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18In this chapter we cover...

Conditions for inference

The t distributions

The one-samplet confidence interval

The one-sample t test

Using technology

Matched pairs t procedures

Robustness of t procedures

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Inference about aPopulation Mean

This chapter describes confidence intervals and significance tests for the mean μ

of a population. We used the z procedures in exactly this setting to introduce theideas of confidence intervals and tests. Now we discard the unrealistic conditionthat we know the population standard deviation σ and present procedures forpractical use. We also pay more attention to the real-data setting of our work. Thedetails of confidence intervals and tests change only slightly when you don’t knowσ . More important, you can interpret your results exactly as before. To emphasizethis, Examples 18.2 and 18.3 repeat the most important examples of Chapters 14and 15.

Conditions for inferenceConfidence intervals and tests of significance for the mean μ of a Normal popu-lation are based on the sample mean x . Confidence levels and P-values are prob-abilities calculated from the sampling distribution of x . Here are the conditionsneeded for realistic inference about a population mean.

433



434 C H A P T E R 18 • Inference about a Population Mean

CONDITIONS FOR INFERENCE ABOUT A MEAN

• We can regard our data as a simple random sample (SRS) from thepopulation. This condition is very important.

• Observations from the population have a Normal distribution withmean μ and standard deviation σ . In practice, it is enough that thedistribution be symmetric and single-peaked unless the sample is verysmall. Both μ and σ are unknown parameters.

There is another condition that applies to all of the inference methods inthis book: the population must be much larger than the sample, say at least 20 times as

CAUTIONUTION

large.1 All of our examples and exercises satisfy this condition. Practical settingsin which the sample is a large part of the population are rather special, and we willnot discuss them.

When the conditions for inference are satisfied, the sample mean x has theNormal distribution with mean μ and standard deviation σ/

√n. Because we don’t

know σ , we estimate it by the sample standard deviation s . We then estimate thestandard deviation of x by s/

√n. This quantity is called the standard error of the

sample mean x .

STANDARD ERROR

When the standard deviation of a statistic is estimated from data, the resultis called the standard error of the statistic. The standard error of the samplemean x is s/

√n.

A P P L Y Y O U R K N O W L E D G E

18.1 Travel time to work. A study of commuting times reports the travel timesto work of a random sample of 20 employed adults in New York State.The mean is x = 31.25 minutes and the standard deviation is s = 21.88minutes. What is the standard error of the mean?

Tony Freeman/PhotoEdit

18.2 Rats eating oat bran. In a study of the effect of diet on cholesterol, ratswere fed several different diets.2 One diet had 5% added fiber from oat bran.The study report gives results in the form “mean plus or minus the standarderror of the mean.” This form is very common in scientific publications.For the 6 rats fed this diet, blood cholesterol levels (in milligrams perdeciliter of blood) were 89.01 ± 5.36. What are x and s for these6 rats?



The t distributions 435

The t distributionsIf we knew the value of σ , we would base confidence intervals and tests for μ onthe one-sample z statistic

z = x − μ

σ/√

n

This z statistic has the standard Normal distribution N(0, 1). In practice, we don’tknow σ , so we substitute the standard error s/

√n of x for its standard deviation

σ/√

n. The statistic that results does not have a Normal distribution. It has a dis-tribution that is new to us, called a t distribution.

THE ONE-SAMPLE t STATISTIC AND THE t DISTRIBUTIONS

Draw an SRS of size n from a large population that has the Normaldistribution with mean μ and standard deviation σ . The one-samplet statistic

t = x − μ

s/√

n

has the t distribution with n − 1 degrees of freedom.

The t statistic has the same interpretation as any standardized statistic: it sayshow far x is from its mean μ in standard deviation units. There is a different tdistribution for each sample size. We specify a particular t distribution by giv-ing its degrees of freedom. The degrees of freedom for the one-sample t statistic degrees of freedomcome from the sample standard deviation s in the denominator of t . We saw inChapter 2 (page 49) that s has n − 1 degrees of freedom. There are other t statis-tics with different degrees of freedom, some of which we will meet later. We willwrite the t distribution with n − 1 degrees of freedom as t(n − 1) for short.

Figure 18.1 compares the density curves of the standard Normal distributionand the t distributions with 2 and 9 degrees of freedom. The figure illustrates thesefacts about the t distributions:

• The density curves of the t distributions are similar in shape to the standardNormal curve. They are symmetric about 0, single-peaked, and bell-shaped.

• The spread of the t distributions is a bit greater than that of the standardNormal distribution. The t distributions in Figure 18.1 have more probabilityin the tails and less in the center than does the standard Normal. This is truebecause substituting the estimate s for the fixed parameter σ introduces morevariation into the statistic.

• As the degrees of freedom increase, the t density curve approaches theN(0, 1) curve ever more closely. This happens because s estimates σ moreaccurately as the sample size increases. So using s in place of σ causes littleextra variation when the sample is large.




0

t, 2 degrees of freedomt, 9 degrees of freedomstandard Normal

t distributions have morearea in the tails than thestandard Normal distribution.

F I G U R E 1 8 . 1 Density curves for the t distributions with 2 and 9 degrees offreedom and the standard Normal distribution. All are symmetric with center 0.The t distributions are somewhat more spread out.

Table C in the back of the book gives critical values for the t distributions.Each row in the table contains critical values for the t distribution whose degreesof freedom appear at the left of the row. For convenience, we label the table entriesboth by the confidence level C (in percent) required for confidence intervals andby the one-sided and two-sided P-values for each critical value. You have alreadyused the standard Normal critical values z∗ in the bottom row of Table C. Bylooking down any column, you can check that the t critical values approach theNormal values as the degrees of freedom increase. As in the case of the Normaltable, statistical software makes Table C unnecessary.

Better statistics, better beer

The t distribution and the tinference procedures were inventedby William S. Gosset (1876–1937).Gosset worked for the Guinnessbrewery, and his goal in life was tomake better beer. He used his new tprocedures to find the best varietiesof barley and hops. Gosset’sstatistical work helped him becomehead brewer, a more interesting titlethan professor of statistics. BecauseGosset published under the penname “Student,” you will often seethe t distribution called “Student’st” in his honor.

E X A M P L E 1 8 . 1 t critical values

Figure 18.1 shows the density curve for the t distribution with 9 degrees of freedom.What point on this distribution has probability 0.05 to its right? In Table C, look inthe df = 9 row above one-sided P-value .05 and you will find that this critical value ist∗ = 1.833. To use software, enter the degrees of freedom and the probability you wantto the left, 0.95 in this case. Here is Minitab’s output:

Student's t distribution with 9 DFP(X<=x) x

0.95 1.83311


18.3 Critical values. Use Table C or software to find

(a) the critical value for a one-sided test with level α = 0.05 based on thet(5) distribution.

(b) the critical value for a 98% confidence interval based on thet(21) distribution.



The one-sample t confidence interval 437

18.4 More critical values. You have an SRS of size 25 and calculate the one-samplet statistic. What is the critical value t∗ such that

(a) t has probability 0.025 to the right of t∗?

(b) t has probability 0.75 to the left of t∗?

The one-sample t confidence intervalTo analyze samples from Normal populations with unknown σ , just replace thestandard deviation σ/

√n of x by its standard error s/

√n in the z procedures of

Chapters 14 and 15. The confidence interval and test that result are one-sample tprocedures. Critical values and P-values come from the t distribution with n − 1degrees of freedom. The one-sample t procedures are similar in both reasoning andcomputational detail to the z procedures.

THE ONE-SAMPLE t CONFIDENCE INTERVAL

Draw an SRS of size n from a large population having unknown mean μ. Alevel C confidence interval for μ is

x ± t∗ s√n

where t∗ is the critical value for the t(n − 1) density curve with area Cbetween −t∗ and t∗. This interval is exact when the population distributionis Normal and is approximately correct for large n in other cases.

E X A M P L E 1 8 . 2 Healing of skin wounds

Let’s look again at the biological study we met in Example 14.3. We follow the four-step

4STEPSTEP

process for a confidence interval, outlined on page 350.

STATE: Biologists studying the healing of skin wounds measured the rate at which newcells closed a razor cut made in the skin of an anesthetized newt. Here are data from18 newts, measured in micrometers (millionths of a meter) per hour:3

29 27 34 40 22 28 14 35 2635 12 30 23 18 11 22 23 33

This is one of several sets of measurements made under different conditions. We wantto estimate the mean rate for comparison with rates under other conditions.

David A. Northcott/CORBIS

FORMULATE: We will estimate the mean rate μ for all newts of this species by givinga 95% confidence interval.

SOLVE: We must first check the conditions for inference.

• As in Chapter 14 (page 351), we are willing to regard these newts as an SRS fromtheir species.

• The stemplot in Figure 18.2 does not suggest any strong departures from Normality.




We can proceed to calculation. For these data,

x = 25.67 and s = 8.324

The degrees of freedom are n − 1 = 17. From Table C we find that for 95% confidencet∗ = 2.110. The confidence interval is

x ± t∗ s√n

= 25.67 ± 2.1108.324√

18= 25.67 ± 4.14

= 21.53 to 29.81 micrometers per hour

CONCLUDE: We are 95% confident that the mean healing rate for all newts of thisspecies is between 21.53 and 29.81 micrometers per hour.

1

1

2

2

3

3

4

2 4

8

2 2 3 3

6 7 8 9

0 3 4

5 5

0

1

F I G U R E 1 8 . 2 Stemplot ofthe healing rates in Example18.2.

Our work in Example 18.2 is very similar to what we did in Example 14.3on page 350. To make the inference realistic we replaced the assumed σ = 8 bys = 8.324 calculated from the data and replaced the standard Normal critical valuez∗ = 1.960 by the t critical value t∗ = 2.110.

The one-sample t confidence interval has the form

estimate ± t∗SEestimate

where “SE” stands for “standard error.” We will meet a number of confidence in-tervals that have this common form. In Example 18.2, the estimate is the samplemean x , and its standard error is

SEx = s√n

= 8.324√

18= 1.962

Software will find x , s , SEx , and the confidence interval from the data. Figure 18.5(page 442) displays typical software output for Example 18.2.


18.5 Critical values. What critical value t∗ from Table C would you use for aconfidence interval for the mean of the population in each of the followingsituations?

(a) A 95% confidence interval based on n = 10 observations.

(b) A 99% confidence interval from an SRS of 20 observations.

(c) An 80% confidence interval from a sample of size 7.

18.6 The conductivity of glass. How well materials conduct heat matters whendesigning houses, for example. Conductivity is measured in terms of watts of heatpower transmitted per square meter of surface per degree Celsius of temperaturedifference on the two sides of the material. In these units, glass has conductivityabout 1. The National Institute of Standards and Technology provides exact dataon properties of materials. Here are 11 measurements of the heat conductivity of aparticular type of glass:4

1.11 1.07 1.11 1.07 1.12 1.08 1.08 1.18 1.18 1.18 1.12



The one-sample t test 439

(a) We can consider this an SRS of all specimens of glass of this type. Make astemplot. Is there any sign of major deviation from Normality?

(b) Give a 95% confidence interval for the mean conductivity.

(c) Use your interval to do a test: is there significant evidence at the 5% levelthat the mean conductivity of this type of glass is not 1?

18.7 Ancient air. The composition of the earth’s atmosphere may have changed over 4STEPSTEP

time. To try to discover the nature of the atmosphere long ago, we can examinethe gas in bubbles inside ancient amber. Amber is tree resin that has hardenedand been trapped in rocks. The gas in bubbles within amber should be a sample ofthe atmosphere at the time the amber was formed. Measurements on specimens ofamber from the late Cretaceous era (75 to 95 million years ago) give thesepercents of nitrogen:5

63.4 65.0 64.4 63.3 54.8 64.5 60.8 49.1 51.0

Assume (this is not yet agreed on by experts) that these observations are an SRSfrom the late Cretaceous atmosphere. Use a 90% confidence interval to estimatethe mean percent of nitrogen in ancient air. Follow the four-step process asillustrated in Example 18.2. David Sanger Photography/Alamy

The one-sample t testLike the confidence interval, the t test is very similar to the z test we met earlier.

THE ONE-SAMPLE t TEST

Draw an SRS of size n from a large population having unknown mean μ. Totest the hypothesis H0: μ = μ0, compute the one-sample t statistic

t = x − μ0

s/√

n

In terms of a variable T having the t(n − 1) distribution, the P-value for atest of H0 against

Ha: μ > μ0 is P (T ≥ t)t

Ha: μ < μ0 is P (T ≤ t)t

Ha: μ �= μ0 is 2P (T ≥ |t |)|t|

These P-values are exact if the population distribution is Normal and areapproximately correct for large n in other cases.




E X A M P L E 1 8 . 3 Sweetening colas

Here is a more realistic analysis of the cola-sweetening example from Chapter 15. We4STEPSTEP

follow the four-step process for a significance test, outlined on page 372.

STATE: Cola makers test new recipes for loss of sweetness during storage. Trained tastersrate the sweetness before and after storage. Here are the sweetness losses (sweetnessbefore storage minus sweetness after storage) found by 10 tasters for one new cola recipe:

2.0 0.4 0.7 2.0 −0.4 2.2 −1.3 1.2 1.1 2.3

Are these data good evidence that the cola lost sweetness?

FORMULATE: Tasters vary in their perception of sweetness loss. So we ask the questionin terms of the mean loss μ for a large population of tasters. The null hypothesis is “noloss,” and the alternative hypothesis says “there is a loss.”

H0: μ = 0

Ha: μ > 0

SOLVE: First check the conditions for inference. As before, we are willing to regardthese 10 carefully trained tasters as an SRS from a large population of all trained tasters.Figure 18.3 is a stemplot of the data. We can’t judge Normality from just 10 observations;there are no outliers but the data are somewhat skewed. P-values for the t test may beonly approximately accurate.

−1 3

4

4 7

−0

0

1 1

2 0 0 2 3

2

F I G U R E 1 8 . 3 Stemplot ofthe sweetness losses in Example18.3.

The basic statistics arex = 1.02 and s = 1.196

The one-sample t statistic is

t = x − μ0

s/√

n= 1.02 − 0

1.196/√

10

= 2.697

The P-value for t = 2.697 is the area to the right of 2.697 under the t distribution curvewith degrees of freedom n − 1 = 9. Figure 18.4 shows this area. Software (see Figure18.6) tells us that P = 0.0123.

Values of t

t distribution,9 degrees of freedom

P-value = 0.0123

t = 2.697

F I G U R E 1 8 . 4 The P-valuefor the one-sided t test inExample 18.3.



Using technology 441

Without software, we can pin P between two values by using Table C. Search thedf = 9 row of Table C for entries that bracket t = 2.697. The observed t lies between

df = 9

t∗ 2.398 2.821

P .02 .01the critical values for one-sided P-values 0.01 and 0.02.

CONCLUDE: There is quite strong evidence (P < 0.02) for a loss of sweetness.


18.8 Is it significant? The one-sample t statistic for testing

H0: μ = 0

Ha: μ > 0

from a sample of n = 15 observations has the value t = 1.82.

(a) What are the degrees of freedom for this statistic?

(b) Give the two critical values t∗ from Table C that bracket t . What are theone-sided P-values for these two entries?

(c) Is the value t = 1.82 significant at the 5% level? Is it significant at the 1%level?

18.9 Is it significant? The one-sample t statistic from a sample of n = 25observations for the two-sided test of

H0: μ = 64

Ha: μ �= 64

has the value t = 1.12.

(a) What are the degrees of freedom for t?

(b) Locate the two critical values t∗ from Table C that bracket t . What are thetwo-sided P-values for these two entries?

(c) Is the value t = 1.12 statistically significant at the 10% level? At the 5%level?

18.10 Ancient air, continued. Do the data of Exercise 18.7 give good reason to think 4STEPSTEP

that the percent of nitrogen in the air during the Cretaceous era was quitedifferent from the present 78.1%? Carry out a test of significance, following thefour-step process as illustrated in Example 18.3.

Using technologyAny technology suitable for statistics will implement the one-sample t procedures.As usual, you can read and use almost any output now that you know what to lookfor. Figure 18.5 displays output for the 95% confidence interval of Example 18.2




Texas Instruments TI-83 Plus

Excel

Minitab

CrunchIt!

Rate123456789

10111213

EDCB F

Sheet1 Sheet2

Microsoft Excel This is the estimate x.

This is the margin oferror ±t* SE.

Rate

Sheet3

One sample T statistics

Mean 25.6667

1.9621

26.5

8.3243

69.2941

29

11

40

18

4.1396

Standard Error

Median

Standard Deviation

Sample Variance

Range

Minimum

Maximum

Count

Confidence Level(95.0%)

Session

One-Sample T: Rate

Variable N Mean StDev SE Mean 95.0% CIRate 18 25.67 8.32 1.96 (21.53, 29.81)

95% confidence interval results:

Variable Sample Mean Std. Err DF L. Limit U. Limit

rate 25.666666 1.9620584 17 21.527084 29.806248

μ : mean of Variable

F I G U R E 1 8 . 5 The t confidence interval of Example 18.2: output from a graphingcalculator, two statistical programs, and a spreadsheet program.



Excel

Minitab

CrunchIt!

Texas Instruments TI-83 Plus

One sample T statistics

Session

Rate123456789

10111213

DCBA E

Sheet1

Microsoft Excel

This is the P-value.

This is the t statistic.

MeanStandard ErrorMedianStandard DeviationSample VarianceRangeMinimumMaximumCountConfidence Level(95.0%)

1.02000.3782

(B3.0)/B4TDIST (D3,9,1)

1.151.19611.4307

2.696690.01226

3.6-1.32.30

0.8556

Sweetness loss

μ : mean of Variable

H0 : μ = 0

Hypothesis test results:

One-Sample T: Loss

Test of mu = 0 vs mu > 0

Variable N Mean StDev SE MeanLoss 10 1.020 1.196 0.378

Variable 95.0% Lower Bound T PLoss 0.327 2.70 0.012

Variable Sample Mean Std. Err. DF T-Stat P-value

loss 1.02 0.37824154 9 2.6966896 0.0123

HA : μ > 0

F I G U R E 1 8 . 6 The t test of Example 18.3: output from a graphing calculator, twostatistical programs, and a spreadsheet program.

443




from a graphing calculator, two statistical programs, and a spreadsheet program.The TI-83, CrunchIt!, and Minitab outputs are straightforward. All three givethe estimate x and the confidence interval plus a clearly labeled selection of otherinformation. The confidence interval agrees with our hand calculation in Example18.2. In general, software results are more accurate because of the rounding inhand calculations. Excel gives several descriptive measures but does not give theconfidence interval. The entry labeled “Confidence Level (95.0%)” is the marginof error. You can use this together with x to get the interval using either a calculatoror the spreadsheet’s formula capability.

Figure 18.6 displays output for the t test in Example 18.3. The TI-83,CrunchIt!, and Minitab all give the sample mean x , the t statistic, and its P-value.Accurate P-values are the biggest advantage of software for the t procedures. Ex-cel is as usual more awkward than software designed for statistics. It lacks a one-sample t test menu selection but does have a function named TDIST for tail areasunder t density curves. The Excel output shows functions for the t statistic and itsP-value to the right of the main display, along with their values t = 2.69669 andP = 0.01226.

Matched pairs t proceduresThe study of healing in Example 18.2 estimated the mean healing rate fornewts under natural conditions, but the researchers then compared results un-der several conditions. The taste test in Example 18.3 was a matched pairsstudy in which the same 10 tasters rated before-and-after sweetness. Compara-tive studies are more convincing than single-sample investigations. For that rea-son, one-sample inference is less common than comparative inference. One com-mon design to compare two treatments makes use of one-sample procedures. Ina matched pairs design, subjects are matched in pairs and each treatment ismatched pairs designgiven to one subject in each pair. Another situation calling for matched pairsis before-and-after observations on the same subjects, as in the taste test ofExample 18.3.

MATCHED PAIRS t PROCEDURES

To compare the responses to the two treatments in a matched pairs design,find the difference between the responses within each pair. Then apply theone-sample t procedures to these differences.

The parameter μ in a matched pairs t procedure is the mean difference inthe responses to the two treatments within matched pairs of subjects in the entirepopulation.



Matched pairs t procedures 445

T A B L E 1 8 . 1 Average time (seconds) to complete a maze

Subject Unscented Scented Difference Subject Unscented Scented Difference

1 30.60 37.97 −7.37 12 58.93 83.50 −24.572 48.43 51.57 −3.14 13 54.47 38.30 16.173 60.77 56.67 4.10 14 43.53 51.37 −7.844 36.07 40.47 −4.40 15 37.93 29.33 8.605 68.47 49.00 19.47 16 43.50 54.27 −10.776 32.43 43.23 −10.80 17 87.70 62.73 24.977 43.70 44.57 −0.87 18 53.53 58.00 −4.478 37.10 28.40 8.70 19 64.30 52.40 11.909 31.17 28.23 2.94 20 47.37 53.63 −6.26

10 51.23 68.47 −17.24 21 53.67 47.00 6.6711 65.40 51.10 14.30

Image State Royalty Free/Alamy

E X A M P L E 1 8 . 4 Floral scents and learning

STATE: We hear that listening to Mozart improves students’ performance on tests. Inthe EESEE case study “Floral Scents and Learning,”investigators asked whether pleasantodors have a similar effect. Twenty-one subjects worked a paper-and-pencil maze whilewearing a mask. The mask was either unscented or carried a floral scent. The responsevariable is their average time on three trials. Each subject worked the maze with bothmasks, in a random order. The randomization is important because subjects tend to im-prove their times as they work a maze repeatedly. Table 18.1 gives the subjects’ averagetimes with both masks. Is there evidence that subjects worked the maze faster wearingthe scented mask?

FORMULATE: Take μ to be the mean difference (time unscented minus time scented)in the population of healthy adults. The null hypothesis says that the scents have noeffect, and Ha says that unscented times are longer than scented times on the average.

4STEPSTEP

So we test the hypotheses

H0: μ = 0

Ha: μ > 0

SOLVE: The subjects are not an actual SRS from the population of all healthy adults.But we are willing to regard them as an SRS in their performance on a maze. To ana-lyze the data, subtract the scented time from the unscented time for each subject. The21 differences form a single sample from the population with unknown mean μ. Theyappear in the “Difference” column in Table 18.1. Positive differences show that a sub-ject’s scented time was shorter than the unscented time. Figure 18.7 is a stemplot ofthe differences, rounded to the nearest whole second. The distribution is symmetric andreasonably Normal in shape.

The 21 differences have

5

7 1 1

8 7 6 4 4 3 1

3 4 7 9 9

2 4 6 9

5

−2

− 1

−00

1

2

F I G U R E 1 8 . 7 Stemplot ofthe time differences inExample 18.4.x = 0.9567 and s = 12.5479




The one-sample t statistic is therefore

t = x − 0s/

√n

= 0.9567 − 0

12.5479/√

21

= 0.349

Find the P-value from the t(20) distribution. (Remember that the degrees of freedomare 1 less than the sample size.) Table C shows that 0.349 is less than critical value forone-sided P = 0.25. The P-value is therefore greater than 0.25. Software gives the valueP = 0.3652.

df = 20

t∗ 0.687 0.860

P .25 .20

CONCLUDE: The data do not support the claim that floral scents improve perfor-mance. The average improvement is small, just 0.96 seconds over the 50 seconds thatthe average subject took when wearing the unscented mask. This small improvement isnot statistically significant at even the 25% level.

Example 18.4 illustrates how to turn matched pairs data into single-sampledata by taking differences within each pair. We are making inferences about a sin-gle population, the population of all differences within matched pairs. It is incorrect

CAUTIONUTION

to ignore the matching and analyze the data as if we had two samples, one from subjectswho wore unscented masks and a second from subjects who wore scented masks. Infer-ence procedures for comparing two samples assume that the samples are selectedindependently of each other. This condition does not hold when the same subjectsare measured twice. The proper analysis depends on the design used to produce thedata.


Many exercises from this point on ask you to give the P-value of a t test. If you havesuitable technology, give the exact P-value. Otherwise, use Table C to give two valuesbetween which P lies.

18.11 Does nature heal better? Our bodies have a natural electrical field that is4STEPSTEP

known to help wounds heal. Might higher or lower levels speed healing? A seriesof experiments with newts investigated this question. In one experiment, the twohind limbs of 14 newts were assigned at random to either experimental or controlgroups. This is a matched pairs design. The electrical field in the experimentallimbs was reduced to half its natural value by applying a voltage. The controllimbs were not manipulated. Table 18.2 gives the rates at which new cells closed arazor cut in each limb.6 Is there good evidence that changing the electrical fieldfrom its natural level slows healing?

(a) State hypotheses to be tested. Explain what the parameter μ in yourhypotheses stands for.

(b) We are willing to regard these 14 newts, which were randomly assigned froma larger group, as an SRS of all newts of this species. Because the t test uses thedifferences within matched pairs, these differences must be at least roughlyNormal. Make a stemplot of the 14 differences and comment on its shape.

(c) Complete the Solve and Conclude steps by carrying out the matched pairst test.



Robustness of t procedures 447

T A B L E 1 8 . 2 Healing rates (micrometers per hour) for newts

Experimental Control Experimental ControlNewt limb limb Newt limb limb

1 24 25 8 33 362 23 13 9 28 353 47 44 10 28 384 42 45 11 21 435 26 57 12 27 316 46 42 13 25 267 38 50 14 45 48

18.12 How much better does nature heal? Give a 90% confidence interval for thedifference in healing rates (control minus experimental) in Table 18.2.

Robustness of t proceduresThe t confidence interval and test are exactly correct when the distribution of thepopulation is exactly Normal. No real data are exactly Normal. The usefulness ofthe t procedures in practice therefore depends on how strongly they are affectedby lack of Normality.

Catching cheaters

A certification test for surgeons asks277 multiple-choice questions.Smith and Jones have 193 commonright answers and 53 identicalwrong choices. The computer flagstheir 246 identical answers asevidence of possible cheating. Theysue. The court wants to know howunlikely it is that exams this similarwould occur just by chance. That is,the court wants a P-value.Statisticians offer several P-valuesbased on different models for theexam-taking process. They all saythat results this similar wouldalmost never happen just by chance.Smith and Jones fail the exam.

ROBUST PROCEDURES

A confidence interval or significance test is called robust if the confidencelevel or P-value does not change very much when the conditions for use ofthe procedure are violated.

The condition that the population be Normal rules out outliers, so the pres-ence of outliers shows that this condition is not fulfilled. The t procedures arenot robust against outliers unless the sample is very large, because x and s are notresistant to outliers.

Fortunately, the t procedures are quite robust against non-Normality of thepopulation except when outliers or strong skewness are present. (Skewness is moreserious than other kinds of non-Normality.) As the size of the sample increases, thecentral limit theorem ensures that the distribution of the sample mean x becomesmore nearly Normal and that the t distribution becomes more accurate for criticalvalues and P-values of the t procedures.

Always make a plot to check for skewness and outliers before you use the tprocedures for small samples. For most purposes, you can safely use the one-sample




t procedures when n ≥ 15 unless an outlier or quite strong skewness is present.Here are practical guidelines for inference on a single mean.7

USING THE t PROCEDURES

• Except in the case of small samples, the condition that the data are anSRS from the population of interest is more important than thecondition that the population distribution is Normal.

• Sample size less than 15: Use t procedures if the data appear close toNormal (roughly symmetric, single peak, no outliers). If the data areskewed or if outliers are present, do not use t .

• Sample size at least 15: The t procedures can be used except in thepresence of outliers or strong skewness.

• Large samples: The t procedures can be used even for clearly skeweddistributions when the sample is large, roughly n ≥ 40.

E X A M P L E 1 8 . 5 Can we use t?

Figure 18.8 shows plots of several data sets. For which of these can we safely use thet procedures? 8

• Figure 18.8(a) is a histogram of the percent of each state’s adult residentswho are college graduates. We have data on the entire population of 50 states, soinference is not needed. We can calculate the exact mean for the population.There is no uncertainty due to having only a sample from the population,and no need for a confidence interval or test. If these data were an SRS from alarger population, t inference would be safe despite the mild skewness becausen = 50.

• Figure 18.8(b) is a stemplot of the force required to pull apart 20 pieces of Douglasfir. The data are strongly skewed to the left with possible low outliers, so we cannot trustthe t procedures for n = 20.

• Figure 18.8(c) is a stemplot of the lengths of 23 specimens of the red variety ofthe tropical flower Heliconia. The histogram is mildly skewed to the right and there areno outliers. We can use the t distributions for such data.

• Figure 18.8(d) is a histogram of the heights of the students in a college class.This distribution is quite symmetric and appears close to Normal. We can use thet procedures for any sample size.


18.13 An outlier strikes. Table 18.3 (page 450) gives data for another experimentfrom the study of healing rates in newts. The setup is exactly as in Exercise 18.11,except that the electrical field in the experimental limbs was reduced to zero byapplying a voltage.



Robustness of t procedures 449

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F I G U R E 1 8 . 8 Can we use t procedures for these data? (a) Percent of adult collegegraduates in the 50 states. No, this is an entire population, not a sample. (b) Forcerequired to pull apart 20 pieces of Douglas fir. No, there are just 20 observations andstrong skewness. (c) Lengths of 23 tropical flowers of the same variety. Yes, the sample islarge enough to overcome the mild skewness. (d) Heights of college students. Yes, for anysize sample, because the distribution is close to Normal.

(a) Make a stemplot of the differences between limbs of the same newt (controllimb minus experimental limb). There is a high outlier.

(b) Carry out two t tests to see if the mean healing rate is significantly higher inthe control limbs, one including all 12 newts and another that omits the outlier.What are the test statistics and their P-values? Does the outlier have a stronginfluence on your conclusion?




T A B L E 1 8 . 3 Healing rates (micrometers per hour) for newts

Experimental Control Experimental ControlNewt limb limb Newt limb limb

1 28 36 7 45 392 31 41 8 25 563 27 39 9 28 334 33 42 10 33 205 33 44 11 47 496 38 39 12 23 30

18.14 Reading scores in Atlanta. The Trial Urban District Assessment (TUDA) is agovernment-sponsored study of student achievement in large urban schooldistricts. TUDA gives a reading test scored from 0 to 500. A score of 243 is a“basic” reading level and a score of 281 is “proficient.” Scores for a random sampleof 1470 eighth-graders in Atlanta had x = 240 with standard error 1.1.9

(a) We don’t have the 1470 individual scores, but use of the t procedures is surelysafe. Why?

(b) Give a 99% confidence interval for the mean score of all Atlantaeighth-graders. (Be careful: the report gives the standard error of x , not thestandard deviation s .)

(c) Urban children often perform below the basic level. Is there good evidencethat the mean for all Atlanta eighth-graders is less than the basic level?

David Grossman/The Image Works

C H A P T E R 18 SUMMARY

Tests and confidence intervals for the mean μ of a Normal population are basedon the sample mean x of an SRS. Because of the central limit theorem, theresulting procedures are approximately correct for other population distributionswhen the sample is large.The standardized sample mean is the one-sample z statistic

z = x − μ

σ/√

n

If we knew σ , we would use the z statistic and the standard Normal distribution.In practice, we do not know σ . Replace the standard deviation σ/

√n of x by the

standard error s/√

n to get the one-sample t statistic

t = x − μ

s/√

n

The t statistic has the t distribution with n − 1 degrees of freedom.There is a t distribution for every positive degrees of freedom. All are symmetricdistributions similar in shape to the standard Normal distribution. The



Check Your Skills 451

t distribution approaches the N(0, 1) distribution as the degrees of freedomincrease.A level C confidence interval for the mean μ of a Normal population is

x ± t∗ s√n

The critical value t∗ is chosen so that the t curve with n − 1 degrees of freedomhas area C between −t∗ and t∗.Significance tests for H0: μ = μ0 are based on the t statistic. Use P-values orfixed significance levels from the t(n − 1) distribution.Use these one-sample procedures to analyze matched pairs data by first takingthe difference within each matched pair to produce a single sample.The t procedures are quite robust when the population is non-Normal, especiallyfor larger sample sizes. The t procedures are useful for non-Normal data whenn ≥ 15 unless the data show outliers or strong skewness.

C H E C K Y O U R S K I L L S

18.15 We prefer the t procedures to the z procedures for inference about a populationmean because

(a) z can be used only for large samples.

(b) z requires that you know the population standard deviation σ .

(c) z requires that you can regard your data as an SRS from the population.

18.16 You are testing H0: μ = 10 against Ha: μ < 10 based on an SRS of20 observations from a Normal population. The data give x = 8 and s = 4. Thevalue of the t statistic is

(a) −0.5. (b) −10. (c) −2.24.

18.17 You are testing H0: μ = 10 against Ha: μ < 10 based on an SRS of20 observations from a Normal population. The t statistic is t = −2.25. Thedegrees of freedom for this statistic are

(a) 19. (b) 20. (c) 21.

18.18 The P-value for the statistic in the previous exercise

(a) falls between 0.01 and 0.02.

(b) falls between 0.02 and 0.04.

(c) is greater than 0.25.

18.19 You have an SRS of 15 observations from a Normally distributed population.What critical value would you use to obtain a 98% confidence interval for themean μ of the population?

(a) 2.326 (b) 2.602 (c) 2.624

18.20 You are testing H0: μ = 0 against Ha: μ �= 0 based on an SRS of 15 observationsfrom a Normal population. What values of the t statistic are statisticallysignificant at the α = 0.005 level?

(a) t < −3.326 or t > 3.326 (b) t < −3.286 or t > 3.286 (c) t > 2.977




18.21 You are testing H0: μ = 0 against Ha: μ > 0 based on an SRS of 15 observationsfrom a Normal population. What values of the t statistic are statisticallysignificant at the α = 0.005 level?

(a) t < −3.326 or t > 3.326 (b) t > 2.947 (c) t > 2.977

18.22 Data on the blood cholesterol levels of 6 rats (milligrams per deciliter of blood)give x = 85 and s = 12. A 95% confidence interval for the mean bloodcholesterol of rats under this condition is

(a) 72.4 to 97.6. (b) 73.0 to 97.0. (c) 75.4 to 94.6.

18.23 Which of these settings does not allow use of a matched pairs t procedure?

(a) You interview both the husband and the wife in 64 married couples and askeach about their ideal number of children.

(b) You interview a sample of 64 unmarried male students and another sample of64 unmarried female students and ask each about their ideal number of children.

(c) You interview 64 female students in their freshman year and again in theirsenior year and ask each about their ideal number of children.

18.24 Because the t procedures are robust, the most important condition for their safeuse is that

(a) the population standard deviation σ is known.

(b) the population distribution is exactly Normal.

(c) the data can be regarded as an SRS from the population.

C H A P T E R 18 EXERCISES

18.25 Read carefully. You read in the report of a psychology experiment that,“Separate analyses for our two groups of 12 participants revealed no overallplacebo effect for our student group (mean = 0.08, SD = 0.37, t(11) = 0.49) anda significant effect for our non-student group (mean = 0.35, SD = 0.37, t(11) =3.25, p < 0.01).”10 What are the correct values of the two t statistics based onthe means and standard deviations? (The null hypothesis is that the mean effectis zero.) Compare each correct t-value with the critical values in Table C. Whatcan you say about the two-sided P-value in each case?

18.26 Alcohol in wine. The alcohol content of wine depends on the grape variety, theway in which the wine is produced from the grapes, the weather, and otherinfluences. Here are data on the percent of alcohol in wine produced from thesame grape variety in the same year by 48 winemakers in the same region ofItaly:11

12.86 12.88 12.81 12.70 12.51 12.60 12.25 12.53 13.49 12.8412.93 13.36 13.52 13.62 12.25 13.16 13.88 12.87 13.32 13.0813.50 12.79 13.11 13.23 12.58 13.17 13.84 12.45 14.34 13.4812.36 13.69 12.85 12.96 13.78 13.73 13.45 12.82 13.58 13.4012.20 12.77 14.16 13.71 13.40 13.27 13.17 14.13

(a) Make a histogram of the data, using class width 0.25. The shape of thedistribution is a bit irregular, but there are no outliers or strong skewness. There isno reason to avoid use of t procedures for n = 48.

(b) Give a 95% confidence interval for the mean alcohol content of wine of thistype.



Chapter 18 Exercises 453

(c) Based on your confidence interval, is the mean alcohol content significantlydifferent at the α = 0.05 level from 12%? From 13%?

David Fleetham/Taxi/Getty Images

18.27 Sharks. Great white sharks are big and hungry. Here are the lengths in feet of44 great whites:12

18.7 12.3 18.6 16.4 15.7 18.3 14.6 15.8 14.9 17.6 12.116.4 16.7 17.8 16.2 12.6 17.8 13.8 12.2 15.2 14.7 12.413.2 15.8 14.3 16.6 9.4 18.2 13.2 13.6 15.3 16.1 13.519.1 16.2 22.8 16.8 13.6 13.2 15.7 19.7 18.7 13.2 16.8

(a) Examine these data for shape, center, spread, and outliers. The distribution isreasonably Normal except for one outlier in each direction. Because these are notextreme and preserve the symmetry of the distribution, use of the t procedures issafe with 44 observations.

(b) Give a 95% confidence interval for the mean length of great white sharks.Based on this interval, is there significant evidence at the 5% level to reject theclaim “Great white sharks average 20 feet in length”?

(c) Before accepting the conclusions of (b), you need more information aboutthe data. What would you like to know?

Public Domain

18.28 Learning Blissymbols. Blissymbols are pictographs (think of Egyptianhieroglyphics) sometimes used to help learning-disabled children. In a study ofcomputer-assisted learning, 12 normal-ability schoolchildren were assigned atrandom to each of four computer learning programs. After they used the program,they attempted to recognize 24 Blissymbols. Here are the counts correct for one ofthe programs:13

12 22 9 14 20 15 9 10 11 11 15 6

(a) Make a stemplot (split the stems). Are there outliers or strong skewness thatwould forbid use of the t procedures?

(b) Give a 90% confidence interval for the mean count correct among allchildren of this age who use the program.

18.29 Worker absenteeism. A study of unexcused absenteeism among factory workerslooked at a year’s records for 668 workers in an English factory. The mean numberof days absent was 9.88 and the standard deviation was 17.847 days.14 Regardthese workers in this year as a random sample of all workers in all years as long asthis factory does not change work conditions or worker benefits. What can yousay with 99% confidence about the mean number of unexcused absences for allworkers?

18.30 A big toe problem. Hallux abducto valgus (call it HAV) is a deformation of the 4STEPSTEP

big toe that often requires surgery. Doctors used X-rays to measure the angle (indegrees) of deformity in 38 consecutive patients under the age of 21 who came toa medical center for surgery to correct HAV. The angle is a measure of theseriousness of the deformity. Here are the data:15

28 32 25 34 38 26 25 18 30 26 28 13 2021 17 16 21 23 14 32 25 21 22 20 18 2616 30 30 20 50 25 26 28 31 38 32 21




It is reasonable to regard these patients as a random sample of young patients whorequire HAV surgery. Carry out the Solve and Conclude steps of a 95% confidenceinterval for the mean HAV angle in the population of all such patients.

18.31 Conditions for inference. The number of days absent for a worker in a yearcannot be less than 0. The data in Exercise 18.29 have standard deviation muchgreater than the mean days absent. What does this say about the shape of thedistribution of days absent? Why can we nonetheless use a t confidence intervalfor the population mean?

18.32 An outlier’s effect. The data in Exercise 18.30 follow a Normal distributionquite closely except for one patient with HAV angle 50 degrees, a high outlier.

(a) Find the 95% confidence interval for the population mean based on the37 patients who remain after you drop the outlier.

(b) Compare your interval in (a) with your interval from Exercise 18.30. What isthe most important effect of removing the outlier?

Martin Dohrn/Photo Researchers

18.33 Cockroach metabolism. To study the metabolism of insects, researchers fedcockroaches measured amounts of a sugar solution. After 2, 5, and 10 hours, theydissected some of the cockroaches and measured the amount of sugar in varioustissues.16 Five roaches fed the sugar d-glucose and dissected after 10 hours hadthe following amounts (in micrograms) of d-glucose in their hindguts:

55.95 68.24 52.73 21.50 23.78

The researchers gave a 95% confidence interval for the mean amount ofd-glucose in cockroach hindguts under these conditions. The insects are a randomsample from a uniform population grown in the laboratory. We therefore expectresponses to be Normal. What confidence interval did the researchers give?

18.34 Growing trees faster. The concentration of carbon dioxide (CO2) in theatmosphere is increasing rapidly due to our use of fossil fuels. Because plants useCO2 to fuel photosynthesis, more CO2 may cause trees and other plants to growfaster. An elaborate apparatus allows researchers to pipe extra CO2 to a 30-metercircle of forest. They selected two nearby circles in each of three parts of a pineforest and randomly chose one of each pair to receive extra CO2. The responsevariable is the mean increase in base area for 30 to 40 trees in a circle during agrowing season. We measure this in percent increase per year. Here are one year’sdata:17

Pair Control plot Treated plot

1 9.752 10.5872 7.263 9.2443 5.742 8.675

(a) State the null and alternative hypotheses. Explain clearly why theinvestigators used a one-sided alternative.

(b) Carry out a test and report your conclusion in simple language.

(c) The investigators used the test you just carried out. Any use of thet procedures with samples this size is risky. Why?




18.35 Fungus in the air. The air in poultry-processing plants often contains fungusspores. Inadequate ventilation can affect the health of the workers. The problemis most serious during the summer. To measure the presence of spores, air samplesare pumped to an agar plate and “colony forming units (CFUs)” are counted afteran incubation period. Here are data from two locations in a plant that processes37,000 turkeys per day, taken on four days in the summer. The units are CFUs percubic meter of air.18

Day 1 Day 2 Day 3 Day 4

Kill room 3175 2526 1763 1090Processing 529 141 362 224

(a) Explain carefully why these are matched pairs data.

(b) The spore count is clearly higher in the kill room. Give sample means and a90% confidence interval to estimate how much higher. Be sure to state yourconclusion in plain English.

(c) You will often see the t procedures used for data like these. You should regardthe results as only rough approximations. Why?

18.36 Calcium and blood pressure. In a randomized comparative experiment onthe effect of calcium in the diet on blood pressure, researchers divided 54healthy white males at random into two groups. One group received calcium;the other, a placebo. At the beginning of the study, the researchers measuredmany variables on the subjects. The paper reporting the study gives x = 114.9and s = 9.3 for the seated systolic blood pressure of the 27 members of theplacebo group.

(a) Give a 95% confidence interval for the mean blood pressure in thepopulation from which the subjects were recruited.

(b) What conditions for the population and the study design are required by theprocedure you used in (a)? Which of these conditions are important for thevalidity of the procedure in this case?

18.37 The placebo effect. The placebo effect is particularly strong in patients withParkinson’s disease. To understand the workings of the placebo effect, scientistsmeasure activity at a key point in the brain when patients receive a placebothat they think is an active drug and also when no treatment is given.19 Thesame six patients are measured both with and without the placebo, at differenttimes.

(a) Explain why the proper procedure to compare the mean response to placebowith control (no treatment) is a matched pairs t test.

(b) The six differences (treatment minus control) had x = −0.326 ands = 0.181. Is there significant evidence of a difference between treatment andcontrol?

18.38 How much oil? How much oil wells in a given field will ultimately produce iskey information in deciding whether to drill more wells. Following are theestimated total amounts of oil recovered from 64 wells in the DevonianRichmond Dolomite area of the Michigan basin, in thousands of barrels.20




21.71 53.2 46.4 42.7 50.4 97.7 103.1 51.943.4 69.5 156.5 34.6 37.9 12.9 2.5 31.479.5 26.9 18.5 14.7 32.9 196 24.9 118.282.2 35.1 47.6 54.2 63.1 69.8 57.4 65.656.4 49.4 44.9 34.6 92.2 37.0 58.8 21.336.6 64.9 14.8 17.6 29.1 61.4 38.6 32.512.0 28.3 204.9 44.5 10.3 37.7 33.7 81.112.1 20.1 30.5 7.1 10.1 18.0 3.0 2.0

Take these wells to be an SRS of wells in this area.

(a) Give a 95% t confidence interval for the mean amount of oil recovered fromall wells in this area.

(b) Make a graph of the data. The distribution is very skewed, with several highoutliers. A computer-intensive method that gives accurate confidence intervalswithout assuming any specific shape for the distribution gives a 95% confidenceinterval of 40.28 to 60.32. How does the t interval compare with this? Should thet procedures be used with these data?

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18.39 Weeds among the corn. Velvetleaf is a particularly annoying weed in cornfields. It produces lots of seeds, and the seeds wait in the soil for years untilconditions are right. How many seeds do velvetleaf plants produce? Here arecounts from 28 plants that came up in a corn field when no herbicide was used:21

2450 2504 2114 1110 2137 8015 1623 1531 2008 1716721 863 1136 2819 1911 2101 1051 218 1711 164

2228 363 5973 1050 1961 1809 130 880

We would like to give a confidence interval for the mean number of seedsproduced by velvetleaf plants. Alas, the t interval can’t be safely used for thesedata. Why not?

The following exercises ask you to answer questions from data without having the stepsoutlined as part of the exercise. Follow the Formulate, Solve, and Conclude steps ofthe four-step process. The process is illustrated in Examples 18.2, 18.3, and 18.4. Itmay be helpful to restate in your own words the State information given in the exercise.

18.40 Natural weed control? Fortunately, we aren’t really interested in the number of

4STEPSTEP

seeds velvetleaf plants produce (see Exercise 18.39). The velvetleaf seed beetlefeeds on the seeds and might be a natural weed control. Here are the total seeds,seeds infected by the beetle, and percent of seeds infected for 28 velvetleaf plants:

Seeds 2450 2504 2114 1110 2137 8015 1623 1531 2008 1716Infected 135 101 76 24 121 189 31 44 73 12Percent 5.5 4.0 3.6 2.2 5.7 2.4 1.9 2.9 3.6 0.7

Seeds 721 863 1136 2819 1911 2101 1051 218 1711 164Infected 27 40 41 79 82 85 42 0 64 7Percent 3.7 4.6 3.6 2.8 4.3 4.0 4.0 0.0 3.7 4.3

Seeds 2228 363 5973 1050 1961 1809 130 880Infected 156 31 240 91 137 92 5 23Percent 7.0 8.5 4.0 8.7 7.0 5.1 3.8 2.6




Do a complete analysis of the percent of seeds infected by the beetle. Include a90% confidence interval for the mean percent infected in the population of allvelvetleaf plants. Do you think that the beetle is very helpful in controlling theweed? Follow the four-step process as illustrated in Example 18.2.

18.41 Auto crankshafts. Here are measurements (in millimeters) of a criticaldimension for 16 auto engine crankshafts:

224.120 224.001 224.017 223.982 223.989 223.961223.960 224.089 223.987 223.976 223.902 223.980224.098 224.057 223.913 223.999

The dimension is supposed to be 224 mm and the variability of the manufacturingprocess is unknown. Is there evidence that the mean dimension is not 224 mm?Do a complete analysis, following the four-step process as illustrated inExample 18.3.

18.42 Mutual-funds performance. Mutual funds often compare their performancewith a benchmark provided by an “index” that describes the performance of theclass of assets in which the fund invests. For example, the Vanguard InternationalGrowth Fund benchmarks its performance against the EAFE (Europe, Australasia,Far East) index. Table 18.4 gives the annual returns (percent) for the fund and theindex. Does the fund’s performance differ significantly from that of its benchmark?

(a) Explain clearly why the matched pairs t test is the proper choice to answerthis question.

(b) Do a complete analysis that answers the question posed. Follow the four-stepprocess as illustrated in Example 18.4.

18.43 Right versus left. The design of controls and instruments affects how easilypeople can use them. A student project investigated this effect by asking25 right-handed students to turn a knob (with their right hands) that moved an

T A B L E 1 8 . 4 A mutual fund versus its benchmark index

Fund Index Fund IndexYear return return Year return return

1982 5.27 −1.86 1994 0.76 7.781983 43.08 23.69 1995 14.89 11.211984 −1.02 7.38 1996 14.65 6.051985 56.94 56.16 1997 4.12 1.781986 56.71 69.44 1998 16.93 20.001987 12.48 24.63 1999 26.34 26.961988 11.61 28.27 2000 −8.60 −14.171989 24.76 10.54 2001 −18.92 −21.441990 −12.05 −23.45 2002 −17.79 −15.941991 4.74 12.13 2003 34.45 38.591992 −5.79 −12.17 2004 18.95 20.251993 44.74 32.56 2005 15.00 13.54




T A B L E 1 8 . 5 Performance times (seconds) using right-hand andleft-hand threads

Subject Right thread Left thread Subject Right thread Left thread

1 113 137 14 107 872 105 105 15 118 1663 130 133 16 103 1464 101 108 17 111 1235 138 115 18 104 1356 118 170 19 111 1127 87 103 20 89 938 116 145 21 78 769 75 78 22 100 116

10 96 107 23 89 7811 122 84 24 85 10112 103 148 25 88 12313 116 147

T A B L E 1 8 . 6 Absorption extent for twoversions of a drug

Reference GenericSubject drug drug

15 4108 17553 2526 11389 2779 1613

13 3852 225412 1833 1310

8 2463 212018 2059 185120 1709 187817 1829 1682

2 2594 26134 2344 2738

16 1864 23026 1022 1284

10 2256 30525 938 12877 1339 1930

14 1262 196411 1438 2549

1 1735 334019 1020 3050




indicator by screw action. There were two identical instruments, one with aright-hand thread (the knob turns clockwise) and the other with a left-handthread (the knob turns counterclockwise). Table 18.5 gives the times in secondseach subject took to move the indicator a fixed distance.22

(a) Each of the 25 students used both instruments. Discuss briefly how you woulduse randomization in arranging the experiment.

(b) The project hoped to show that right-handed people find right-hand threadseasier to use. Do an analysis that leads to a conclusion about this issue.

18.44 Comparing two drugs. Makers of generic drugs must show that they do notdiffer significantly from the “reference” drugs that they imitate. One aspect inwhich drugs might differ is their extent of absorption in the blood. Table 18.6gives data taken from 20 healthy nonsmoking male subjects for one pair ofdrugs.23 This is a matched pairs design. Numbers 1 to 20 were assigned at randomto the subjects. Subjects 1 to 10 received the generic drug first, and Subjects 11 to20 received the reference drug first. In all cases, a washout period separated thetwo drugs so that the first had disappeared from the blood before the subject tookthe second. Do the drugs differ significantly in absorption?

18.45 Practical significance? Give a 90% confidence interval for the mean timeadvantage of right-hand over left-hand threads in the setting of Exercise 18.43.Do you think that the time saved would be of practical importance if the taskwere performed many times—for example, by an assembly-line worker? To helpanswer this question, find the mean time for right-hand threads as a percent of themean time for left-hand threads.

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