JERZY NEYMAN: A PRINCIPAL FOUNDER OF MODERN STATISTICAL THEORY
Step 3 Interpret the confidence interval
Note:The confidence interval is exact for normal populations and is approximately correct for large samples from nonnormal populations.
EXAMPLE 10.4 The Pooled t-Interval Procedure
Faculty Salaries Obtain a 95% confidence interval for the difference, μ1−μ2, between the mean salaries of faculty in private and public institutions.
Solution We apply Procedure 10.2.
Step 1 For a confidence level of1−α, use Table IV to findtα/2with df=n1+n2−2.
For a 95% confidence interval,α=0.05. From Table 10.6,n1=35 andn2=30, so df=n1+n2−2=35+30−2=63. In Table IV, we find that with df=63, tα/2 =t0.05/2=t0.025=1.998.
Step 2 The endpoints of the confidence interval forμ1−μ2are ( ¯x1−x¯2)± tα/2ãsp
(1/n1)+(1/n2),
wherespis the pooled sample standard deviation.
From Step 1,tα/2=1.998. Also,n1=35,n2=30, and, from Example 10.3, we know that ¯x1=98.19, ¯x2=83.18, andsp=25.19. Hence the endpoints of the con- fidence interval forμ1−μ2are
(98.19−83.18)±1.998ã25.19
(1/35)+(1/30), or 15.01±12.52. Thus the 95% confidence interval is from 2.49 to 27.53.
Step 3 Interpret the confidence interval.
Interpretation We can be 95% confident that the difference between the mean salaries of faculty in private institutions and public institutions is somewhere be- tween $2,490 and $27,530. In other words (see page 465), we can be 95% confident that the mean salary of faculty in private institutions exceeds that of faculty in public institutions by somewhere between $2,490 and $27,530.
Report 10.2
Exercise 10.51 on page 478
10.2 Inferences for Two Population Means:σs Assumed Equal 475
The Relation between Hypothesis Tests and Confidence Intervals
Hypothesis tests and confidence intervals are closely related. Consider, for example, a two-tailed hypothesis test for comparing two population means at the significance levelα. In this case, the null hypothesis will be rejected if and only if the (1−α)- level confidence interval forμ1−μ2 does not contain 0. You are asked to examine the relation between hypothesis tests and confidence intervals in greater detail in Ex- ercises 10.63–10.65.
THE TECHNOLOGY CENTER
Most statistical technologies have programs that automatically perform pooled t- procedures. In this subsection, we present output and step-by-step instructions for such programs.
EXAMPLE 10.5 Using Technology to Conduct Pooled t-Procedures
Faculty Salaries Table 10.5 on page 471 shows the annual salaries, in thousands of dollars, for independent samples of 35 faculty members in private institutions and 30 faculty members in public institutions. Use Minitab, Excel, or the TI-83/84 Plus to perform the hypothesis test in Example 10.3 and obtain the confidence interval required in Example 10.4.
Solution Letμ1andμ2denote the mean salaries of all faculty in private and public institutions, respectively. The task in Example 10.3 is to perform the hypothesis test
H0: μ1=μ2(mean salaries are the same) Ha: μ1=μ2(mean salaries are different)
at the 5% significance level; the task in Example 10.4 is to obtain a 95% confidence interval forμ1−μ2.
We applied the pooled t-procedures programs to the data, resulting in Out- put 10.1 on this and the next page. Steps for generating that output are presented in Instructions 10.1 on the next page.Note to Excel users:For brevity, we have pre- sented only the essential portions of the actual output.
As shown in Output 10.1, theP-value for the hypothesis test is about 0.02. Be- cause theP-value is less than the specified significance level of 0.05, we rejectH0. Output 10.1 also shows that a 95% confidence interval for the difference between the means is from 2.49 to 27.54.
MINITAB
OUTPUT 10.1 Pooledt-procedures on the salary data
EXCEL
OUTPUT 10.1 (cont.) Pooledt-procedures on the salary data
TI-83/84 PLUS
Using2-SampTTest Using2-SampTInt
INSTRUCTIONS 10.1 Steps for generating Output 10.1 MINITAB
1 Store the two samples of salary data from Table 10.5 in columns named PRIVATE and PUBLIC
2 ChooseStat ➤Basic Statistics➤2-Sample t. . . 3 Press the F3 key to reset the dialog box
4 SelectEach sample is in its own columnfrom the drop-down list box
5 Click in theSample 1text box and specify PRIVATE 6 Click in theSample 2text box and specify PUBLIC 7 Click theOptions. . . button
8 Click in theConfidence leveltext box and type95 9 Click the arrow button at the right of theAlternative
hypothesisdrop-down list box and select Difference=hypothesized difference 10 Check theAssume equal variancescheck box 11 ClickOKtwice
EXCEL
1 Store the two samples of salary data from Table 10.5 in columns named PRIVATE and PUBLIC
2 ChooseXLSTAT➤Parametric tests➤Two-sample t-test and z-test
3 Click the reset button in the lower left corner of the dialog box
4 Click in theSample 1selection box and then select the column of the worksheet that contains the PRIVATE data
5 Click in theSample 2selection box and then select the column of the worksheet that contains the PUBLIC data 6 Click theOptionstab
7 Click the arrow button at the right of theAlternative hypothesisdrop-down list box and selectMean 1 – Mean 2=D
8 Type5in theSignificance level (%)text box 9 ClickOK
10 Click theContinuebutton in theXLSTAT – Selections dialog box
TI-83/84 PLUS
Store the two samples of salary data from Table 10.5 in lists named PRIV and PUBL.
FOR THE HYPOTHESIS TEST:
1 PressSTAT, arrow over toTESTS, and press4 2 HighlightDataand pressENTER
3 Press the down-arrow key
4 Press2nd➤LIST, arrow down to PRIV, and press ENTERtwice
5 Press2nd➤LIST, arrow down to PUBL, and press ENTERtwice
6 Type1forFreq1, pressENTER, type1forFreq2, and pressENTER
7 Highlight=μ2and pressENTER
8 Press the down-arrow key, highlightYes, and press ENTER
9 Arrow down toCalculateand pressENTER FOR THE CI:
1 PressSTAT, arrow over toTESTS, and press0 2 HighlightDataand pressENTER
3 Press the down-arrow key
4 Press2nd➤LIST, arrow down to PRIV, and press ENTERtwice
5 Press2nd➤LIST, arrow down to PUBL, and press ENTERtwice
6 Type1forFreq1, pressENTER, type1forFreq2, and pressENTER
7 Type.95forC-Leveland pressENTER 8 HighlightYes, and pressENTER
9 Press the down-arrow key and pressENTER
10.2 Inferences for Two Population Means:σs Assumed Equal 477 Note to Minitab and Excel users: Although Minitab and Excel simultaneously per- form a hypothesis test and obtain a confidence interval, the type of confidence interval found depends on the type of hypothesis test. Specifically, Minitab and Excel compute a two-sided confidence interval for a two-tailed test and a one-sided confidence inter- val for a one-tailed test. To perform a one-tailed hypothesis test and obtain a two-sided confidence interval, apply Minitab’s or Excel’s pooledt-procedure twice: once for the one-tailed hypothesis test and once for the confidence interval specifying a two-tailed hypothesis test.
Exercises 10.2
Understanding the Concepts and Skills
10.33 Regarding the four conditions required for using the pooled t-procedures:
a. what are they? b. how important is each condition?
10.34 Explain whyspis called the pooled sample standard deviation.
In each of Exercises10.35–10.38, we have provided summary statis- tics for independent simple random samples from two populations.
Preliminary data analyses indicate that the variable under consid- eration is normally distributed on each population. Decide, in each case, whether use of the pooled t-test and pooled t-interval procedure is reasonable. Explain your answer.
10.35 x¯1=468.3,s1=38.2,n1=6, x¯2=394.6,s2=84.7,n2=14 10.36 x¯1=115.1,s1=79.4,n1=51,
¯
x2=24.3,s2=10.5,n2=19 10.37 x¯1=118,s1=12.04,n1=99,
¯
x2=110,s2=11.25,n2=80
10.38 x¯1=39.04,s1=18.82,n1=51,
¯
x2=49.92,s2=18.97,n2=53
In each of Exercises10.39–10.44, we have provided summary statis- tics for independent simple random samples from two populations.
In each case, use the pooled t-test and the pooled t-interval proce- dure to conduct the required hypothesis test and obtain the specified confidence interval.
10.39 x¯1=10,s1=2.1,n1=15, ¯x2=12,s2=2.3,n2=15 a. Two-tailed test,α=0.05
b. 95% confidence interval
10.40 x¯1=10,s1=4,n1=15, ¯x2=12,s2=5,n2=15 a. Two-tailed test,α=0.05
b. 95% confidence interval
10.41 x¯1=20,s1=4,n1=10, ¯x2=18,s2=5,n2=15 a. Right-tailed test,α=0.05
b. 90% confidence interval
10.42 x¯1=20,s1=4,n1=10, ¯x2=23,s2=5,n2=15 a. Left-tailed test,α=0.05
b. 90% confidence interval
10.43 x¯1=20,s1=4,n1=20, ¯x2=24,s2=5,n2=15 a. Left-tailed test,α=0.05
b. 90% confidence interval
10.44 x¯1=23,s1=10,n1=25, ¯x2=24,s2=12,n2=50 a. Right-tailed test,α=0.10
b. 95% confidence interval
Applying the Concepts and Skills
Preliminary data analyses indicate that you can reasonably con- sider the assumptions for using pooled t-procedures satisfied in Exer- cises10.45–10.50. For each exercise, perform the required hypothesis test by using either the critical-value approach or the P-value approach.
10.45 Doing Time. TheFederal Bureau of Prisonspublishes data inPrison Statisticson the times served by prisoners released from federal institutions for the first time. Independent random samples of released prisoners in the fraud and firearms offense categories yielded the following information on time served, in months.
Fraud Firearms
3.6 17.9 25.5 23.8
5.3 5.9 10.4 17.9
10.7 7.0 18.4 21.9
8.5 13.9 19.6 13.3
11.8 16.6 20.9 16.1
At the 5% significance level, do the data provide sufficient evidence to conclude that the mean time served for fraud is less than that for firearms offenses? (Note:x¯1=10.12,s1=4.90, ¯x2=18.78, and s2=4.64.)
10.46 Gender and Humor. Sense of humor of 200 male and 200 female students was examined in a paper. The researchers studied measures of abstract reasoning, verbal intelligence, humor production ability, and mating success. The students’ humor ability was tested on the basis of the number of captionswritten for funny images. Follow- ing are the number of captions given by a few participants.
Men Women
10 12 15 9 10 14 8 20 3 7
13 5 18 3 11 25 17 12 13 3
4 13 23 5 9 11 5 18 24 23
14 5 27 22 33 9 22 27 14 13
19 3 4 15 30 2 21 8 3 6
8 20 37 26 19 1 5 19 4 15
At the 1% significance level, do the data provide sufficient evidence to conclude that, on average, males have a better sense of humor and, in particular, a better frame of reference than females? (Note:
¯
x1=11.39,s1=4.14, ¯x2=9.85, ands2=3.67.)
10.47 Fortified Juice and PTH. V. Tangpricha et al. did a study to determine whether fortifying orange juice with Vitamin D would result in changes in the blood levels of five biochemical variables.
One of those variables was the concentration of parathyroid hor- mone (PTH), measured in picograms/milliliter (pg/mL). The re- searchers published their results in the paper “Fortification of Orange Juice with Vitamin D: A Novel Approach for Enhancing Vitamin D Nutritional Health” (American Journal of Clinical Nutrition, Vol. 77, pp. 1478–1483). Concentration levels were recorded at the beginning of the experiment and again at the end of 12 weeks. The following data, based on the results of the study, provide the decrease (negative values indicate increase) in PTH levels, in pg/mL, for those drinking the fortified juice and for those drinking the unfortified juice.
Fortified Unfortified
−7.7 11.2 65.8 −45.6 65.1 0.0 40.0
−4.8 26.4 55.9 −15.5 −48.8 15.0 8.8 34.4 −5.0 −2.2 13.5 −6.1 29.4
−20.1 −40.2 73.5 −20.5 −48.4 −28.7
At the 5% significance level, do the data provide sufficient evidence to conclude that drinking fortified orange juice reduces PTH level more than drinking unfortified orange juice? (Note:The mean and standard deviation for the data on fortified juice are 9.0 pg/mL and 37.4 pg/mL, respectively, and for the data on unfortified juice, they are 1.6 pg/mL and 34.6 pg/mL, respectively.)
10.48 Driving Distances. Data on household vehicle miles of travel (VMT) are compiled annually by theFederal Highway Administra- tionand are published inNational Household Travel Survey, Sum- mary of Travel Trends. Independent random samples of 15 midwest- ern households and 14 southern households provided the following data on last year’s VMT, in thousands of miles.
Midwest South
16.2 12.9 17.3 22.2 19.2 9.3 14.6 18.6 10.8 24.6 20.2 15.8 11.2 16.6 16.6 18.0 12.2 20.1 24.4 20.3 20.9 16.0 17.5 18.2
9.6 15.1 18.3 22.8 11.5
At the 5% significance level, does there appear to be a difference in last year’s mean VMT for midwestern and southern households?
(Note:x¯1=16.23,s1=4.06, ¯x2=17.69, ands2=4.42.)
10.49 Indian Spleen Length.Bhavan Dhingra et al. researchers in India, were interested in ultra-sonographic measurement of liver and spleen size in healthy Indian children. They published their findings in the article ”Normal values of liver and spleen size by Ultrasonogra- phy in Indian children” (Indian Pediatrics, Vol. 47, pp. 487–492). The researchers randomly sampled 347 male and 250 female healthy chil- dren in India. The mean and standard deviation of the spleen lengths for the male children were 7.9 cm and 0.94 cm, respectively, and those for the female children were 7.6 cm and 0.99 cm, respectively. At the 1% significance level, do the data provide sufficient evidence to conclude that a difference exists in mean spleen lengths of male and female Indian children?
10.50 Recess and Wasted Food.Wendy Bounds et al. conducted a study to determine, among other things, scheduling recess before lunch is one way to increase children’s food and nutrient consump- tion at lunch and reduce plate waste. Results were published in the
online article “Investigation of the School Professionals’perceptions and Practices Regarding Issues Influencing Recess Placement in Ele- mentary Schools” (National Food Service Management Institute, The University of Mississippi, 2008). Summary statistics for the score on the opinion of recess placement issues by randomly selected students are presented in the following table.
Lunch Before Recess Lunch After Recess
¯
x1=3.49 x¯2=2.73 s1=0.57 s2=0.54 n1=314 n2=314
At the 1% significance level, do the data provide sufficient evidence to conclude that the mean score for food wasted for lunches before recess exceeds that for lunches after recess?
In Exercises10.51–10.56, apply Procedure 10.2 on page 474 to obtain the required confidence interval. Interpret your result in each case.
10.51 Doing Time. Refer to Exercise 10.45 and obtain a 90% con- fidence interval for the difference between the mean times served by prisoners in the fraud and firearms offense categories.
10.52 Gender and Humor. Refer to Exercise 10.46 and obtain a 98% confidence interval for the difference between the mean number of captions for males and females.
10.53 Fortified Juice and PTH. Refer to Exercise 10.47 and find a 90% confidence interval for the difference between the mean reduc- tions in PTH levels for fortified and unfortified orange juice.
10.54 Driving Distances. Refer to Exercise 10.48 and determine a 95% confidence interval for the difference between last year’s mean VMTs by midwestern and southern households.
10.55 Indian Spleen Length. Refer to Exercise 10.49 and deter- mine a 99% confidence interval for the difference between mean spleen lengths of Indian male and female children.
10.56 Recess and Wasted Food. Refer to Exercise 10.50 and find a 98% confidence interval for the difference between the mean score for food wasted for lunches before recess and that for lunches after recess.
Working with Large Data Sets
10.57 Vegetarians and Omnivores. Philosophical and health is- sues are prompting an increasing number of Taiwanese to switch to a vegetarian lifestyle. In the paper “LDL of Taiwanese Vegetarians Are Less Oxidizable than Those of Omnivores” (Journal of Nutri- tion, Vol. 130, pp. 1591–1596), S. Lu et al. compared the daily intake of nutrients by vegetarians and omnivores living in Taiwan. Among the nutrients considered was protein. Too little protein stunts growth and interferes with all bodily functions; too much protein puts a strain on the kidneys, can cause diarrhea and dehydration, and can leach calcium from bones and teeth. Independent random samples of 51 fe- male vegetarians and 53 female omnivores yielded the data, in grams, on daily protein intake presented on the WeissStats site. Use the tech- nology of your choice to do the following.
a. Obtain normal probability plots, boxplots, and the standard devi- ations for the two samples.
b. Do the data provide sufficient evidence to conclude that the mean daily protein intakes of female vegetarians and female omnivores differ? Perform the required hypothesis test at the 1% significance level.
10.2 Inferences for Two Population Means:σs Assumed Equal 479 c. Find a 99% confidence interval for the difference between the mean
daily protein intakes of female vegetarians and female omnivores.
d. Are your procedures in parts (b) and (c) justified? Explain your answer.
10.58 Children of Diabetic Mothers. The paper “Correlations Be- tween the Intrauterine Metabolic Environment and Blood Pressure in Adolescent Offspring of Diabetic Mothers” (Journal of Pediatrics, Vol. 136, Issue 5, pp. 587–592) by N. Cho et al. presented findings of research on children of diabetic mothers. Past studies have shown that maternal diabetes results in obesity, blood pressure, and glucose- tolerance complications in the offspring. The WeissStats site provides data on systolic blood pressure, in mm Hg, from independent random samples of 99 adolescent offspring of diabetic mothers (ODM) and 80 adolescent offspring of nondiabetic mothers (ONM).
a. Obtain normal probability plots, boxplots, and the standard devi- ations for the two samples.
b. At the 5% significance level, do the data provide sufficient evi- dence to conclude that the mean systolic blood pressure of ODM children exceeds that of ONM children?
c. Determine a 95% confidence interval for the difference between the mean systolic blood pressures of ODM and ONM children.
d. Are your procedures in parts (b) and (c) justified? Explain your answer.
10.59 A Better Golf Tee? An independent golf equipment testing facility compared the difference in the performance of golf balls hit off a regular 2-3/4wooden tee to those hit off a 3Stinger Competi- tion golf tee. A Callaway Great Big Bertha driver with 10 degrees of loft was used for the test, and a robot swung the club head at approx- imately 95 miles per hour. Data on total distance traveled (in yards) with each type of tee, based on the test results, are provided on the WeissStats site.
a. Obtain normal probability plots, boxplots, and the standard devi- ations for the two samples.
b. At the 1% significance level, do the data provide sufficient evi- dence to conclude that, on average, the Stinger tee improves total distance traveled?
c. Find a 99% confidence interval for the difference between the mean total distance traveled with the regular and Stinger tees.
d. Are your procedures in parts (b) and (c) justified? Why or why not?
Extending the Concepts and Skills
10.60 In this section, we introduced the pooledt-test, which pro- vides a method for comparing two population means. In deriving the pooledt-test, we stated that the variable
z= ( ¯x1−x¯2)−(μ1−μ2) σ√
(1/n1)+(1/n2)
cannot be used as a basis for the required test statistic becauseσ is unknown. Why can’t that variable be used as a basis for the required test statistic?
10.61 The formula for the pooled variance,sp2, is given on page 469.
Show that, if the sample sizes,n1 andn2, are equal, thens2pis the mean ofs12ands22.
10.62 Simulation. In this exercise, you are to perform a computer simulation to illustrate the distribution of the pooledt-statistic, given in Key Fact 10.2 on page 469.
a. Simulate 1000 random samples of size 4 from a normally dis- tributed variable with a mean of 100 and a standard deviation of 16. Then obtain the sample mean and sample standard deviation of each of the 1000 samples.
b. Simulate 1000 random samples of size 3 from a normally dis- tributed variable with a mean of 110 and a standard deviation of 16. Then obtain the sample mean and sample standard deviation of each of the 1000 samples.
c. Determine the value of the pooled t-statistic for each of the 1000 pairs of samples obtained in parts (a) and (b).
d. Obtain a histogram of the 1000 values found in part (c).
e. Theoretically, what is the distribution of all possible values of the pooledt-statistic?
f. Compare your results from parts (d) and (e).
10.63 Two-Tailed Hypothesis Tests and CIs. As we mentioned on page 475, the following relationship holds between hypothesis tests and confidence intervals: For a two-tailed hypothesis test at the sig- nificance levelα, the null hypothesis H0:μ1=μ2 will be rejected in favor of the alternative hypothesisHa:μ1=μ2if and only if the (1−α)-level confidence interval forμ1−μ2does not contain 0. In each case, illustrate the preceding relationship by comparing the re- sults of the hypothesis test and confidence interval in the specified exercises.
a. Exercises 10.48 and 10.54 b. Exercises 10.49 and 10.55
10.64 Left-Tailed Hypothesis Tests and CIs. If the assumptions for a pooledt-interval are satisfied, the formula for a (1−α)-level upper confidence bound for the difference, μ1−μ2, between two population means is
( ¯x1−x¯2)+tαãsp
(1/n1)+(1/n2).
For a left-tailed hypothesis test at the significance levelα, the null hypothesis H0:μ1=μ2 will be rejected in favor of the alternative hypothesis Ha:μ1< μ2 if and only if the (1−α)-level upper con- fidence bound forμ1−μ2 is less than or equal to 0. In each case, illustrate the preceding relationship by obtaining the appropriate up- per confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise 10.45 b. Exercise 10.46
10.65 Right-Tailed Hypothesis Tests and CIs. If the assumptions for a pooledt-interval are satisfied, the formula for a (1−α)-level lower confidence bound for the difference, μ1−μ2, between two population means is
( ¯x1−x¯2)−tαãsp
(1/n1)+(1/n2).
For a right-tailed hypothesis test at the significance levelα, the null hypothesis H0:μ1=μ2 will be rejected in favor of the alternative hypothesisHa:μ1> μ2if and only if the (1−α)-level lower confi- dence bound forμ1−μ2is greater than or equal to 0. In each case, il- lustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.
a. Exercise 10.47 b. Exercise 10.50
10.66 Permutation Tests.With the advent of high-speed comput- ing, new procedures have been developed that permit statistical infer- ences to be performed under less restrictive conditions than those of classical procedures.Permutation testsconstitute one such collec- tion of new procedures. To perform a permutation test to compare two population means using independent samples, proceed as follows.
1. Combine the two samples.
2. Randomly selectn1members from the combined sample. Now treat thesen1members as the first sample and the remainingn2mem- bers as the second sample.