Chapter 1 – Overview and Descriptive Statistics
SHORT ANSWER
1. Give one possible sample of size 4 from each of the following populations:
a. All daily newspapers published in the United States
b. All companies listed on the New York Stock Exchange
c. All students at your college or university
d. All grade point averages of students at your college or university
ANS:
a. Houston Chronicle, Des Moines Register, Chicago Tribune, Washington Post
b. Capital One, Campbell Soup, Merrill Lynch, Pulitzer
c. John Anderson, Emily Black, Bill Carter, Kay Davis
d. 2.58. 2.96, 3.51, 3.69
PTS: 1
2. A Southern State University system consists of 23 campuses. An administrator wishes to make an inference about
the average distance between the hometowns of students and their campuses. Describe and discuss several different
sampling methods that might be employed. Would this be an enumerative or an analytic study? Explain your
reasoning.
ANS:
One could take a simple random sample of students from all students in the California State University system and
ask each student in the sample to report the distance from their hometown to campus. Alternatively, the sample
could be generated by taking a stratified random sample by taking a simple random sample from each of the 23
campuses and again asking each student in the sample to report the distance from their hometown to campus.
Certain problems might arise with self reporting of distances, such as recording error or poor recall. This study is
enumerative because there exists a finite, identifiable population of objects from which to sample.
PTS: 1
3. A Michigan city divides naturally into ten district neighborhoods. How might a real estate appraiser select a sample
of single-family homes that could be used as a basis for developing an equation to predict appraised value from
characteristics such as age, size, number of bathrooms, and distance to the nearest school, and so on? Is the study
enumerative or analytic?
ANS:
One could generate a simple random sample of all single family homes in the city or a stratified random sample by
taking a simple random sample from each of the 10 district neighborhoods. From each of the homes in the sample
the necessary variables would be collected. This would be an enumerative study because there exists a finite,
identifiable population of objects from which to sample.
, PTS: 1
4. An experiment was carried out to study how flow rate through a solenoid valve in an automobile‘s pollution-control
system depended on three factors: armature lengths, spring load, and bobbin depth. Two different levels (low and
high) of each factor were chosen, and a single observation on flow was made for each combination of levels.
a. The resulting data set consisted of how many observations?
b. Is this an enumerative or analytic study? Explain your reasoning.
ANS:
a. Number observations equal 2 2 2=8
b. This could be called an analytic study because the data would be collected on an existing process.
There is no sampling frame.
PTS: 1
5. The accompanying data specific gravity values for various wood types used in construction .
.41 .41 .42 .42. .42 .42 .42 .43 .44
.54 .55 .58 .62 .66 .66 .67 .68 .75
.31 .35 .36 .36 .37 .38 .40 .40 .40
.45 .46 .46 .47 .48 .48 .48 .51 .54
Construct a stem-and-leaf display using repeated stems and comment on any interesting features of the display.
ANS:
One method of denoting the pairs of stems having equal values is to denote the stem by L, for ‗low‘ and the second
stem by H, for ‗high‘. Using this notation, the stem-and-leaf display would appear as follows:
3L 1 stem: tenths
3H 56678 leaf: hundredths
4L 000112222234
5L 144
5H 58
6L 2
6H 6678
7L
7H 5
The stem-and-leaf display on the previous page shows that .45 is a good representative value for the data. In
addition, the display is not symmetric and appears to be positively skewed. The spread of the data is .75 - .31 = .44,
which is .44/.45 = .978 or about 98% of the typical value of .45. This constitutes a reasonably large amount of
variation in the data. The data value .75 is a possible outlier.
PTS: 1
6. Temperature transducers of a certain type are shipped in batches of 50. A sample of 60 batches was selected, and
the number of transducers in each batch not conforming to design specifications was determined, resulting in the
following data:
0 4 2 1 3 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1
2 1 2 4 0 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
5 0 2 3 2 1 0 6 4 2 1 6 0 3 3 3 6 1 2 3
, a. Determine nfrequencies nand nrelative nfrequencies nfor nthe nobserved nvalues nof nx n= nnumber nof nonconforming
ntransducers nin na nbatch.
b. What nproportion nof nbatches nin nthe nsample nhas nat nmost nfour nnonconforming ntransducers? nWhat nproportion
nhas nfewer nthan nfour? nWhat nproportion nhas nat nleast nfour nnonconforming nunits?
ANS:
a.
Number nNonconforming Frequency Relative nFrequency
0 7 0.117
1 12 0.200
2 13 0.217
3 14 0.233
4 6 0.100
5 3 0.050
6 3 0.050
7 1 0.017
8 1 0.017
1.001
The nrelative n frequencies ndon’t nadd nup nexactly nto n1because nthey nhave nbeen nrounded
b. nThe nnumber nof nbatches nwith nat nmost n4 nnonconforming nitems nis n7+12+13+14+6=52, nwhich nis na nproportion nof
n52/60=.867. nThe nproportion nof nbatches nwith n(strictly) nfewer nthan n4 nnonconforming nitems nis n46/60=.767.
PTS: n n 1
7. The nnumber nof ncontaminating nparticles non na nsilicon nwafer nprior nto na ncertain nrinsing nprocess nwas ndetermined nfor
n each nwafer nin na nsample nsize n100, nresulting nin nthe nfollowing nfrequencies:
Number nof n particles Frequency Number nof n particles Frequency
0 1 8 12
1 2 9 4
2 3 10 5
3 12 11 3
4 11 12 1
5 15 13 2
6 18 14 1
7 10
a. What nproportion nof nthe nsampled nwafers nhad nat nleast ntwo nparticles? nAt nleast nsix nparticles?
b. What nproportion nof nthe nsampled nwafers nhad nbetween nfour nand nnine nparticles, ninclusive? nStrictly nbetween nfour
nand nnine nparticles?
ANS:
a. From nthis nfrequency ndistribution, nthe nproportion nof nwafers nthat ncontained nat nleast ntwo nparticles nis n(100-1-2)/100 n=
.97, nor n97%. nIn na nsimilar nfashion, nthe nproportion ncontaining nat nleast n6 nparticles nis n(100 n– n1-2-3-12-11-
15)/100 n= n56/100 n= n.56, nor n56%.
b. The nproportion ncontaining nbetween n4 nand n9 nparticles ninclusive nis n(11+15+18+10+12+4)/100 n= n70/100 n=
n.70, nor n70%. nThe nproportion nthat ncontain nstrictly nbetween n4 nand n9 n(meaning nstrictly nmore nthan n4 nand
nstrictly nless nthan n9) nis n(15+ n18+10+12)/100= n55/100 n= n.55, nor n55%.
, PTS: n n 1
8. The ncumulative nfrequency nand ncumulative nrelative nfrequency nfor na nparticular nclass ninterval nare nthe nsum nof
nfrequencies nand nrelative nfrequencies, nrespectively, nfor nthat ninterval nand nall nintervals nlying nbelow nit.
nCompute nthe ncumulative nfrequencies nand ncumulative nrelative nfrequencies nfor nthe nfollowing ndata:
75 89 80 93 64 67 72 70 66 85
89 81 81 71 74 82 85 63 72 81
81 95 84 81 80 70 69 66 60 83
85 98 84 68 90 82 69 72 87 88
ANS:
Class Frequency Relative Cumulative Cumulative
nFrequency nFrequency nRelative
nFrequency
60 n– nunder n65 3 .075 3 .075
65 n– nunder n70 6 .15 9 .225
70 n– nunder n75 7 .175 16 .40
75 n– nunder n80 1 .025 17 .425
80 n– nunder n85 12 .30 29 .725
85 n– nunder n90 7 .175 36 .90
90 n– nunder n95 2 .05 38 .95
95 n– nunder n100 2 .05 40 1.0
PTS: n n 1
9. Consider nthe nfollowing nobservations non nshear nstrength nof na njoint nbonded nin na nparticular nmanner:
30.0 4.4 33.1 66.7 81.5 22.2 40.4 16.4 73.7 36.6 109.9
a. Determine nthe nvalue nof nthe nsample nmean.
b. Determine nthe nvalue nof nthe nsample nmedian. nWhy nis nit nso ndifferent nfrom nthe nmean?
c. Calculate na ntrimmed nmean nby ndeleting nthe nsmallest nand nlargest nobservations. nWhat nis nthe
corresponding ntrimming npercentage? nHow ndoes nthe nvalue nof nthis
n n n compare nto nthe nmean nand
n median?
ANS:
a. The nsum nof nthe nn n= n11 ndata npoints nis n514.90, nso n n = n514.90/11 n= n46.81.
b. The nsample nsize n(n n= n11) nis nodd, nso nthere nwill nbe na nmiddle nvalue. nSorting nfrom nsmallest nto nlargest: n4.4 n 16.4
n 22.2
30.0 n 33.1 n 36.6 n 40.4 n 66.7 n 73.7 n 81.5 n 109.9. nThe nsixth nvalue, n36.6 nis nthe nmiddle, nor nmedian, nvalue.
nThe nmean ndiffers nfrom nthe nmedian nbecause nthe nlargest nsample nobservations nare nmuch nfurther nfrom nthe
nmedian nthan nare nthe nsmallest nvalues.
c. Deleting n the nsmallest n(x n= n4.4) nand nlargest n(x n= n109.9) nvalues, nthe nsum nof nthe nremaining n9 nobservations nis
n400.6. nThe ntrimmed nmean n n is n400.6/9 n= n44.51. nThe ntrimming npercentage nis n100(1/11) n= n9.1%. n
n lies nbetween nthe nmean nand nmedian.
PTS: n n 1
10. A nsample nof n26 noffshore noil nworkers ntook npart nin na nsimulated nescape nexercise, nresulting nin nthe naccompanying
n data non ntime n(sec) nto ncomplete nthe nescape:
373 370 364 366 364 325 339 393
356 359 363 375 424 325 394 402
SHORT ANSWER
1. Give one possible sample of size 4 from each of the following populations:
a. All daily newspapers published in the United States
b. All companies listed on the New York Stock Exchange
c. All students at your college or university
d. All grade point averages of students at your college or university
ANS:
a. Houston Chronicle, Des Moines Register, Chicago Tribune, Washington Post
b. Capital One, Campbell Soup, Merrill Lynch, Pulitzer
c. John Anderson, Emily Black, Bill Carter, Kay Davis
d. 2.58. 2.96, 3.51, 3.69
PTS: 1
2. A Southern State University system consists of 23 campuses. An administrator wishes to make an inference about
the average distance between the hometowns of students and their campuses. Describe and discuss several different
sampling methods that might be employed. Would this be an enumerative or an analytic study? Explain your
reasoning.
ANS:
One could take a simple random sample of students from all students in the California State University system and
ask each student in the sample to report the distance from their hometown to campus. Alternatively, the sample
could be generated by taking a stratified random sample by taking a simple random sample from each of the 23
campuses and again asking each student in the sample to report the distance from their hometown to campus.
Certain problems might arise with self reporting of distances, such as recording error or poor recall. This study is
enumerative because there exists a finite, identifiable population of objects from which to sample.
PTS: 1
3. A Michigan city divides naturally into ten district neighborhoods. How might a real estate appraiser select a sample
of single-family homes that could be used as a basis for developing an equation to predict appraised value from
characteristics such as age, size, number of bathrooms, and distance to the nearest school, and so on? Is the study
enumerative or analytic?
ANS:
One could generate a simple random sample of all single family homes in the city or a stratified random sample by
taking a simple random sample from each of the 10 district neighborhoods. From each of the homes in the sample
the necessary variables would be collected. This would be an enumerative study because there exists a finite,
identifiable population of objects from which to sample.
, PTS: 1
4. An experiment was carried out to study how flow rate through a solenoid valve in an automobile‘s pollution-control
system depended on three factors: armature lengths, spring load, and bobbin depth. Two different levels (low and
high) of each factor were chosen, and a single observation on flow was made for each combination of levels.
a. The resulting data set consisted of how many observations?
b. Is this an enumerative or analytic study? Explain your reasoning.
ANS:
a. Number observations equal 2 2 2=8
b. This could be called an analytic study because the data would be collected on an existing process.
There is no sampling frame.
PTS: 1
5. The accompanying data specific gravity values for various wood types used in construction .
.41 .41 .42 .42. .42 .42 .42 .43 .44
.54 .55 .58 .62 .66 .66 .67 .68 .75
.31 .35 .36 .36 .37 .38 .40 .40 .40
.45 .46 .46 .47 .48 .48 .48 .51 .54
Construct a stem-and-leaf display using repeated stems and comment on any interesting features of the display.
ANS:
One method of denoting the pairs of stems having equal values is to denote the stem by L, for ‗low‘ and the second
stem by H, for ‗high‘. Using this notation, the stem-and-leaf display would appear as follows:
3L 1 stem: tenths
3H 56678 leaf: hundredths
4L 000112222234
5L 144
5H 58
6L 2
6H 6678
7L
7H 5
The stem-and-leaf display on the previous page shows that .45 is a good representative value for the data. In
addition, the display is not symmetric and appears to be positively skewed. The spread of the data is .75 - .31 = .44,
which is .44/.45 = .978 or about 98% of the typical value of .45. This constitutes a reasonably large amount of
variation in the data. The data value .75 is a possible outlier.
PTS: 1
6. Temperature transducers of a certain type are shipped in batches of 50. A sample of 60 batches was selected, and
the number of transducers in each batch not conforming to design specifications was determined, resulting in the
following data:
0 4 2 1 3 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1
2 1 2 4 0 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
5 0 2 3 2 1 0 6 4 2 1 6 0 3 3 3 6 1 2 3
, a. Determine nfrequencies nand nrelative nfrequencies nfor nthe nobserved nvalues nof nx n= nnumber nof nonconforming
ntransducers nin na nbatch.
b. What nproportion nof nbatches nin nthe nsample nhas nat nmost nfour nnonconforming ntransducers? nWhat nproportion
nhas nfewer nthan nfour? nWhat nproportion nhas nat nleast nfour nnonconforming nunits?
ANS:
a.
Number nNonconforming Frequency Relative nFrequency
0 7 0.117
1 12 0.200
2 13 0.217
3 14 0.233
4 6 0.100
5 3 0.050
6 3 0.050
7 1 0.017
8 1 0.017
1.001
The nrelative n frequencies ndon’t nadd nup nexactly nto n1because nthey nhave nbeen nrounded
b. nThe nnumber nof nbatches nwith nat nmost n4 nnonconforming nitems nis n7+12+13+14+6=52, nwhich nis na nproportion nof
n52/60=.867. nThe nproportion nof nbatches nwith n(strictly) nfewer nthan n4 nnonconforming nitems nis n46/60=.767.
PTS: n n 1
7. The nnumber nof ncontaminating nparticles non na nsilicon nwafer nprior nto na ncertain nrinsing nprocess nwas ndetermined nfor
n each nwafer nin na nsample nsize n100, nresulting nin nthe nfollowing nfrequencies:
Number nof n particles Frequency Number nof n particles Frequency
0 1 8 12
1 2 9 4
2 3 10 5
3 12 11 3
4 11 12 1
5 15 13 2
6 18 14 1
7 10
a. What nproportion nof nthe nsampled nwafers nhad nat nleast ntwo nparticles? nAt nleast nsix nparticles?
b. What nproportion nof nthe nsampled nwafers nhad nbetween nfour nand nnine nparticles, ninclusive? nStrictly nbetween nfour
nand nnine nparticles?
ANS:
a. From nthis nfrequency ndistribution, nthe nproportion nof nwafers nthat ncontained nat nleast ntwo nparticles nis n(100-1-2)/100 n=
.97, nor n97%. nIn na nsimilar nfashion, nthe nproportion ncontaining nat nleast n6 nparticles nis n(100 n– n1-2-3-12-11-
15)/100 n= n56/100 n= n.56, nor n56%.
b. The nproportion ncontaining nbetween n4 nand n9 nparticles ninclusive nis n(11+15+18+10+12+4)/100 n= n70/100 n=
n.70, nor n70%. nThe nproportion nthat ncontain nstrictly nbetween n4 nand n9 n(meaning nstrictly nmore nthan n4 nand
nstrictly nless nthan n9) nis n(15+ n18+10+12)/100= n55/100 n= n.55, nor n55%.
, PTS: n n 1
8. The ncumulative nfrequency nand ncumulative nrelative nfrequency nfor na nparticular nclass ninterval nare nthe nsum nof
nfrequencies nand nrelative nfrequencies, nrespectively, nfor nthat ninterval nand nall nintervals nlying nbelow nit.
nCompute nthe ncumulative nfrequencies nand ncumulative nrelative nfrequencies nfor nthe nfollowing ndata:
75 89 80 93 64 67 72 70 66 85
89 81 81 71 74 82 85 63 72 81
81 95 84 81 80 70 69 66 60 83
85 98 84 68 90 82 69 72 87 88
ANS:
Class Frequency Relative Cumulative Cumulative
nFrequency nFrequency nRelative
nFrequency
60 n– nunder n65 3 .075 3 .075
65 n– nunder n70 6 .15 9 .225
70 n– nunder n75 7 .175 16 .40
75 n– nunder n80 1 .025 17 .425
80 n– nunder n85 12 .30 29 .725
85 n– nunder n90 7 .175 36 .90
90 n– nunder n95 2 .05 38 .95
95 n– nunder n100 2 .05 40 1.0
PTS: n n 1
9. Consider nthe nfollowing nobservations non nshear nstrength nof na njoint nbonded nin na nparticular nmanner:
30.0 4.4 33.1 66.7 81.5 22.2 40.4 16.4 73.7 36.6 109.9
a. Determine nthe nvalue nof nthe nsample nmean.
b. Determine nthe nvalue nof nthe nsample nmedian. nWhy nis nit nso ndifferent nfrom nthe nmean?
c. Calculate na ntrimmed nmean nby ndeleting nthe nsmallest nand nlargest nobservations. nWhat nis nthe
corresponding ntrimming npercentage? nHow ndoes nthe nvalue nof nthis
n n n compare nto nthe nmean nand
n median?
ANS:
a. The nsum nof nthe nn n= n11 ndata npoints nis n514.90, nso n n = n514.90/11 n= n46.81.
b. The nsample nsize n(n n= n11) nis nodd, nso nthere nwill nbe na nmiddle nvalue. nSorting nfrom nsmallest nto nlargest: n4.4 n 16.4
n 22.2
30.0 n 33.1 n 36.6 n 40.4 n 66.7 n 73.7 n 81.5 n 109.9. nThe nsixth nvalue, n36.6 nis nthe nmiddle, nor nmedian, nvalue.
nThe nmean ndiffers nfrom nthe nmedian nbecause nthe nlargest nsample nobservations nare nmuch nfurther nfrom nthe
nmedian nthan nare nthe nsmallest nvalues.
c. Deleting n the nsmallest n(x n= n4.4) nand nlargest n(x n= n109.9) nvalues, nthe nsum nof nthe nremaining n9 nobservations nis
n400.6. nThe ntrimmed nmean n n is n400.6/9 n= n44.51. nThe ntrimming npercentage nis n100(1/11) n= n9.1%. n
n lies nbetween nthe nmean nand nmedian.
PTS: n n 1
10. A nsample nof n26 noffshore noil nworkers ntook npart nin na nsimulated nescape nexercise, nresulting nin nthe naccompanying
n data non ntime n(sec) nto ncomplete nthe nescape:
373 370 364 366 364 325 339 393
356 359 363 375 424 325 394 402
