Chapter 16 EPS28 points

• A confidence interval uses __________ data to estimate an unknown ________________ ______________ with

an indication of how accurate the estimate is and of how confident we are that the result is correct.

• Any confidence interval has two parts: an interval calculated from the data and a _________ __________ _____. ____

The confidence interval often has the form

_______________ ± __________ ______ ________

• The confidence level is the success rate of the method that produces the interval. That is, C is

the ____________ that the method will give a correct answer. If you use 95% confidence intervals often, in the

long run ________ of your intervals will contain the true parameter value. You do not know whether or not a 95%

confidence interval calculated from a particular set of data contains the true parameter value.

• A level C confidence interval for the mean μ of a Normal population with known standard deviation σ, based

on an SRS of size n, is given by:

•The critical value z* is chosen so that the standard Normal curve has area C between _____ and ________.

•Other things being equal, the margin of error of a confidence interval gets smaller as

• the confidence level C (increases / decreases)

• the population standard deviation σ (increases / decreases)

• the sample size n (increases / decreases)

CHECK YOUR SKILLS ( 1 point each - 8 points total)

16.11 To give a 99.9% confidence interval for a population mean μ, you would use the critical value

(a)z* = 1.960.

(b)z* = 2.576.

(c)z* = 3.291.

Use the following information for Exercise 16.12, 16.13, and 16.14. A laboratory scale is known to have a standard deviation of σ = 0.001 gram in repeated weighings. Scale readings in repeated weighings are Normally distributed, with mean equal to the true weight of the specimen. Three weighings of a specimen on this scale give 3.412, 3.416, and 3.414 grams.

16.12 A 95% confidence interval for the true weight of this specimen is

(a)3.414 ± 0.00113.(b)3.414 ± 0.00065.(c)3.414 ± 0.00196.

16.13 You want a 99% confidence interval for the true weight of this specimen. The margin of error for this interval will be

(a)smaller than the margin of error for 95% confidence.(b)greater than the margin of error for 95% confidence.(c)about the same as the margin of error for 95% confidence.

16.14 Another specimen is weighed eight times on this scale. The average weight is 4.1602 grams. A 99% confidence interval for the true weight of this specimen is

(a)4.1602 ± 0.00032.(b)4.1602 ± 0.00069.(c)4.1602 ± 0.00091.

Use the following information for Exercise 16.15, 16.16, 16.17, and 16.18. The National Assessment of Educational Progress (NAEP) includes a mathematical test for eighth-grade students. Scores on the test range from 0 to 500. Suppose that you give the NAEP test to an SRS of 900 eighth-graders from a large population in which the scores have mean μ = 285 and standard deviation σ = 125. The mean will vary if you take repeated samples.

16.15 The sampling distribution of is approximately Normal. It has mean μ = 285. What is its standard deviation?

(a)125.(b)4.167.(c)0.139.

16.16 Suppose that an SRS of 900 eighth-graders has = 288. Based on this sample, a 95% confidence interval for μ is

(a)8.17 ± 0.27.(b)288 ± 8.17.(c)285 ± 8.17.

16.17 In Exercise 16.16, suppose that we computed a 99% confidence interval for μ.

(a)This 99% confidence interval would have a smaller margin of error than the 95% confidence interval.(b)This 99% confidence interval would have a larger margin of error than the 95% confidence interval.(c)This 99% confidence interval could have either a smaller or a larger margin of error than the 95% confidence interval. This varies from sample to sample.

16.18 Suppose that we took an SRS of 1600 eighth-graders and found = 288. Compared with an SRS of 900 eighth-graders, the margin of error for a 95% confidence interval for μ is

(a)smaller.(b)larger.(c)either smaller or larger but we can’t say which.

16.19 Student study times. A class survey in a large class for first-year college students asked, “About how many hours do you study during a typical week?” The mean response of the 463 students was = 15.3 hours.5 Suppose that we know that the study time follows a Normal distribution with standard deviation σ = 8.5 hours in the population of all first-year students at this university. (4 points)

(a)Use the survey result to give a 99% confidence interval for the mean study time of all first-year students.

(b)What condition not yet mentioned must be met for your confidence interval to be valid?

16.21 An outlier strikes. There were actually 464 responses to the class survey in Exercise 16.19. One student claimed to study 10,000 hours per week (10,000 is more than the number of hours in a year). We know he’s joking, so we left out this value. If we did a calculation without looking at the data, we would get = 36.8 hours for all 464 students. Now what is the 99% confidence interval for the population mean? (Continue to use σ = 8.5.) Compare the new interval with that in Exercise 16.19. The message is clear: always look at your data, because outliers can greatly change your result. (2 points)

16.24 Explaining confidence. Here is an explanation from the Associated Press concerning one of its opinion polls. Explain briefly but clearly in what way this explanation is incorrect. ( 2 points)

For a poll of 1,600 adults, the variation due to sampling error is no more than three percentage points either way. The error margin is said to be valid at the 95 percent confidence level. This means that, if the same questions were repeated in 20 polls, the results of at least 19 surveys would be within three percentage points of the results of this survey.

Exercises 16.25 to 16.27 ask you to answer questions from data. Assume that the “simple conditions” hold in each case. The exercise statements give you the State step of the four-step process. In your work, follow the Plan, Solve, and Conclude steps, illustrated in Example 16.3 for a confidence interval.

16.25 Pulling wood apart. How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Here are data from students doing a laboratory exercise: (6 points)

(a)We are willing to regard the wood pieces prepared for the lab session as an SRS of all similar pieces of Douglas fir. Engineers also commonly assume that characteristics of materials vary Normally. Make a graph to show the shape of the distribution for these data. Does it appear safe to assume that the Normality condition is satisfied? Suppose that the strength of pieces of wood like these follows a Normal distribution with standard deviation 3000 pounds.

(b)Give a 95% confidence interval for the mean load required to pull the wood apart.

16.27 This wine stinks. Sulfur compounds cause “off-odors” in wine, so winemakers want to know the odor threshold, the lowest concentration of a compound that the human nose can detect. The odor threshold for dimethyl sulfide (DMS) in trained wine tasters is about 25 micrograms per liter of wine (μg/L). The untrained noses of consumers may be less sensitive, however. Here are the DMS odor thresholds for 10 untrained students: (6 points)

(a)Assume that the standard deviation of the odor threshold for untrained noses is known to be σ = 7 μg/L. Briefly discuss the other two “simple conditions,” using a stemplot to verify that the distribution is roughly symmetric with no outliers.

(b)Give a 95% confidence interval for the mean DMS odor threshold among all students.

16.28 Why are larger samples better? Statisticians prefer large samples. Describe briefly the effect of increasing the size of a sample on the margin of error of a 95% confidence interval. (2 points)