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Chapter 1: Q14E (page 25)

The accompanying data set consists of observations,on shower-flow rate (L/min) for a sample of n=129,houses in Perth, Australia (鈥淎n Application of Bayes,Methodology to the Analysis of Diary Records in a

Water Use Study,鈥� J. Amer. Stat. Assoc., 1987: 705鈥�711):

4.6 12.3 7.1 7.0 4.0 9.2 6.7 6.9 11.5 5.1

11.2 10.5 14.3 8.0 8.8 6.4 5.1 5.6 9.6 7.5

7.5 6.2 5.8 2.3 3.4 10.4 9.8 6.6 3.7 6.4

8.3 6.5 7.6 9.3 9.2 7.3 5.0 6.3 13.8 6.2

5.4 4.8 7.5 6.0 6.9 10.8 7.5 6.6 5.0 3.3

7.6 3.9 11.9 2.2 15.0 7.2 6.1 15.3 18.9 7.2

5.4 5.5 4.3 9.0 12.7 11.3 7.4 5.0 3.5 8.2

8.4 7.3 10.3 11.9 6.0 5.6 9.5 9.3 10.4 9.7

5.1 6.7 10.2 6.2 8.4 7.0 4.8 5.6 10.5 14.6

10.8 15.5 7.5 6.4 3.4 5.5 6.6 5.9 15.0 9.6

7.8 7.0 6.9 4.1 3.6 11.9 3.7 5.7 6.8 11.3

9.3 9.6 10.4 9.3 6.9 9.8 9.1 10.6 4.5 6.2

8.3 3.2 4.9 5.0 6.0 8.2 6.3 3.8 6.0

  1. Construct a stem-and-leaf display of the data.
  2. What is a typical, or representative, flow rate?
  3. Does the display appear to be highly concentrated or spread out?
  4. Does the distribution of values appear to be reasonably symmetric? If not, how would you describe the departure from symmetry?
  5. Would you describe any observation as being far from the rest of the data (an outlier)?

Short Answer

Expert verified

a.

The stem and leaf display is:

2

23

3

2344567789

4

01356889

5

00001114455666789

6

0000122223344456667789999

7

00012233455555668

8

02233448

9

012233335666788

10

2344455688

11

2335999

12

37

13

8

14

36

15

0035

16


17


18

9

b. A typical or representative flow rate is 7 L/min.

c. The display is concentrated.

d. The distribution of values does not appear to be reasonably symmetric. The departure from symmetry implies non-identical sides of the graph from reference mid point.

e. Observation 18.9 is the outlier., far from rest of the points.

Step by step solution

01

Given information

Theobservations on shower-flow rate (L/min) for a sample of n=129houses in Perth, Australia is provided.

02

Construct a stem and leaf diagram

a.

A stem-and-leaf display provides a visual representation of the dataset.

The steps to construct a stem-and-leaf display are as follows,

1) Select the leading digit for the stem and trailing digits for the leaves(only numbers at unit鈥檚 place).

2) Represent the stem digits vertically and similarly the trailing digits corresponding to the stem digits.

3) Mention the units for the display.

The stem and leaf display for the provided scenario is,

2

23

3

2344567789

4

01356889

5

00001114455666789

6

0000122223344456667789999

7

00012233455555668

8

02233448

9

012233335666788

10

2344455688

11

2335999

12

37

13

8

14

36

15

0035

16


17


18

9

Unit: 10|2=10.2

03

Provide a representative flow rate

b.

Referring to part a. the stem and leaf plot is,

2

23

3

2344567789

4

01356889

5

00001114455666789

6

0000122223344456667789999

7

00012233455555668

8

02233448

9

012233335666788

10

2344455688

11

2335999

12

37

13

8

14

36

15

0035

16


17


18

9

A representative flow rate is the middle flow rate; that is, 7 L/min.

Thus, a typical or representative flow rate is 7 L/min.

04

Check whether the data is highly concentrated or spread out

c.

From the stem and leaf plot display, the data appears to be clustered around the stem values of 5 to 10, with only one extreme value of 18.9.

Thus, it can be concluded by the observations in the plot that the display is concentrated.

05

Check whether the distribution is symmetric

d.

Referring to (a) part of the question,

From the stem and leaf display, it can be observed that the distribution appears to be positively skewed and not symmetric as the shape is not bell-shaped and most of the values lie at the top of the display.

A symmetric graph must have the almost same type of appearance on either end of the reference point(mostly the central point of the graph).

In this case, assuming the central value close to 9, the data is concentrated on the left end as compared to the right side.

06

Check if there is any outlier

e.

Referring to (a) part of the question,

From the stem and leaf display, it can be observed that observation 18.9 lies far away from other observations. This is an outlier that is present in the dataset.

Thus, observation 18.9 is the outlier.

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Most popular questions from this chapter

Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) be independent \({\rm{rv's}}\) with mean values \({{\rm{\mu }}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{\mu }}_{\rm{n}}}\) and variances \({\rm{\sigma }}_{\rm{1}}^{\rm{2}}{\rm{, \ldots ,\sigma }}_{{{\rm{\sigma }}^{\rm{*}}}}^{\rm{2}}\)Consider a function\({\rm{h}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}} \right)\), and use it to define a \({\rm{rv}}\)\({\rm{Y = h}}\left( {{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}} \right)\). Under rather general conditions on the \({\rm{h}}\)function, if the \({{\rm{\sigma }}_{\rm{i}}}\)'s are all small relative to the corresponding \({{\rm{\mu }}_{\rm{i}}}\)'s, it can be shown that (\({\rm{E(Y)\gg h}}\left( {{{\rm{\mu }}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{\mu }}_{\rm{n}}}} \right)\) and\({\rm{V(Y)\hat U }}{\left( {\frac{{{\rm{露h }}}}{{{\rm{露 }}{{\rm{x}}_{\rm{1}}}}}} \right)^{\rm{2}}}{\rm{ \times \sigma }}_{\rm{1}}^{\rm{2}}{\rm{ + L + }}{\left( {\frac{{{\rm{露h }}}}{{{\rm{露 }}{{\rm{x}}_{\rm{n}}}}}} \right)^{\rm{2}}}{\rm{ \times \sigma}}_{\rm{n}}^{\rm{2}}\))where each partial derivative is evaluated at \(\left( {{{\rm{x}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}} \right){\rm{ = }}\)\(\left( {{{\rm{\mu }}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{\mu }}_{\rm{n}}}} \right)\). Suppose three resistors with resistances \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{,}}{{\rm{X}}_{\rm{3}}}\)are connected in parallel across a battery with voltage\({{\rm{X}}_{\rm{4}}}\). Then by Ohm's law, the current is

\({\rm{Y = }}{{\rm{X}}_{\rm{4}}}\left( {\frac{{\rm{1}}}{{{{\rm{X}}_{\rm{1}}}}}{\rm{ + }}\frac{{\rm{1}}}{{{{\rm{X}}_{\rm{2}}}}}{\rm{ + }}\frac{{\rm{1}}}{{{{\rm{X}}_{\rm{3}}}}}} \right)\)

Let \({{\rm{\mu }}_{\rm{1}}}{\rm{ = 10}}\)ohms, \({{\rm{\sigma }}_{\rm{1}}}{\rm{ = 1}}{\rm{.0ohm,}}\quad {{\rm{\mu }}_{\rm{2}}}{\rm{ = 15}}\)ohms, \({{\rm{\sigma }}_{\rm{2}}}{\rm{ = 1}}{\rm{.0ohm,}}{{\rm{\mu }}_{\rm{3}}}{\rm{ = 20}}\)ohms, \({{\rm{\sigma }}_{\rm{3}}}{\rm{ = 1}}{\rm{.5ohms,}}{{\rm{\mu }}_{\rm{4}}}{\rm{ = 120\;V}}\),

\({{\rm{\sigma }}_{\rm{4}}}{\rm{ = 4}}{\rm{.0 \backslash mathrm\{ \;V}}\)}. Calculate the approximate expected value and standard deviation of the current (suggested by "6andom Samplings," CHEMTECH, \({\rm{1984: 696 - 697}}\)).

Consider the following sample of observations on coating thickness for low-viscosity paint ("Achieving a Target Value for a Manufacturing Process: \({\rm{A}}\)Case Study, \({\rm{97}}\)J. of Quality Technology, \({\rm{1992:22 - 26}}\)):

\(\begin{array}{*{20}{r}}{{\rm{.83}}}&{{\rm{.88}}}&{{\rm{.88}}}&{{\rm{1}}{\rm{.04}}}&{{\rm{1}}{\rm{.09}}}&{{\rm{1}}{\rm{.12}}}&{{\rm{1}}{\rm{.29}}}&{{\rm{1}}{\rm{.31}}}\\{{\rm{1}}{\rm{.48}}}&{{\rm{1}}{\rm{.49}}}&{{\rm{1}}{\rm{.59}}}&{{\rm{1}}{\rm{.62}}}&{{\rm{1}}{\rm{.65}}}&{{\rm{1}}{\rm{.71}}}&{{\rm{1}}{\rm{.76}}}&{{\rm{1}}{\rm{.83}}}\end{array}\)

Assume that the distribution of counting thickness is normal (a normal probability plot strongly supports this assumption).

a. Calculate a point estimate of the mean value of coating thickness, and state which estimator you used.

b. Calculate a point estimate of the median of the coating thickness distribution, and state which estimator you used.

C. Calculate a point estimate of the value that separates the largest of\({\rm{10\% }}\) all values in the thickness distribution from the remaining \({\rm{90\% }}\)and state which estimator you used. (Hint Express what you are trying to estimate in terms of \({\rm{\mu }}\)and \({\rm{\sigma }}\)

d. Estimate \({\rm{P}}\left( {{\rm{X < 1}}{\rm{.5}}} \right){\rm{,}}\)i.e., the proportion of all thickness values less than 1.5. (Hint: If you knew the values of \({\rm{\mu }}\)and \({\rm{\sigma }}\), you could calculate this probability. These values are not available, but they can be estimated.)

e. What is the estimated standard error of the estimator that you used in part (b)?

A transformation of data values by means of some mathematical function, such as\(\sqrt x \)or\(1/x\), can often yield a set of numbers that has 鈥渘icer鈥� statistical properties than the original data. In particular, it may be possible to find a function for which the histogram of transformed values is more symmetric (or, even better, more like a bell-shaped curve) than the original data. As an example, the article 鈥淭ime Lapse Cinematographic Analysis of Beryllium鈥�Lung FibroblastInteractions鈥� (Environ. Research,1983: 34鈥�43) reported the results of experiments designed to study the behavior of certain individual cells that had been exposed to beryllium. An important characteristic of such an individual cell is its interdivision time (IDT). IDTs were determined for a large number of cells, both in exposed (treatment) and unexposed(control) conditions. The authors of the articleused a logarithmic transformation, that is, transformed value=log(original value). Consider the following representative IDT data:

IDT log10(IDT) IDT log10(IDT) IDT log10(IDT)

28.1 1.45 60.1 1.78 21.0 1.32

31.2 1.49 23.7 1.37 22.3 1.35

13.7 1.14 18.6 1.27 15.5 1.19

46.0 1.66 21.4 1.33 36.3 1.56

25.8 1.41 26.6 1.42 19.1 1.28

16.8 1.23 26.2 1.42 38.4 1.58

34.8 1.54 32.0 1.51 72.8 1.86

62.3 1.79 43.5 1.64 48.9 1.69

28.0 1.45 17.4 1.24 21.4 1.33

17.9 1.25 38.8 1.59 20.7 1.32

19.5 1.29 30.6 1.49 57.3 1.76

21.1 1.32 55.6 1.75 40.9 1.61

31.9 1.50 25.5 1.41

28.9 1.46 52.1 1.72

Use class intervals 10-<20, 20-<30,鈥� to construct a histogram of the original data. Use intervals1.1-<1.2, 1.2-<1.3,鈥� to do the same for the transformed data. What is the effect of the transformation?

A sample of 77 individuals working at a particular office wasselected and the noise level (dBA) experienced by each individual was determined, yielding the followingdata (鈥淎cceptable Noise Levels for Construction Site Offices,鈥� Building Serv. Engr. Research and Technology, 2009: 87鈥�94).

55.3

55.3

55.3

55.9

55.9

55.9

55.9

56.1

56.1

56.1

56.1

56.1

56.1

56.8

56.8

57.0

57.0

57.0

57.8

57.8

57.8

57.9

57.9

57.9

58.8

58.8

58.8

59.8

59.8

59.8

62.2

62.2

63.8

63.8

63.8

63.9

63.9

63.9

64.7

64.7

64.7

65.1

65.1

65.1

65.3

65.3

65.3

65.3

67.4

67.4

67.4

67.4

68.7

68.7

68.7

68.7

69.0

70.4

70.4

71.2

71.2

71.2

73.0

73.0

73.1

73.1

74.6

74.6

74.6

74.6

79.3

79.3

79.3

79.3

83.0

83.0

83.0

Use various techniques discussed in this chapter to organize, summarize, and describe the data.

Two airplanes are flying in the same direction in adjacent parallel corridors. At time \({\rm{t = 10}}\), the first airplane is \({\rm{10}}\)km ahead of the second one. Suppose the speed of the first plane (km/hr.) is normally distributed with mean \({\rm{520\;}}\)and standard deviation \({\rm{10}}\) and the second plane鈥檚 speed is also normally distributed with mean and standard deviation \({\rm{500\; and\; 10}}\), respectively.

a. What is the probability that after \({\rm{2hr}}{\rm{. }}\)of flying, the second plane has not caught up to the first plane?

b. Determine the probability that the planes are separated by at most \({\rm{10km\; after\; 2hr}}{\rm{. }}\)
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