Tutorial work 1, Questions and answers - Tutorial Exercise 1 PDF

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STATS 7053: Statistics in Engineering Tutorial Exercise 1 Week of 4 August, 2014

1. Consider a numerical data set x1 , x2 , . . . , xn where each xi consists of at least two digits. If n is “small”, a quick way to get an informative visual representation of the data by hand is by constructing a stemplot, also called a stem-and-leaf plot. (See Devore: Sect. 1.2.) To do this, (a) Select one of more leading digits for the stem values. The trailing digits become the leaves. (b) List the possible stem values in a vertical column. (c) Record the leaves for each observation betside the corresponding stem value. (d) Indicate the units for stems and leaves somewhere in the display. For example, if a data set consists of exam scores between 0 and 100, a score of 92 would have a stem of 9 and a leaf of 2. The possible stems are the integers 0, 1, . . . , 10. (Devore: Sect.1.2, Exercise 10) The flexural strength of concrete is a measure of ability to resist failure in bending. The data below gives the flexural strength (in MegaPascal, MPa, where 1 Pa (Pascal)= 1.45−4 psi) of 27 beams of high-performance concrete: 5.9 8.2

7.2 8.7

7.3 7.8

6.3 9.7

8.1 7.4

6.8 7.7

7.0 9.7

7.6 7.8

6.8 7.7

6.5 11.6

7.0 11.3

6.3 11.8

7.9 10.7

9.0

(a) Construct a stem-and-leaf display of the data. What appears to be a representative (typical) strength value? Do the observations (measurements) appear to be highly concentrated about the representative value or rather spread out? Solution: The stem and leaf plot is (with the stems being “ones” and the leaves being “tenths”) 5 6 7 8 9 10 11

. . . . . . .

9 3 0 1 0 7 3

3 0 2 7

5 2 7 7

6

8

8 3

8 4

6

7

7

8

8

9

A representative value might be taken to be somewhere in the middle, perhaps about between 7.6 and 8. (For these data, the mean is 8.3 and the median is 7.7.) The observations appear reasonably spread out, mainly due to the three observations with 11 as the stem. (What constitutes large or small variation usually depends on the particular application. An often-used rule of thumb compares the range (largest value minus smallest value) with a representative value. Taking 8 as a representative value and 12-6=6 as the (approximate) range, the range is 6/8 = 0.75 or 75% of the representative value. Most researchers would call this a larige amount of variation.) (b) Does the display appear to be reasonably symmetric about a representative value, or would you describe the shape in some other way? Solution:No, the display does not appear symmetric. It seems to be right-skewed (positive skewness). (c) Do there appear to be any outlying strength values? Solution:There are no observations that are wildly different from the rest. (That said, the observations 11.3, 11.6, and 11.8 might be investigated as suspected outliers.) (d) What proportion of observations in this sample exceed (i.e., are strictly greater than) 10 MPa? Solution:4/27 = 0.148

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2. For data x1 , x2 , . . . , xn , prove that: (a)

n X

(xi − x ¯) = 0;

i=1

Solution: n n n X X X x ¯ xi − (xi − x ¯) =

=

(b)

n X

(xi − x ¯)2 =

i=1

n X

i=1

i=1

i=1

n¯ x − n¯ x = 0.

x2 . x2i − n¯

i=1

Solution: n X (xi − x ¯)2

n X xxi + x ¯2 ) (x2i − 2¯

=

i=1 n

i=1

X

=

x x2i − 2¯

n X

xi +

=

X

=

n X

x ¯2

i=1

i=1

i=1 n

n X

x2i − 2¯ x(n¯ x) + n¯ x2

i=1

x2i − n¯ x2

i=1

3. The following data are carbon contents (%) for 20 samples of coal. 87 86

86 84

85 83

87 83

86 82

87 84

86 83

81 79

77 82

85 73

(a) Calculate the sample mean, sample variance and sample standard deviation. Solution:¯ x = 83.3, s2 = 13.168, s = 3.629. (b) Calulate the sample median, lower quartile, upper quartile and IQR. Solution:The sorted data are: 73 84

77 85

79 85

The position of the median is m =

21 2

81 86

82 86

82 86

83 86

83 87

83 87

84 87

= 10.5 and hence the sample median is M =

The position of the lower quartile is q =

21 4

= 5.25 and hence LQ =

The position of the upper quartile is 3q = 3 × The interquartile range is IQR = 86 − 82 = 4. (c) Construct a boxplot of the data. Solution:

2

21 4

3×82+82 4

= 15.75 and hence U Q =

84+84 2

= 84.

= 82.

86+3×86 4

= 86.

74

76

78

80

82

84

86

The coal data

% Carbon

Note, U Q + 1.5IQR = 92. Since the maximum value in the data set is 87, the upper whisker extends up to that value. On the other hand, LQ − 1.5I QR = 76. The lower whisker extends to the smallest point ≥ 76 which is in this case 77. There is also one observation below 76, namely 73, and it is plotted separately.

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4. The yield point, X, for 120 sample lengths of a given cable, measured in N mm−2 , have been grouped as follows: Yield point 50 < X ≤ 80 80 < X ≤ 90 90 < X ≤ 100 100 < X ≤ 110 110 < X ≤ 120 120 < X ≤ 130 130 < X ≤ 150 150 < X ≤ 180 180 < X ≤ 240

Frequency 3 6 13 25 24 21 18 7 3

Construct a histogram for these data and comment on its shape. Note that the class intervals are not all of equal length. Solution:

Frequency 3 6 13 25 24 21 18 7 3

Relative frequency 0.025 0.05 0.1083 0.2083 0.2 0.175 0.15 0.0583 0.025

Density 0.000833 0.005000 0.010833 0.020833 0.020000 0.017500 0.007500 0.001944 0.000417

0.010 0.000

0.005

Density

0.015

0.020

Yield point 50 < X ≤ 80 80 < X ≤ 90 90 < X ≤ 100 100 < X ≤ 110 110 < X ≤ 120 120 < X ≤ 130 130 < X ≤ 150 150 < X ≤ 180 180 < X ≤ 240

50

100

150

200

Yield Point

5. For each of ten streets with bike lanes, investigators measured the distance X between the centre line and a cylist in the bike lane. They used photography to determine the distance Y between the 4

cyclist and a passing car on those same ten streets, recording all distances in feet. The data and a scatter plot are shown below.

car

7

8

9

10

11

Car 5.5 6.2 6.3 7.0 7.8 8.3 7.1 10.0 10.8 11.0

6

Centre 12.8 12.9 12.9 13.6 14.5 14.6 15.1 17.5 19.5 20.8

14

16

18

20

centre

(a) Find x ¯, y¯, sx , sy , sxy and r for these data. Solution:¯ x = 15.42, y¯ = 8, sx = 2.875, sy = 1.977, sxy = 5.46 and r = 0.961. (b) Suppose the measurements were converted to metres using the rule 1ft = 0.3048m. Find the sample covariance and sample correlation coefficient. Solution:The sample covariance would be 5.46 × 0.30482 = 0.50725. The sample correlation coefficient would be unchanged. ⋆

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