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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.

Short Answer

Expert verified

The histogram is nearly unimodal and most of the data values lies in the class interval of\(56 - < 60\).

The five-number summary are Smallest\({x_i}\): 55.3, lower fourth: 57.8, median: 64.7, upper fourth: 70.8, and largest\({x_i}\): 83.

The boxplot shows that the data, is somewhat, negatively skewed. Because there are no outliers, there is less variation, and the upper whisker is larger than the lower one.

Step by step solution

01

Given information

The data are provided that consists of 77 observations on the noise level experienced by each individual.

02

Construct frequency distribution for the given data

The class width should be equal to 5. The following table shows three columns namely class, frequency and relative frequency:

Use this formula to calculate relative frequency of each data set,

\(relative{\rm{ }}frequency{\rm{ }}of{\rm{ }}a{\rm{ }}value = \frac{{number{\rm{ }}of{\rm{ }}times{\rm{ }}the{\rm{ }}value{\rm{ }}occurs}}{{number{\rm{ }}of{\rm{ }}observations{\rm{ }}in{\rm{ }}the{\rm{ }}data{\rm{ }}set}}\)

Class

Frequency

Relative Frequency

52 - < 56

7

\(0.0910\)

56 - < 60

23

0.2987

60 - < 64

8

0.1039

64 - < 68

14

0.1818

68 - < 72

10

0.1299

72 - < 76

8

0.1039

76 - < 80

4

0.0519

80 - < 84

3

0.0390

03

Construct histogram using a frequency distribution table

Following are the steps to make a histogram:

1. On the vertical axis, place the frequencies and label this axis as 鈥淧ercent鈥.

2. On the horizontal axis, place the lower value of each interval. Label this axis as 鈥淣oise Level (dBA).

3. Draw a bar extending from the lower value of each interval to the lower value of the next interval. The height of each bar should be equal to the frequency of its corresponding interval.

With a positive skew, the histogram is bimodal but near to unimodal. Almost one-third of the data values are concentrated in the class interval 50-60.

04

Compute five-number summary

The five-number summary is smallest \({x_i}\), lower fourth, median, upper fourth and largest\({x_i}\).Since sample size of test is odd, the median is the\({\left( {\frac{{n + 1}}{2}} \right)^{th}}\)ordered value when the data are in ascending order as below.

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

The smallest value is: 55.3 and the largest value is: 83.

Let \(\tilde x\) be the required median. Use the formula to calculate median when n is odd,

\(\tilde x = {\left( {\frac{{n + 1}}{2}} \right)^{th}}\,ordered\,value\)

\(\begin{aligned}\tilde x &= {\left( {\frac{{77 + 1}}{2}} \right)^{th}}\\ &= {39^{th\,}}observation\\ &= 64.7\end{aligned}\)

The lower fourth is the median of smallest half of the data as the median of the data is \(\tilde x = 64.7\)so lower half contains 41 values.

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

Since \(n = 41\)is odd, calculate the median using the formula:

\(\tilde x = {\left( {\frac{{n + 1}}{2}} \right)^{th}}\,ordered\,value\)

\(\begin{aligned}\tilde x &= {\left( {\frac{{41 + 1}}{2}} \right)^{th}}\,ordered\,value\\ &= {21^{th\,}}observation\\ &= 57.8\end{aligned}\)

Similarly, the upper fourth is the median of largest half of the data as the median of the data is\(\tilde x = 64.7\) so it contains 36 values.

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

Since \(n = 36\)is even, calculate the median using the formula:

\(\tilde x = \frac{{{{\left( {\frac{n}{2}} \right)}^{th}} + {{\left( {\frac{n}{2} + 1} \right)}^{th}}\,ordered\,value}}{2}\)

\(\begin{aligned}\tilde x &= \frac{{{{\left( {\frac{{36}}{2}} \right)}^{th}} + {{\left( {\frac{{36}}{2} + 1} \right)}^{th}}ordered\,value}}{2}\\ &= \frac{{\left( {70.4 + 71.2} \right)}}{2}\\ &= 70.8\end{aligned}\)

Thus, the five-number summary to construct boxplot are as follows:

Smallest \({x_i}\): 55.3, lower fourth: 57.8, median: 64.7, upper fourth: 70.8,

largest \({x_i}\): 83.

05

Construct box plot of measurements for the given data and draw conclusion

Following are the steps to make boxplot by hand:

1. Draw a plot line of range 55 to 85.

2. Draw three horizontal lines that consists of first quartile, second quartile and third quartile and make two vertical lines to make it in rectangular form like a box for given data.

3. Draw whiskers on both sides of a boxplot and set the minimum and maximum value with respect to the obtained lower fence and upper fence.

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Most popular questions from this chapter

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)?

The three measures of center introduced in this chapter are the mean, median, and trimmed mean. Two additional measures of center that are occasionally used are the midrange,which is the average of the smallest and largest observations, and the midfourth,which is the average of the two fourths. Which of these five measures of center are resistant to the effects of outliers and which are not? Explain your reasoning.

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

Consider the strength data for beams given in Example

1.2.

a. Construct a stem-and-leaf display of the data. What appears to be a representative strength value? Do the observations appear to be highly

concentrated about the representative value or rather spread out?

b. Does the display appear to be reasonably symmetric about a representative value, or would you describe its shape in some other way?

c. Do there appear to be any outlying strength values?

d. What proportion of strength observations in this sample exceeds 10 MPa?

The accompanying data set consists of observations on shear strength (lb) of ultrasonic spot welds made on a certain type of alclad sheet. Construct a relative frequency histogram based on ten equal-width classes with boundaries 4000, 4200, 鈥. [The histogram will agree with the one in 鈥淐omparison of Properties of Joints Prepared by Ultrasonic Welding and Other Means鈥 (J. of Aircraft, 1983: 552鈥556).] Comment on its features.

5434

4948

4521

4570

4990

5702

5241

5112

5015

4659

4806

4637

5670

4381

4820

5043

4886

4599

5288

5299

4848

5378

5260

5055

5828

5218

4859

4780

5027

5008

4609

4772

5133

5095

4618

4848

5089

5518

5333

5164

5342

5069

4755

4925

5001

4803

4951

5679

5256

5207

5621

4918

5138

4786

4500

5461

5049

4974

4592

4173

5296

4965

5170

4740

5173

4568

5653

5078

4900

4968

5248

5245

4723

5275

5419

5205

4452

5227

5555

5388

5498

4681

5076

4774

4931

4493

5309

5582

4308

4823

4417

5364

5640

5069

5188

5764

5273

5042

5189

4986

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