a
standard deviation of 3.2.
(a) Obtain a 95% confidence interval for μ, the mean number of songs downloaded by the
population of all students.
(b) Does μ lie in your interval obtained in Part(a)?
(c) In a long series of repeated experiments, with new samples of 39 students collected for
each experiment, what proportion of the resulting confidence intervals will contain the
true population mean? Explain your reasoning.
8.23
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The freshness of produce at a super-store is rated on a scale of 1 to 5 with 5 being very fresh.
From a random sample of 49 customers, the average score was 3.8 with a standard deviation
of .7.
(a) Obtain a 95% confidence interval for the population mean, μ, the mean score for the
distribution of all possible customers.
(b) Does μ lie in your interval obtained in Part(a)? Explain.
(c) In a long series of repeated experiments, with new random samples of 49 customers
each day, what proportion of the resulting confidence intervals will contain the true
population mean? Explain your reasoning.
Answer:
(a) (3.604, 3.996)
8.24 Referring to Example 7, where the 158 times to complete the firefighter test have mean
307.77 and standard deviation 51.852, obtain a 99% confidence interval for the mean time of
all possible recruits who would complete the test.
8.25 Based on a survey of 140 employed persons in a city, the mean and standard deviation of the
commuting distances between home and the principal place of business are found to be 8.6
and 4.3 miles, respectively. Determine a 90% confidence interval for the mean commuting
distance for the population of all employed persons in the city.
Answer:
(8.00, 9.20) miles
8.26 A manager at a power company monitored the employee time required to process high-
efficiency lamp bulb rebates. A random sample of 40 applications gave a sample mean time
of 3.8 minutes and a standard deviation of 1.2 minutes. Construct a 90% confidence interval
for the mean time to process μ.
8.27 A credit company randomly selected 50 contested items and recorded the dollar amount being
contested. These contested items had a sample mean dollars and s = 24.63 dollars.
Construct a 95% confidence interval for the mean amount contested, μ.
Answer:
(88.91, 102.57)
8.28 In a study to determine whether a certain stimulant produces hyperactivity, 55 mice were
injected with 10 micrograms of the stimulant. Afterward, each mouse was given a
hyperactivity rating score. The mean score was and s = 2.8. Give a 95% confidence
interval for the population mean score μ.
8.29 Refer to the Statistics in Context section of Chapter 7 concerning monthly changes in the
Canadian to U.S. exchange rate. A computer calculation gives and s = .0207 for
the n = 154 monthly changes. Find a 95% confidence interval for the mean monthly change.
Answer:
(−.00497, .00157)
8.30 An employee of an on-campus copy center wants to determine the mean number of copies
before a cartridge needs to be replaced. She records the life length in thousands of copies for
43 cartridges and obtains thousand copies. Obtain a 90%
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confidence interval for the population mean, μ, number of copies in thousands before a
cartridge should be replaced.
8.31 Refer to the 40 height measurements given in Exercise 8.4, which have
centimeter. Calculate a 99% confidence interval for the
population mean height.
Answer:
(1.521, 1.909) centimeters
8.32 Radiation measurements on a sample of 65 microwave ovens produced and s = .06
Determine a 95% confidence interval for the mean radiation.
8.33 Refer to the data on the growth of female salmon in the marine environment in Table D.7 of
the Data Bank. A computer calculation gives a 95% confidence interval.
One-Sample Z: Fmarine
Variable N Mean StDev 95.0% CI
Fmarine 40 429.15 41.05 (416.43, 441.87)
(a) Does the 95% confidence interval cover the true mean growth of all female salmon in
that marine environment?
(b) Why are you 95% confident that the interval (416.43, 441.87) covers the true mean?
Answer:
(b) In long run, covers with proportion .95
8.34 Refer to the data on the girth, in centimeters, of grizzly bears in Table D.8 of the Data Bank.
A computer calculation gives
One-Sample Z: Girth
Variable bf N Mean StDev 95.0% CI
Girth 61 93.39 21.79 ( 87.93, 98.86)
(a) Does the 95% confidence interval cover the true mean girth of all grizzly bears in the
area of the study? Explain.
(b) Why are you 95% confident that the interval (87.93, 98.86) covers the true mean?
8.35 The amount of PCBs (polychlorinated biphenyls) was measured in 40 samples of soil that
were treated with contaminated sludge. The following summary statistics were obtained.
(a) Obtain a 95% confidence interval for the population mean μ, amount of PCBs in the
soil.
Answer parts (b), (c), and (d) Yes, No, or Cannot tell. Explain your answer.
(b) Does the sample mean PCB content lie in your interval obtained in part (a)?
(c) Does the population mean PCB content lie in your interval obtained in part (a)?
(d) It is likely that 95% of the data lie in your interval obtained in part (a)?
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Answer:
(b) Yes, in the middle (d) No
8.36 A national fast food chain, with thousands of franchise locations, needed to audit the books at
each location. They first selected a sample of 50 locations and performed the audit. They
determined that a 95% confidence interval for the mean time to complete an audit is
Answer the following questions “Yes,” “No,” or “Cannot tell” and justify your answer.
(a) Does the population mean lie in the interval
(28.4, 52.7)?
(b) Does the sample mean lie in the interval (28.4, 52.7)?
(c) For a future sample of 50 franchise locations, will the sample mean fall in the interval
(28.4, 52.7)?
(d) Does 95% of the sample data lie in the interval (28.4, 52.7)?
Copyright © 2010 John Wiley & Sons, Inc. All rights reserved.
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