Devore J.L., Berk K.N. Modern Mathematical Statistics with Applications

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Thus a sample size of 381 should be used. The expression for n based on the

traditional CI gives a slightly larger value of 385.

■

One-Sided Conﬁdence Intervals (Conﬁdence Bounds)

The confidence intervals discussed thus far give both a lower confidence bound and

an upper confidence bound for the parameter being estimated. In some circum-

stances, an investigator will want only one of these two types of bounds. For

example, a psychologist may wish to calculate a 95% upper confidence bound for

true average reaction time to a particular stimulus, or a surgeon may want only a

lower confidence bound for true average remission time after colon cancer

surgery. Because the cumulative area under the standard normal curve to the left

of 1.645 is .95,

X  m

ﬃﬃﬃ

< 1:645



 :95

Manipulating the inequality inside the parentheses to isolate m on one side and

replacing rv’s by calculated values gives the inequality m >

x  1:645s=

ﬃﬃﬃ

; the

expression on the right is the desired lower confidence bound. Starting with

P(1.645 < Z)  .95 and manipulating the inequality results in the upper confi-

dence bound. A similar argument gives a one-sided bound associated with any

other confidence level.

PROPOSITION

A large-sample upper confidence bound for m is

m <

x þ z



ﬃﬃﬃ

and a large-sample lower confidence bound for m is

m >

x  z



ﬃﬃﬃ

A one-sided confidence bound for p results from replacing z

a/2

by z

and 

by either + or – in the CI formula (8.10) for p. In all cases the confidence

level is approximately 100(1  a)%.

Example 8.10 A random sample of 50 patients who had been seen at an outpatient clinic was

selected, and the waiting time to see a physician was determined for each one,

resulting in a sample mean time of 40.3 min and a sample standard deviation of

28.0 min (suggested by the article “An Example of Good but Partially Successful

OR Engagement: Improving Outpatient Clinic Operations”, Interfaces 28, #5).

An upper confidence bound for tru e average waiting time with a confidence level

of roughly 95% is

398 CHAPTER 8 Statistical Intervals Based on a Single Sample

40:3 þð1: 645Þð28:0Þ=

ﬃﬃﬃﬃﬃ

¼ 40:3 þ 6:5 ¼ 46:8

That is, with a confidence level of about 95%, m < 46.8. Note that the sample

standard deviation is quite large relative to the sample mean. If these were the

values of s and m, respectively, then population normality would not be sensible

because there would then be quite a large probability of obtaining a negative

waiting time. But because n is large here, our confiden ce b ound is valid even

though the population distribution is probably positively skewed.

■

Exercises Section 8.2 (12–28)

12. A random sample of 110 lightning flashes in a

region resulted in a sample average radar echo

duration of .81 s and a sample standard deviation

of .34 s (“Lightning Strikes to an Airplane in a

Thunderstorm,” J. Aircraft, 1984: 607–611).

Calculate a 99% (two-sided) confidence interval

for the true average echo duration m, and inter-

pret the resulting interval.

13. The article “Extravisual Damage Detection?

Defining the Standard Normal Tree” (Photogram-

metric Engrg. Remote Sensing, 1981: 515–522)

discusses the use of color infrared photography in

identification of normal trees in Douglas fir

stands. Among data reported were summary sta-

tistics for green-filter analytic optical densitomet-

ric measurements on samples of both healthy and

diseased trees. For a sample of 69 healthy trees,

the sample mean dye-layer density was 1.028, and

the sample standard deviation was .163.

a. Calculate a 95% (two-sided) CI for the true

average dye-layer density for all such trees.

b. Suppose the investigators had made a rough

guess of .16 for the value of s before collect-

ing data. What sample size would be neces-

sary to obtain an interval width of .05 for a

confidence level of 95%?

14. The article “Evaluating Tunnel Kiln Perfor-

mance” (Amer. Ceramic Soc. Bull., Aug. 1997:

59–63) gave the following summary information

for fracture strengths (MPa) of n ¼ 169 ceramic

bars fired in a particular kiln:

x ¼ 89:10;

s ¼ 3:73.

a. Calculate a (two-sided) confidence interval

for true average fracture strength using a con-

fidence level of 95%. Does it appear that true

average fracture strength has been precisely

estimated?

b. Suppose the investigators had believed a

priori that the population standard deviation

was about 4 MPa. Based on this supposition,

how large a sample would have been required

to estimate m to within .5 MPa with 95%

confidence?

15. Determine the confidence level for each of the

following large-sample one-sided confidence

bounds:

a. Upper bound:

x þ :84s=

ﬃﬃﬃ

b. Lower bound:

x  2:05s=

ﬃﬃﬃ

c. Upper bound:

x þ :67s=

ﬃﬃﬃ

16. A sample of 66 obese adults was put on a low-

carbohydrate diet for a year. The average weight

loss was 11 lb and the standard deviation was

19 lb. Calculate a 99% lower confidence bound

for the true average weight loss. What does the

bound say about confidence that the mean weight

loss is positive?

17. A study was done on 41 first-year medical stu-

dents to see if their anxiety levels changed during

the first semester. One measure used was the

level of serum cortisol, which is associated with

stress. For each of the 41 students the level was

compared during finals at the end of the semester

against the level in the first week of classes. The

average difference was 2.08 with a standard

deviation of 7.88. Find a 95% lower confidence

bound for the population mean difference m.

Does the bound suggest that the mean population

stress change is necessarily positive?

18. The article “Ultimate Load Capacities of Expan-

sion Anchor Bolts” (J. Energy Engrg., 1993:

139–158) gave the following summary data on

shear strength (kip) for a sample of 3/8-in. anchor

bolts: n ¼ 78;

x ¼ 4:25; s ¼ 1:30. Calculate a

lower confidence bound using a confidence

level of 90% for true average shear strength.

19. The article “Limited Yield Estimation for Visual

Defect Sources” (IEEE Trans. Semicon. Manuf.,

1997: 17–23) reported that, in a study of a

8.2 Large-Sample Conﬁdence Intervals for a Population Mean and Proportion 399

particular wafer inspection process, 356 dies

were examined by an inspection probe and 201

of these passed the probe. Assuming a stable

process, calculate a 95% (two-sided) confidence

interval for the proportion of all dies that pass the

probe.

20. The Associated Press (October 9, 2002) reported

that in a survey of 4722 American youngsters aged

6–19, 15% were seriously overweight (a body

mass index of at least 30; this index is a measure

of weight relative to height). Calculate and inter-

pret a confidence interval using a 99% confidence

level for the proportion of all American youngsters

who are seriously overweight.

21. A random sample of 539 households from a mid-

western city was selected, and it was determined

that 133 of these households owned at least one

firearm (“The Social Determinants of Gun Own-

ership: Self-Protection in an Urban Environment,”

Criminology, 1997: 629–640). Using a 95% con-

fidence level, calculate a lower confidence bound

for the proportion of all households in this city that

own at least one firearm.

22. In a sample of 1000 randomly selected consumers

who had opportunities to send in a rebate claim

form after purchasing a product, 250 of these

people said they never did so (“Rebates: Get

What You Deserve”, Consumer Reports, May

2009: 7). Reasons cited for their behavior included

too many steps in the process, amount too small,

missed deadline, fear of being placed on a mailing

list, lost receipt, and doubts about receiving the

money. Calculate an upper confidence bound at

the 95% confidence level for the true proportion of

such consumers who never apply for a rebate.

Based on this bound, is there compelling evidence

that the true proportion of such consumers is smal-

ler than 1/3? Explain your reasoning.

23. The article “An Evaluation of Football Helmets

Under Impact Conditions” (Amer. J. Sports Med.,

1984: 233–237) reports that when each football

helmet in a random sample of 37 suspension-type

helmets was subjected to a certain impact test, 24

showed damage. Let p denote the proportion of all

helmets of this type that would show damage

when tested in the prescribed manner.

a. Calculate a 99% CI for p.

b. What sample size would be required for the

width of a 99% CI to be at most .10, irrespec-

tive of

24. A sample of 56 research cotton samples resulted

in a sample average percentage elongation of 8.17

and a sample standard deviation of 1.42 (“An

Apparent Relation Between the Spiral Angle f,

the Percent Elongation E

, and the Dimensions of

the Cotton Fiber,” Textile Res. J., 1978: 407–410).

Calculate a 95% large-sample CI for the true aver-

age percentage elongation m. What assumptions

are you making about the distribution of percent-

age elongation?

25. A state legislator wishes to survey residents of her

district to see what proportion of the electorate is

aware of her position on using state funds to pay

for abortions.

a. What sample size is necessary if the 95% CI

for p is to have width of at most .10 irrespective

of p?

b. If the legislator has strong reason to believe

that at least

of the electorate know of her

position, how large a sample size would you

recommend?

26. The superintendent of a large school district, hav-

ing once had a course in probability and statistics,

believes that the number of teachers absent on any

given day has a Poisson distribution with parame-

ter l. Use the accompanying data on absences for

50 days to derive a large-sample CI for l.[Hint:

The mean and variance of a Poisson variable both

equal l,so

Z ¼

X  l

ﬃﬃﬃﬃﬃﬃﬃﬃ

l=n

has approximately a standard normal distri-

bution. Now proceed as in the derivation of

the interval for p by making a probability

statement (with probability 1  a) and solv-

ing the resulting inequalities for l (see the

argument just after (8.10))].

Number of

absences 0 123 45678910

Frequency 1 48108753211

27. Reconsider the CI (8.10) for p, and focus on a

confidence level of 95%. Show that the confi-

dence limits agree quite well with those of the

traditional interval (8.11) once two successes and

two failures have been appended to the sample

[i.e., (8.11) based on (x+2) S’s in (n+4) trials].

[Hint: 1.96  2.] [Note: Agresti and Coull showed

that this adjustment of the traditional interval

also has actual confidence level close to the nomi-

nal level.]

400

CHAPTER 8 Statistical Intervals Based on a Single Sample

28. Young people may feel they are carrying the

weight of the world on their shoulders, when

what they are actually carrying too often is an

excessively heavy backpack. The article “Effec-

tiveness of a School-Based Backpack Health Pro-

motion Program” ( Work, 2003: 113–123) reported

the following data for a sample of 131 sixth

graders: for backpack weight lbðÞ;

x ¼ 13:83;

s ¼ 5:05; for backpack weight as a percentage

of body weight, a 95% CI for the population

mean was (13.62, 15.89).

a. Calculate and interpret a 99% CI for population

mean backpack weight.

b. Obtain a 99% CI for population mean weight

as a percentage of body weight.

c. The American Academy of Orthopedic Sur-

geons recommends that backpack weight be at

most 10% of body weight. What does your

calculation of (b) suggest, and why?

8.3

Intervals Based on a Normal Population

Distribution

The CI for m presented in Section 8.2 is valid provided that n is large. The resulting

interval can be used whatever the nature of the population distribution. The CLT

cannot be invoked, however, when n is small. In this case, one way to proceed is to

make a specific assumption about the form of the population distribution and then

derive a CI tailored to that assumption. For example, we could develop a CI for m

when the population is described by a gamma distribution, another interval for the

case of a Weibull population, and so on. Statisticians have indeed carried out this

program for a number of different distributional families. Because the normal

distribution is more frequently appropriate as a population model than is any

other type of distribution, we will focus here on a CI for this situation.

ASSUMPTION

The population of interest is normal, so that X

, ..., X

constitutes a random

sample from a normal distribution with both m and s unknown.

The key result underlying the interval in Section 8.2 is that for large n, the rv

Z ¼ð

X  mÞ=ðS=

ﬃﬃﬃ

Þ has approximately a standard normal distribution. When n is

small, S is no longer likely to be close to s, so the variability in the distribution of Z

arises from randomness in both the numerator and the denominator. This implies

that the probability distribution of ð

X  mÞ=ðS=

ﬃﬃﬃ

Þ will be more spread out than

the standard normal distribution. Inferences are based on the following result from

Section 6.4 using the family of t distributions:

THEOREM

When

X is the mean of a random sample of size n from a normal distribution

with mean m, the rv

T ¼

X  m

ﬃﬃﬃ

ð8:13Þ

has the t distribution with n  1 degre es of freedom (df ).

8.3 Intervals Based on a Normal Population Distribution 401

Properties of t Distributions

Before applying this theorem, a review of properties of t distributions is in order.

Although the variable of interest is still ð

X  mÞ=ðS=

ﬃﬃﬃ

Þ, we now denote it by T to

emphasize that it does not have a standard normal distribution when n is small.

Recall that a normal distribution is governed by two parameters, the mean m and the

standard deviation s.At distribution is governed by only one parameter, the

number of degrees of freedom of the distribution, abbreviated df and denoted

by n. Possible values of n are the positive integer s 1, 2, 3, .... Each different value

of n corresponds to a different t distribution.

The density function for a random variable having a t distribution was derived

in Section 6.4. It is quite complicated, but fortunately we need concern ourselves only

with several of the more important features of the corresponding density curves.

PROPERTIES

OF T DISTRI-

BUTIONS

1. Each t

curve is bell-shaped and centered at 0.

2. Each t

curve is more spread out than the standard normal (z) curve.

3. As n increas es, the spread of the t

curve decreases.

4. As n !1, the sequence of t

curves approaches the standard normal

curve (so the z curve is often called the t curve with df ¼1).

Recall the notation for values that capture p articular upper-tail t-curve areas.

NOTATION

Let t

a,n

¼ the number on the measurement axis for which the area under the

t curve with n df to the right of t

a,n

,isa; t

a,n

is called a t critical value.

This notation is illustrated in Figure 8.7. Appendix Table A.5 gives t

a,n

for selected

values of a and n. The colu mns of the table correspond to different values of a.

To obtain t

.05,15

, go to the a ¼ .05 column, look down to the n ¼ 15 row, and

read t

.05,15

¼ 1.753. Similarly, t

.05,22

¼ 1.717 (.05 column, n ¼ 22 row), and

.01,22

¼ 2.508.

The values of t

a,n

exhibit regular behavior as we move across a row or down a

column. For fixed n, t

a,n

increases as a decreases, since we must move farther to the

Shaded area =

a,n

curve

Figure 8.7 A pictorial deﬁnition of t

a,n

402 CHAPTER 8 Statistical Intervals Based on a Single Sample

right of zero to capture area a in the tail. For fixed a,asn is increased (i.e., as we

look down any particular colu mn of the t table) the value of t

a,n

decreases. This is

because a larger value of n implies a t distribution with smaller spread, so it is not

necessary to go so far from zero to capture tail area a. Furthermore, t

a,n

, decreases

more slowly as n increases. Consequently, the table values are shown in increments

of 2 between 30 and 40 df and then jump to n ¼ 50, 60, 120, and finally 1.

Because t

is the standard normal curve , the familiar z

values appear in the last

row of the table. The rule of thumb suggested earlier for use of the large-sample CI

(if n > 40) comes from the approximate equality of the standard normal and

t distributions for n  40.

The One-Sample t Conﬁdence Interval

The standardized variable T has a t distribution with n  1 df, and the area

under the corresponding t density curve between t

a/2,n1

and t

a/2,n1

is 1  a

(area a/2 lies in each tail), so

Pðt

a=2;n1

< T < t

a=2;n1

Þ¼1  a ð8:14Þ

Expression (8.14) differs from expressions in previous sections in that T and t

a/2,n1

are used in place of Z and z

a/2

, but it can be manipulated in the same manner to

obtain a confidence interval for m.

PROPOSITION

Let

x and s be the sample mean and sample standar d deviation computed

from the results of a random sample from a normal population with mean m.

Then a 100(1  a)% confidence interval for m, the one-sample t CI,is

x  t

a=2;n1



ﬃﬃﬃ

;

x þ t

a=2;n1



ﬃﬃﬃ



ð8:15Þ

or, more compactly,

x  t

a=2;n1

 s=

ﬃﬃﬃ

An upper confidence bound for m is

x þ t

a;n1



ﬃﬃﬃ

and replacing + by  in this latter expression gives a lower confidence

bound for m; both have confidence level 100(1  a)%.

Example 8.11 Here are the alcohol percentages for a sample of 16 beers (light beers excluded):

4.68 4.13 4.80 4.63 5.08 5.79 6.29 6.79

4.93 4.25 5.70 4.74 5.88 6.77 6.04 4.95

Figure 8.8 shows a normal probability plot obtained from SAS. The plot is

sufficiently straight for the percentage to be assumed appro ximately normal.

8.3 Intervals Based on a Normal Population Distribution 403

The mean is x ¼ 5:34 and the standard deviation is s ¼ .8483. The sample

size is 16, so a confidence interval for the popul ation mean percentage is based on

15 df. A confidence level of 95% for a two-sided interval requires the t critical

value of 2.131. The resulting interval is

x  t

:025;15



ﬃﬃﬃ

¼ 5:34 ð2:131Þ

:8483

ﬃﬃﬃﬃﬃ

¼ 5:34  :45 ¼ð4:89; 5:79Þ

A 95% lower bound would use 1.753 in place of 2.131. It is interesting that the

95% confidence interval is consistent with the usual statemen t about the equiva-

lence of wine and beer in terms of alcohol content. That is, assuming an alcohol

percentage of 13% for wine, a 5-oz serving yields .65 oz of alcohol, while,

assuming 5.34% alcohol, a 12-oz serving of beer has .64 oz of alcohol.

Unfortunately, it is not easy to select n to control the width of the t interval.

This is because the width involves the unknown (before data collection) s and

because n enters not only through 1=

ﬃﬃﬃ

but also through t

a/2,n1

. As a result, an

appropriate n can be obtained only by trial and error.

In Chapter 14, we will discuss a small-sample CI for m that is valid provided

only that the population distribution is symmetric, a weaker assumption than

normality. However, when the population distribution is normal, the t interval

tends to be shorter than would be any other interval with the same confidence level.

A Prediction Interval for a Single Future Value

In many applications, an investigator wishes to predict a single value of a variable to

be observed at some future time, rather than to estimate the mean value of that

variable.

Example 8.12 Consider the following sample of fat content (in percentage) of n ¼ 10 randomly

selected hot dogs (“Sensory and Mechanical Assessment of the Quality of Frank-

furters,” J. Texture Stud., 1990: 395–409):

25.2 21.3 22.8 17.0 29.8 21.0 25.5 16.0 20.9 19.5

−2 −10 1 2

7.0

6.5

6.0

5.5

5.0

4.5

4.0

Normal Quantiles

Figure 8.8 A normal probability plot of the alcohol percentage data ■

404

CHAPTER 8 Statistical Intervals Based on a Single Sample

Assuming that these were selected from a normal population distribution, a 95% CI

for (interval estimate of) the population mean fat content is

x  t

:025;9



ﬃﬃﬃ

¼ 21:90  2:262 

4:134

ﬃﬃﬃﬃﬃ

¼ 21:90  2:96 ¼ð18:94; 24:86Þ

Suppose, however, you are going to eat a single hot dog of this type and want a

prediction for the resulting fat content. A point prediction, analogous to a point

estimate, is just

x ¼ 21:90. This prediction unfortunately gives no information

about reliability or precision. ■

The general setup is as follows: We will have available a random sample X

, ..., X

from a normal population distribution, and we wish to predict the value

of X

n+1

, a single future observation. A point predictor is X, and the resulting

prediction error is

X  X

nþ

. The expected value of the prediction error is

Eð

X  X

nþ

Þ¼EðXÞEX

nþ

ðÞ¼m  m ¼ 0

Since X

n+1

is independent of X

, ..., X

, it is independent of X, so the variance of

the prediction error is

Vð

X  X

nþ

Þ¼VðXÞþVX

nþ

ðÞ¼

þ s

¼ s

1 þ



The prediction error is a linear combination of independent normally distributed

rv’s, so itself is normally distributed. Thus

Z ¼

ðX  X

nþ1

Þ0

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 þ



X  X

nþ1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 þ



has a standard normal distribution. As in the derivation of the distribution of

X  mÞ=ðS=

ﬃﬃﬃ

Þ in Section 6.4, it can be shown (Exercise 43) that replacing s by

the sample standard deviation S (of X

, ... , X

) results in

T ¼

X  X

nþ1

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 þ

 t distribut ion with n  1df

Manipulating this T variable as T ¼ð

X  mÞ=ðS=

ﬃﬃﬃ

Þ was manipulated in the

development of a CI gives the following resu lt.

PROPOSITION

A prediction interval (PI) for a single observation to be selected from a

normal population distribution is

x  t

a=2;n1

 s

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 þ

ð8:16Þ

The prediction level is 100(1  a)%.

The interpretation of a 95% prediction level is similar to that of a 95% confidence

level; if the interval (8.16) is calculated for sample after sample, in the long run

95% of these intervals will include the corresponding future values of X.

8.3 Intervals Based on a Normal Population Distribution 405

Example 8.13

(Example 8.12

continued)

With n ¼ 10, x ¼ 21:90, s ¼ 4.134, and t

.025,9

¼ 2.262, a 95% PI for the fat

content of a single hot dog is

21:90 ð2:262Þð4: 134Þ

ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ

1 þ

¼ 21:90  9:81 ¼ 12:09; 31:71Þð

This interval is quite wide, indicating substantial uncertainty about fat content.

Notice that the width of the PI is more than three times that of the CI.

■

The error of prediction is X  X

nþ

, a difference between two random vari-

ables, whereas the estimation error is

X  m, the difference between a random

variable and a fixed (but unknown) value. The PI is wider than the CI because

there is more variability in the prediction error (due to X

n+1

) than in the estimation

error. In fact, as n gets arbitrarily large, the CI shrinks to the single value m, and

the PI approaches m  z

a/2

·s. There is uncertainty about a single X value even

when there is no need to estimate.

Tolerance Intervals

In addition to confidence intervals and prediction intervals, statisticians are some-

times called upon to obtain a third type of interval called a tolerance interval (TI).

A TI is an interval that with a high degree of reliability captures at least a specified

percentage of the x values in a population distribution. For example, if the popula-

tion distribution of fuel efficienc y is normal, then the interval from m  1.645s to

m + 1.645s captures 90% of the fuel efficiency values in the population. It can then

be shown that if m and s are replaced by their natural estimates

x and s based on a

sample of size n ¼ 20 and the z critical value 1.645 is replaced by a tolerance

critical value 2.310, the resulting interval contains at least 90% of the popu lation

values with a confiden ce level of 95%.

Please consult one of the chapter references for more information on TIs. And

before you calculate a particular statistical interval, be sure that it is the correct type

of interval to fulfill your objective!

Intervals Based on Nonnormal Population Distributions

The one-sample t CI for m is robust to small or even moderate departures from

normality unless n is quite small. By this we mean that if a critical value for 95%

confidence, for example, is used in calculating the interval, the actual confidence

level will be reasonably close to the nominal 95% level. If, however, n is small and

the population distribution is highly nonnormal, then the actual confidence level

may be considerably different from the one you think you are using when you

obtain a particular critical value from the t table. It would certainly be distressing to

believe that your confidence level is about 95% when in fact it was really more like

88%! The bootstrap technique, discussed in the last section of this chapter, has been

found to be quite successf ul at estimating parameters in a wide variety of non-

normal situations.

In contrast to the confidence interval, the validity of the prediction intervals

described in this section is closely tied to the normality assumption. These latter

intervals should not be used in the absence of compelling evidence for normality.

The excellent reference Statistical Intervals, cited in the bibliography at the end of

this chapter, discusses alternative procedures of this sort for various other situations.

406 CHAPTER 8 Statistical Intervals Based on a Single Sample

Exercises Section 8.3 (29–43)

29. Determine the values of the following quantities:

a. t

.1,15

b. t

.05,15

c. t

.05,25

d. t

.05,40

e. t

.005,40

30. Determine the t critical value that will capture

the desired t curve area in each of the following

cases:

a. Central area ¼ .95, df ¼ 10

b. Central area ¼ .95, df ¼ 20

c. Central area ¼ .99, df ¼ 20

d. Central area ¼ .99, df ¼ 50

e. Upper-tail area ¼ .01, df ¼ 25

f. Lower-tail area ¼ .025, df ¼ 5

31. Determine the t critical value for a two-sided

confidence interval in each of the following

situations:

a. Confidence level ¼ 95%, df ¼ 10

b. Confidence level ¼ 95%, df ¼ 15

c. Confidence level ¼ 99%, df ¼ 15

d. Confidence level ¼ 99%, n ¼ 5

e. Confidence level ¼ 98%, df ¼ 24

f. Confidence level ¼ 99%, n ¼ 38

32. Determine the t critical value for a lower or an

upper confidence bound for each of the situations

described in Exercise 31.

33. A sample of ten guinea pigs yielded the follow-

ing measurements of body temperature in

degrees Celsius (Statistical Exercises in Medical

Research, New York: Wiley, 1979, p. 26):

38.1 38.4 38.3 38.2 38.2 37.9 38.7 38.6

38.0 38.2

a. Verify graphically that it is reasonable to

assume the normal distribution.

b. Compute a 95% confidence interval for the

population mean temperature.

c. What is the CI if temperature is re-expressed

in degrees Fahrenheit? Are guinea pigs

warmer on average than humans?

34. Here is a sample of ACT scores (average of the

Math, English, Social Science, and Natural Sci-

ence scores) for students taking college freshman

calculus:

24.00 28.00 27.75 27.00 24.25 23.50 26.25

24.00 25.00 30.00 23.25 26.25 21.50 26.00

28.00 24.50 22.50 28.25 21.25 19.75

a. Using an appropriate graph, see if it is plausible

that the observations were selected from a

normal distribution.

b. Calculate a two-sided 95% confidence inter-

val for the population mean.

c. The university ACT average for entering

freshmen that year was about 21. Are the

calculus students better than average, as

measured by the ACT?

35. A sample of 14 joint specimens of a particular

type gave a sample mean proportional limit

stress of 8.48 MPa and a sample standard devia-

tion of .79 MPa (“Characterization of Bearing

Strength Factors in Pegged Timber Connec-

tions,” J. Struct. Engrg., 1997: 326–332).

a. Calculate and interpret a 95% lower confi-

dence bound for the true average proportional

limit stress of all such joints. What, if any,

assumptions did you make about the distribu-

tion of proportional limit stress?

b. Calculate and interpret a 95% lower predic-

tion bound for the proportional limit stress of

a single joint of this type.

36. Even as traditional markets for sweetgum lumber

have declined, large section solid timbers tradi-

tionally used for construction bridges and mats

have become increasingly scarce. The article

“Development of Novel Industrial Laminated

Planks from Sweetgum Lumber” (J. of Bridge

Engr., 2008: 64–66) described the manufacturing

and testing of composite beams designed to add

value to low-grade sweetgum lumber. Here is

data on the modulus of rupture (psi; the article

contained summary data expressed in MPa):

6807.99 7637.06 6663.28 6165.03 6991.41 6992.23

6981.46 7569.75 7437.88 6872.39 7663.18 6032.28

6906.04 6617.17 6984.12 7093.71 7659.50 7378.61

7295.54 6702.76 7440.17 8053.26 8284.75 7347.95

7422.69 7886.87 6316.67 7713.65 7503.33 7674.99

a. Verify the plausibility of assuming a normal

population distribution.

b. Estimate the true average modulus of rupture

in a way that conveys information about pre-

cision and reliability.

c. Predict the modulus for a single beam in a

way that conveys information about precision

and reliability. How does the resulting predic-

tion compare to the estimate in (b).

8.3 Intervals Based on a Normal Population Distribution 407