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ENGC 6310

Assignment 1- Answers

Problem 1:

Determine whether the given value is a statistic or a parameter.

1. The current Palestinian Legislative Council consists of 114 men and 17 women.

Answer: parameter

2. A sample of university students is selected and the average (mean) amount of money

spent on transportation is 80 NIS per month.

Answer: statistic

3. In a study of all 3568 school teachers in local public schools, it was found that 800

teachers have master's degree or higher.

Answer: parameter

Problem 2:

Determine which of the four levels of measurement (nominal, ordinal, interval, ratio) is most

appropriate.

1. Weights of students in the fifth grade of a school.

Answer: levels of measurement is interval because the data are differences but no

natural zero.

2. Performance ratings of a school teacher: Excellent, good, average, or poor.

Answer: levels of measurement is ordinal because the data is categories and can be

arranged in order.

3. National ID numbers.

Answer: levels of measurement is interval because the data are differences but no

natural zero.

4. Survey responses of yes, no, and undecided about a decision made by a class professor

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Answer: levels of measurement is nominal because the data are categories but

cannot be arranged in order.

5. A group of students develops a scale for rating the quality of the cafeteria food, with 0

representing “neutral: not good and not bad.” Bad meals are given negative numbers and

good meals are given positive numbers, with the magnitude of the number corresponding

to the severity of badness or goodness. The first three meals are rated as 2, 4, and _5.

What is the level of measurement for such ratings? Explain your choice.

Answer: levels of measurement is ratio because the data contain a natural zero

starting point.

Problem 3:

Identify the (a) sample and (b) population. Also, determine whether the sample is likely to be

representative of the population. Provide an explanation for your answer.

1. A local journalist stands on one famous street and asks 10 adults if they feel that the current

Palestinian Legislative Council is doing a good job.

Answer:

- Population: adult persons in Palestine.

- Sample: 10 persons, the sample is not likely representing the population because

it is small.

2. In a poll of 2000 randomly selected adults run in Gaza Strip, 61% said they own a solar

system for water heating when asked whether they own such system or not.

Answer:

- Population: adult persons in Gaza Strip.

- Sample: random sample, the sample is likely representing the population because

it is a random selection.

3. A graduate student at the University of Newport conducts a research project about how

adult Americans communicate. She begins with a survey mailed to 500 of the adults that

she knows. She asks them to mail back a response to this question: “Do you prefer to use

e-mail or snail mail (the U.S. Postal Service)?” She gets back 65 responses, with 42 of

them indicating a preference for snail mail.

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Answer:

- Population: adult American's.

- Sample: 65 persons, but the sample is not likely representing the population

because it is small and not randomly selected.

4. You were hired to poll a sample of adults about the models of cars they currently own. You

decided to use people with phone numbers listed in directories as the population from

which the sample is drawn. What is wrong with using people with telephone numbers listed

in directories as the population from which the sample is drawn?

Answer:

- Population: Models of cars that adult persons own.

- Sample: people that have phone numbers, but the sample is not likely representing

the population because the sample selection criteria is not accepted since the

people selected may have not cars.

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Problem 4:

Problem 6-8 (Montgomery): Preventing fatigue crack propagation in aircraft structures is an

important element of aircraft safety. An engineering study to investigate fatigue crack in n = 9

cyclically loaded wing boxes reported the following crack lengths (in mm):

2.13, 2.96, 3.02, 1.82, 1.15, 1.37, 2.04, 2.47, 2.60.

(a) Calculate the sample mean.

(b) Calculate the sample variance and sample standard deviation.

(c) Prepare a dot diagram of the data.

Answer:

- The Sample mean

�̅� = ∑𝑥𝑖

𝑛

𝑛

𝑖=1

=19.56

9= 𝟐. 𝟏𝟕𝟑𝟑𝟑𝟑

- The Sample variance

𝑠2 = ∑(𝑥𝑖 − �̅�)2

𝑛 − 1=

3.4428

9 − 1= 𝟎. 𝟒𝟑𝟎𝟑𝟓

𝑛

𝑖=1

- The Sample standard deviation

𝑠 = √𝑠2 = √0.43035 = 𝟎. 𝟔𝟓𝟔𝟎𝟏

- The dot diagram of the data

11.21.41.61.822.22.42.62.833.2

Dot Diagram �̅� = 𝟐. 𝟏𝟕

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Problem 5:

Problem 6-17 (Montgomery, 3rd ed.): 6-17. The following data represent the yield on 90

consecutive batches of ceramic substrate to which ………..

Stem Leaf Frequency

78 3 1

82 6 9 2

83 0 1 6 7 4

84 0 1 1 1 2 5 6 9 8

85 0 1 1 1 4 4 6

86 1 1 1 4 4 4 4 6 6 7 10

87 3 3 3 5 6 6 7 7

88 2 2 3 6 8 5

89 1 1 4 6 6 7 6

90 0 0 1 1 3 4 5 6 6 6 10

91 1 2 4 7 4

92 1 4 4 3

93 1 1 2 2 7 5

94 1 1 1 3 3 4 6 7 8

95 1 2 3 6 4

96 1 3 4 8 4

97 3 8 2

98 0 1

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Problem 6:

Problem 6-33 (Montgomery, 3rd ed.). Construct a frequency distribution and histogram for the

yield data in Exercise 6-17.………..

- Answer:

The number of interval is:

𝑲 = 𝟏 + 𝟑. 𝟑 𝐥𝐨𝐠𝟏𝟎(𝒏) = 𝟏 + 𝟑. 𝟑 𝐥𝐨𝐠𝟏𝟎(𝟗𝟎) = 𝟕. 𝟒𝟓 ≈ 𝟖

May be take 8 interval.

98≤x<101 95≤x<98 92≤x<95 89≤x<92 86≤x<89 83≤x<86 80≤x<83 77≤x<80 Class

1 10 16 20 22 18 2 1 Frequency

Relative

0.011111 0.111111 0.177778 0.222222 0.244444 0.2 0.022222 0.011111 Frequency

90 89 79 63 43 21 3 1 Cumulative

Frequency

1 0.988889 0.877778 0.7 0.477778 0.233333 0.033333 0.011111 Cumulative

Relative

Frequency

- Histogram for Frequency

0

5

10

15

20

25

77-79.99 80-82.99 83-85.99 86-88.99 89-91.99 92-94.99 95-97.99 98-101

Fre

qu

en

cy

yield

Histogram

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- Histogram for Cumulative Frequency

Problem 7:

Problem 6-47 (Montgomery, 3rd ed.). The following data are the temperatures of effluent at

discharge from a sewage treatment.………..

Answer:

(a) The sample mean and median.

- The Sample mean

�̅� = ∑𝒙𝒊

𝒏

𝒏

𝒊=𝟏

=𝟏𝟏𝟓𝟓

𝟐𝟒= 𝟒𝟖. 𝟏𝟐𝟓

- The median

𝒒𝟐 =𝟐(𝒏 + 𝟏)

𝟒=

𝟐(𝟐𝟒 + 𝟏)

𝟒= 𝟏𝟐. 𝟓

The median between value #12 and value #12, where the tow values are equal 49, then the

median equal 49.

(b) The sample variance and sample standard deviation.

- The Sample variance

𝒔𝟐 = ∑(𝒙𝒊 − �̅�)𝟐

𝒏 − 𝟏=

𝟏𝟔𝟔. 𝟔𝟐𝟓

𝟐𝟒 − 𝟏= 𝟕. 𝟐𝟒𝟓

𝒏

𝒊=𝟏

0

10

20

30

40

50

60

70

80

90

100

98-10195-97.9992-94.9989-91.9986-88.9983-85.9980-82.9977-79.99

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- The Sample standard deviation

𝒔 = √𝒔𝟐 = √𝟕. 𝟐𝟒𝟓 = 𝟐. 𝟔𝟗𝟐

(c) Construct a box plot of the data and comment on the information in this display.

𝒒𝟏 =(𝒏 + 𝟏)

𝟒=

(𝟐𝟒 + 𝟏)

𝟒= 𝟔. 𝟐𝟓 = 𝟒𝟔

𝒒𝟑 =𝟑(𝒏 + 𝟏)

𝟒=

𝟐(𝟐𝟒 + 𝟏)

𝟒= 𝟏𝟖. 𝟕𝟓 = 𝟓𝟎

- Box Plot Figure

𝒒𝟏 = 𝟒𝟔

𝒒𝟐 = 𝟒𝟗

𝒒𝟑 = 𝟓𝟎

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Problem 8:

By using the data of yearly rainfall measurements of the 8 stations in Gaza Strip:

1. Present the data as bar Chart (S1, S2), scatter plots (S2, S3), line charts (S3, S5) and

make interpretation on each chart.

- Bar Chart (S1, S2):

- In this chart it shown the value of yearly rainfall in two station is closed approximately

in all year.

- The maximum yearly rainfall recorded in station #1 about 660 mm in 1982, 1986 and

1991, either in the station #2 about 850 mm in 1991.

- The minimum yearly rainfall recorded in station #1 about 216 mm in 1985, either in

the station #2 about 250 mm in 1983,1984 and 1985.

0

100

200

300

400

500

600

700

800

900

1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994

Station #1 Station #2

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- Scatter Plots (S2, S3):

- In this chart it shown the value of yearly rainfall in two station is not closed

approximately in all year.

- The maximum yearly rainfall recorded in station #2 about 850 mm and in the station

#3 about 670 mm in 1991.

- The minimum yearly rainfall recorded in station #2 about 250 mm in 1983,1984 and

1985 and in the station #3 about 210 mm in 1980 and 1993.

- Line Charts (S3, S5):

0

100

200

300

400

500

600

700

800

1980 1981 1982 1983 1984 1985 1986 1987 1988 1989 1990 1991 1992 1993 1994

Station #3 Station #5

0

100

200

300

400

500

600

700

800

900

1978 1980 1982 1984 1986 1988 1990 1992 1994 1996

Station #2 Station #3

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- In this chart it shown the value of yearly rainfall in two station is not closed between

years 1987 and 1991.

- The maximum yearly rainfall recorded in station #3 about 670 mm in 1991 and in the

station #5 about 600 mm in 1986.

- The minimum yearly rainfall recorded in station #3 about 210 mm in 1980 and 1993

and in the station #5 about 160 mm in 1984.

2. Compute the mean, median, mode and the variance of all stations.

Station #1 Station

#2 Station

#3 Station

#4 Station

#5 Station

#6 Station

#7 Station

#8

Mean 427.69 454.71 412.66 358.68 351.22 320.5 316.82 238.16

Median 437.3 486 424 322 344.2 293.5 302.1 238

Mode #N/A 491 #N/A #N/A #N/A #N/A #N/A #N/A

Variance 22362.13067 30727.88 24795.06 21531.05 19035.94 15620.04 17113.09 8054.757

- No value of mode in all station because the value of rainfall didn’t repeat in two years

in any station.

3. Setup the histogram and the best model representing the distribution of the rainfall values

with time in S8.

- Answer:

The number of interval is:

𝐾 = 1 + 3.3 log10(𝑛) = 1 + 3.3 log10(15) = 4.88 ≈ 5

May be take 6 interval.

365≤x<415 315≤x<365 265≤x<315 215≤x<265 165≤x<215 110≤x<165 Class

1 2 2 3 4 3 Frequency Relative

0.066667 0.133333 0.133333 0.2 0.266667 0.2 Frequency

15 14 12 10 7 3 Cumulative Frequency

1 0.933333 0.8 0.666667 0.466667 0.2 Cumulative Relative Frequency

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- Histogram for Frequency

- Histogram for Cumulative Frequency

0

1

2

3

4

5

110-164.99 165-214.99 215-264.99 265-314.99 315-364.99 365-415

Fre

qu

en

cy

rainfall

Histogram

0

2

4

6

8

10

12

14

16

365-415315-364.99265-314.99215-264.99165-214.99110-164.99