Module: Statistics (STS 201) BSc Year 2, Semester 4 Royal University of Bhutan COLLEGE OF NATURAL...

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Module: Statistics (STS 201)

BSc Year 2, Semester 4

Assignment No. 1

Date of submission: 21st April, 2014

Module Tutor: Dr. D. B. Gurung, Lecturer

Sangay Tshering

B.Sc. Sustainable Development

Year 2, Semester 4

Royal University of Bhutan COLLEGE OF NATURAL RESOURCES

Lobesa, Punakha

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Contents Question 1 : ................................................................................................................................................... 3

1. B.Sc. Animal Science students (Worked out in excel). ........................................................................ 3

2. Manual calculations of central tendency and dispersion of B.Sc. AS Student’s (Age) ........................ 4

3. Manual calculations of central tendency and dispersion of B.Sc. AS Student’s (Height) .................... 8

4. Manual calculations of central tendency and dispersion of B.Sc. AS Student’s (Weight) ................. 12

5. Age, Height and Weight data of B.Sc. Agriculture (Worked out in excel). ....................................... 16

6. Manual calculations of central tendency and dispersion of B.Sc. AG Student’s (Age) ...................... 17

7. Manual calculations of central tendency and dispersion of B.Sc. AG Student’s (Height) ................. 20

8. Manual calculations of central tendency and dispersion of B.Sc. AG Student’s (Weight) ................ 24

9. Age, Height and Weight data of B.Sc. Forestry (Worked out in excel). ............................................. 27

Manual calculations of central tendency and dispersion of B.Sc. Forestry Student’s (Age) .................. 28

10. Manual calculations of central tendency and dispersion of B.Sc. FO Student’s (Height) ................ 30

11. Manual calculations of central tendency and dispersion of B.Sc. FO Student’s (Weight) ............... 34

12. Age, Height and Weight data of B.Sc. SD student (Worked out in excel). ...................................... 37

13. Manual calculations of central tendency and dispersion of B.Sc. SD Student’s (Age) .................... 37

13. Manual calculations of central tendency and dispersion of B.Sc. SD Student’s (Height) ................ 41

14. Manual calculations of central tendency and dispersion of B.Sc. SD Student’s (Weight) ............... 44

Question 2: .................................................................................................................................................. 48

1. Age of 30 students from the statistics class (randomly selected) ........................................................ 48

2. Age interval of 30 students ................................................................................................................. 49

3. Report outliers if any .......................................................................................................................... 50

Question3. ................................................................................................................................................... 51

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Question 1 : For the age, height and weight of your B.Sc. Statistics class taking a sample size

of 10 each from the four departments of SD, AG, FO, & AS; compute the mean, median, mode,

deviation, sum of squares, variance, standard deviation, 95% confidence interval, and construct a

box whisker plots. Use SPSS software for the assignment (20 marks).

Solutions:

Computing mean, median, mode, deviation, sum of squares, variance, standard deviations, 95 %

confidence interval, and box whisker plots for:

1. B.Sc. Animal Science students (Worked out in excel).

Age Height (cm) Weight (kg)

x d=(x-x̄)

d²=

(x-x̄)² x d=(x-x̄) d²=(x-x̄)² x d=(x-x̄) d²=(x-x̄)²

35 2.1 4.41 177 12 144 76 8.9 79.21

33 0.1 0.01 176 11 121 74 6.9 47.61

34 1.1 1.21 165 0 0 87 19.9 396.01

33 0.1 0.01 168 3 9 70 2.9 8.41

31 -1.9 3.61 169 4 16 56 -11.1 123.21

31 -1.9 3.61 162 -3 9 64 -3.1 9.61

31 -1.9 3.61 150 -15 225 53 -14.1 198.81

33 0.1 0.01 156 -9 81 69 1.9 3.61

36 3.1 9.61 165 0 0 52 -15.1 228.01

32 -0.9 0.81 162 -3 9 70 2.9 8.41

Sum

329 0

26.9

1650 0 614 671 0 1102.9

Mean 32.9

165

67.1

Mode 33

165

70

Median 33

165

69.5

Table 1: Computation of central tendencies for age, height and weight of B.Sc. Animal Science

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2. Manual calculations of central tendency and dispersion of B.Sc. AS Student’s (Age)

1. Arithmetic Mean:

(𝑿) = ∑𝒙

𝒏=

𝑥1+𝑥2+⋯𝑥𝑛

𝑛

Where n= 10, Σx=329

Therefore 329

10= 32.9

B.Sc. AS students’ age ranges at around 32.9. So, the mean age is 32.9

2. Median (M): The Median is the 'middle value' in the data set.

So, M = n+1

2, Where n=10

Therefore: 10+1

2=

11

2= 𝟓.5

(position)

31 31 31 32 33 33 33 34 35 36

33(Median)

𝟑𝟑+𝟑𝟑

𝟐=

𝟔𝟔

𝟐= 𝟑𝟑

The position of the median of the class is 33. Therefore, the median is 33.

3. Mode :

Rearranging the data in ascending order

31 31 31 32 33 33 33 34 35 36

The mode is the value that occurs the most frequently in a data set or a probability distribution. So,

there is existence of bimodal in this data set so, smallest is shown here. Therefore mode is 31.

4.To find the Deviation and Sum of Squares (using MS Excel Worksheet)

http://en.wikipedia.org/wiki/Data_set

http://en.wikipedia.org/wiki/Probability_distribution

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Table 1.1 Deviation and Sum of Squares

x d=(x-x̄) d²=(x-x̄)²

35 2.1 4.41

33 0.1 0.01

34 1.1 1.21

33 0.1 0.01

31 -1.9 3.61

31 -1.9 3.61

31 -1.9 3.61

33 0.1 0.01

36 3.1 9.61

32 -0.9 0.81

Sum Σx=329 Σd=0 Σd²=26.9

Mean

(x̄) 32.9

mode 33

median 33

5.Deviation (d) = (𝑥 − 𝑥 ) = 0

6.Sum of Squares (SS)

The Squared of the deviation (d2) or Sum of Squares (SS) = 26.9 (from the above Excel Worksheet)

(OR) SS can be calculated using the formula:

𝑠𝑠 = 𝛴 (𝑥 − 𝑥 )2 = 𝟐𝟔. 𝟗

7.Variance = (S2) = SS

𝑛−1, where SS=26.9 and n=10

= 26.9

10−1 = 2.988 (for the sample population 𝑥 )

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8.Standard Deviation (S/SD): √𝑺𝑺

𝒏−𝟏 where SS=26.9 and n=10

= √𝟐𝟔.𝟗

𝟏𝟎−𝟏 = √2.988=1.728

This means that the “mean” is expected to be within the range of 32.9 ± 1.728, i.e. 31.172 to 34.628 of

sample mean.

9.Standard Error: SE = σ �̅� = 𝛔

√𝒏 where σ =1.729 and n=10

1.729

√10 =

1.729

3.162 = 0.5468

This means that the “mean” is estimated to be within the sample mean of 32.9 ± 0.5468 or) the mean of the

data ranges between lower limit of 32.353 to upper limit of 33.446.

10.Confidence Interval

CI 0.95 = x̅ ± t (0.05, df ) x SE

x̅ = sample mean, t (0.05, df) = value from t distribution table at 95% and df (degree of freedom) and SE=

standard error. 𝐂𝐈 = 𝒙 ± 𝐳 𝐱 𝐒𝐄 can be used for calculating 95% confidence interval using z value 2.262

for 95%

Upper Limit (UL) of CI at 95%:

(CI) = �̅� + ZxSE

Where 𝑥 =32.9, Z=2.262 and SE=0.5468

= 32.9 + (2.262x0.5468)

= 32.9+1.24 = 34.14

Lower Limit (LL) of CI at 95%:

(CI) = �̅� − ZxSE

Where 𝑥 =32.9, Z=2.262 and SE=0.5468

= 32.9 - (2.262x0.5468)

= 32.9-1.24 = 31.66

So, P (31.66< 𝒙 ̅ < 34.14)

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Fig1. SPSS generated Box Plot for the age of B.Sc. Animal Science students

The average age of the B.Sc. 4th cohort Animal Science students is 32.9 years. The above

whisker box plot indicates that, the minimum age in the class is 31 and the tallest is 36years.

There is no outlier in the age of B.Sc. 4th cohort Animal Science.

Therefore, at 95% CI for mean, the lower boundary lies at 31.66 and upper boundary at 34.14

So, it is 95% confident that the B.Sc. AS students’ age is between 31.66 and 34.14years.

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SPSS generated result showing mean, 95% Confidence Interval for Mean, median, variance and

Standard deviation for the age of B.Sc. AS students.

Fig 1 SPSS generated descriptive statistics for the age of Animal Science

3. Manual calculations of central tendency and dispersion of B.Sc. AS Student’s (Height)

1. Arithmetic mean: (𝑋) = ∑𝒙

𝒏=


𝑛


Therefore, 1650

10= 165


M = n+1

2 , Where n=10

So,10+1

2=

11

2= 𝟓.5 (Position)

150 156 162 162 165 165 168 169 176 177

165(Median)

= 165+165

2

= 330

2= 𝟏𝟔𝟓

Statistic

Std.

Error

Mean 32.9000 .54671

95% Confidence Interval for Mean Lower Bound 31.6633

Upper Bound 34.1367

Median 33.0000

Variance 2.989

Std. Deviation 1.72884

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3. Mode:


150 156 162 162 165 165 168 169 176 177

The mode is the value that occurs the most frequently in a data set. So, there is existence of

bimodal in this data set so, smallest is shown here. Therefore mode is 162.

4) To find the Deviation and Sum of Squares (using MS Excel Worksheet)

Table 1.2.Deviation and Sum of Squares

x d=(x-x̄) d²=(x-x̄)²

177 12 144

176 11 121

165 0 0

168 3 9

169 4 16

162 -3 9

150 -15 225

156 -9 81

165 0 0

162 -3 9

Sum Σx=1650 Σd=0 Σd²=614

Mean

(x̄) 165

mode 165

median 165

5) Deviation (d) = (𝑥 − �̅�) = 0

6) The Squared of the deviation (d2) or Sum of Squares (SS) = 614 (from the above Excel

Worksheet) (OR) SS can be calculated using the formula:


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𝒔𝒔 = 𝜮 (𝒙 − �̅�)𝟐 = 𝟔𝟏𝟒

7. Variance = (S2) = 𝐒𝐒

𝒏−𝟏 Where SS=614, n=10

= 𝟔𝟏𝟒

𝟏𝟎−𝟏 = 68.22 (for the sample population 𝑥 )

8. Standard Deviation (S/SD): √𝑺𝑺

𝒏−𝟏 = √

614

10−1 = √68.22 = 8.259.

This means that the “mean” is expected to be within the range of 165 ± 8.259, i.e. 156.741 to

173.259 of sample mean.

9. Standard Error: SE = σ �̅� = 𝛔

√𝒏

8.259

√10 =

8.259

3.162 = 2.6119

This means that the “mean” is estimated to be within 165 ± 2.6119.

10. Confidence Interval

CI 0.95 = x̅ ± t (0.05, df ) x SE

x̅ = sample mean, t (0.05, df) = value from t distribution table at 95% and df (degree of freedom)

and SE= standard error.So, 𝐂𝐈 = �̅� ± 𝐳 𝐱 𝐒𝐄 Can be used for calculating 95% confidence interval

using z value 2.262 for 95%

Upper Limit (UL) of CI at 95%

Where �̅� =165, z=2.262 and SE =2.6119

= 165 + (2.262x2.6119)

= 165+5.908 =170.908

Lower Limit (LL) of CI at 95%

Where �̅� =165, z=2.262 and SE =2.6119

= 165 – (2.262x2.6119)

= 165-5.908= 159.092

Therefore, it is 99% confident that the 4th batch B.Sc. Animal Science heights are between

159.092cm to 170.908cm.

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So, P (159.092< 𝒙 ̅ < 170.908)

Therefore, it is 95% confident that the B.Sc. AS students’ height is between 159.092 and 170.908

cm

Fig.2 Box Plot showing the height data

The average heights are 165cm. The Box Plot shows that the minimum height is 150cm and the

maximum is 177cm.There is one student whose height is 150cm showing outlier among the 4th

cohort B.Sc. Animal Science students.

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Standard deviation for the Height of B.Sc. AS students.

Fig 2 SPSS generated descriptive statistics for the Height of Animal Science

4. Manual calculations of central tendency and dispersion of B.Sc. AS Student’s (Weight)

1. Arithmetic mean:

(𝑿) = ∑𝒙

𝒏=


𝑛


= 671

10=67.1


M =n+1

2, Where n=10

10+1

2=

11


52 53 56 64 69 70 70 74 76 87

69.5(Median)

Statistic Std. Error

Mean 1.6500E2 2.61194

95% Confidence Interval for Mean Lower Bound 1.5909E2

Upper Bound 1.7091E2

Median 1.6500E2

Variance 68.222


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69+70

2=

139

2= 𝟔𝟗. 𝟓

3) Mode:


52 53 56 64 69 70 70 74 76 87

The mode is the value that occurs the most frequently in a data set or a probability distribution.

So, the mode in the data set is 70.

4. To find the Deviation and Sum of Squares (using MS Excel Worksheet)

Table 1: Deviation and Sum of Squares

x d=(x-x̄) d²=(x-x̄)²

76 8.9 79.21

74 6.9 47.61

87 19.9 396.01

70 2.9 8.41

56 -11.1 123.21

64 -3.1 9.61

53 -14.1 198.81

69 1.9 3.61

52 -15.1 228.01

70 2.9 8.41

Sum Σx=671 Σd=0 Σd²=1102.9

Mean

(x̄) 67.1

mode 70

median 69.5

5. Deviation (d) = (𝑥 − 𝑥 ̅) = 0


http://en.wikipedia.org/wiki/Probability_distribution

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6. The Squared of the deviation (d2) or Sum of Squares (SS) = 1102.9 (from the above Excel

Worksheet) (OR) SS can be calculated using the formula:

𝒔𝒔 = 𝜮 (𝒙 − �̅�)𝟐 =1102.9

7. Variance = (S2) =SS

𝑛−1. Where SS=1102.9 and n=10

= 1102.9

10−1

=122.54 (for the sample population 𝑥 )

8. Standard Deviation(S/SD) = √𝑆𝑆

𝑛−1 where SS=1102.9 and n=10

=√1102.9

10−1 = √122.54 = 11.06

This means that the “mean” of the 4th batch B.Sc. Animal Science (AS)weights are expected

to be within the range of 67.1 ±11.069, i.e. 56.031 to 78.169 of sample mean.

9. Standard Error: SE = σ �̅� = 𝛔

√𝒏 where σ =11.070

11.070

√10 =

11.070

3.162

= 3.5009.

This shows that the “mean” is estimated to be within the sample mean of 67.1 ± 3.5009 i.e.

63.599 to 70.601


CI 0.95 = x̅ ± t (0.05, df ) x SE


and SE= standard error. So 𝐂𝐈 = 𝒙 ± 𝐳 𝐱 𝐒𝐄 Can be used for calculating 95% confidence interval



Where�̅� = 𝟔𝟕. 𝟏, z=2.262 and SE=7.919

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67.1 + (2.262x3.5009)

= 67.1+ 7.919 =75.019


Where �̅� = 𝟔𝟕. 𝟏, z=2.262 and SE=7.919

= 67.1 – (2.262x3.5009)

= 67.1-7.91= 59.181

P (59.181< 𝒙 ̅ < 75.019) Therefore, it is 95% confident that the 4th cohort B.Sc. Animal

Science weights are between 59.181 and 75.019 kg because the lower and upper boundary

lies between 59.181 kg to 75.019kg.

Fig 3.Box Plot for the weight of BSc Animal Science students

The average weight is 67.1kg. The above whisker box plot shows that the minimum weight in

the class is 52kg and the maximum is 87kg. There is no outlier in the weight of B.Sc. 4th cohort

Animal Science.

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Standard deviation for the Weight of B.Sc. AS students.

fig 3 SPSS generated descriptive statistics for the weight of Animal Science

5. Age, Height and Weight data of B.Sc. Agriculture (Worked out in excel).


x d=(x-x̄) d²=(x-x̄)² x d=(x-x̄) d²=(x-x̄)² x d=(x-x̄) d²=(x-x̄)²

32 0.1 0.01 157 -4.6 21.16 59 -5.6 31.36

29 -2.9 8.41 158 -3.6 12.96 68 3.4 11.56

33 1.1 1.21 160 -1.6 2.56 60 -4.6 21.16

32 0.1 0.01 162 0.4 0.16 65 0.4 0.16

34 2.1 4.41 171 9.4 88.36 82 17.4 302.76

30 -1.9 3.61 150 -11.6 134.56 44 -20.6 424.36

30 -1.9 3.61 168 6.4 40.96 68 3.4 11.56

32 0.1 0.01 160 -1.6 2.56 60 -4.6 21.16

36 4.1 16.81 175 13.4 179.56 75 10.4 108.16

31 -0.9 0.81 155 -6.6 43.56 65 0.4 0.16

Sum 319 0 38.9 1616 0 526.4 646 0 932.4

mean 31.9

161.6

64.6

mode 32

160

60

median 32

160

65

Table 2: Computation of central tendencies for age, height and weight of B.Sc. AG students


Mean 67.1000 3.50063

95% Confidence Interval for Mean Lower Bound 59.1810

Upper Bound 75.0190

Median 69.5000

Variance 122.544

Std. Deviation 1.10700E

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6. Manual calculations of central tendency and dispersion of B.Sc. AG Student’s (Age)

1. Mean (𝑿) = ∑𝒙

𝒏

Where ∑𝑥 = 319, 𝑛 = 10

319

10= 31.9

2. Median: The Median is the 'middle value' in the data set.

= 10+1

2=

11


29 30 30 31 32 32 32 33 34 36

32(Median): 𝟑𝟐+𝟑𝟐

𝟐=

𝟔𝟒

𝟐= 𝟑𝟐

3. Mode:


29 30 30 31 32 32 32 33 34 36

The mode is the value that occurs the most frequently in a data set. In this data set, mode is 32



x d=(x-x̄) d²=x-x̄)²

32 0.1 0.01

29 -2.9 8.41

33 1.1 1.21

32 0.1 0.01

34 2.1 4.41

30 -1.9 3.61

30 -1.9 3.61

32 0.1 0.01

36 4.1 16.81

31 -0.9 0.81

sum Σx=319 Σd=0 Σd²=38.9

mean 31.9

mode 32

median 32


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5. Deviation (d) = (𝑥 − 𝑥 ) =0

6. Sum of Squares: The Squared of the deviation (d2) or Sum of Squares (SS) = 38.9 (from

the above Excel Worksheet) (OR) SS can be calculated using the formula:

SS= 𝜮 (𝒙 − 𝒙 )𝟐 = 𝟑𝟖. 𝟗

7. Variance = (S2) = 𝐒𝐒

𝒏−𝟏 =

𝟑𝟖.𝟗

𝟏𝟎−𝟏 = 4.322

8. Standard Deviation(S/SD): √𝑺𝑺

𝒏−𝟏 = √

𝟑𝟖.𝟗

𝟏𝟎−𝟏

= √4.322=2.078

This means that the 4th batch B.Sc. Agriculture students age are expected to be within the range

of 31.9 ± 2.078, i.e. 29.82 to 33.978 of sample mean.

9. Standard Error (SE): σ 𝑥 = 𝛔

√𝒏

𝟐.𝟎𝟕𝟗

√𝟏𝟎 =

𝟐.𝟎𝟕𝟗

𝟑.𝟏𝟔𝟐 = 0.657

This shows that the “mean” is estimated to be within the sample mean of 31.9 ± 0.657.

10. Confidence Interval (CI)

CI 0.95 = x̅ ± t (0.05, df ) x SE





Where x̅=31.9, z=2.262 and SE=0.657

= 31.9 + (2.262x0.657)

= 31.9+1.486 = 33.386


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Where x̅=31.9, z=2.262 and SE=0.657

= 31.9 - (2.262x0.657

= 31.9-1.486 = 30.414

Therefore, 30.414< 𝒙 ̅ < 33.386

At 95% CI for mean, the lower boundary lies at 30.414 and upper boundary at 33.386 so, the

BSc Agriculture ages are between 30.414years and 33.386 years.

Fig.4 SPSS generated Box Plot for the age of BSc 4th cohort Agriculture

The average age in the 4th cohort BSc Agriculture class is 31.9. The minimum age is 29 and the

maximum is 36 years. There is no outlier as their age ranges between 29 and 36 years.

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SPSS generated result showing mean, 95% Confidence Interval for

Mean, median, variance and Standard deviation for the age of B.Sc. AG

students.


Age Mean 31.9000 .65744

95% Confidence Interval for

Mean

Lower Bound 30.4128

Upper Bound 33.3872

Median 32.0000

Variance 4.322


Fig 4 SPSS descriptive statistics for age of B.Sc. AG

7. Manual calculations of central tendency and dispersion of B.Sc. AG Student’s (Height)

1. Mean = (𝑋) = ∑𝑥

𝑛 =

1616

10= 161.6


n+1

2=

10+1

2=

11


150 155 157 158 160 160 162 168 171 175

160(Median) = 160+160

2=

320

2= 𝟏𝟔𝟎

3. Mode: The mode is the value that occurs the most frequently in a data set.

Arranging the data in ascending order:

150 155 157 158 160 160 162 168 171 175

In this data set mode is 160



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Table 1.4 Deviations and Sum of Squares

x d=(x-x̄) d²=(x-x̄)²

157 -4.6 21.16

158 -3.6 12.96

160 -1.6 2.56

162 0.4 0.16

171 9.4 88.36

150 -11.6 134.56

168 6.4 40.96

160 -1.6 2.56

175 13.4 179.56

155 -6.6 43.56

sum Σx=1616 Σd=0 Σd²=526.4

mean

(x̄) 161.6

mode 160

median 160

5. Deviation (d) = (𝑥 − 𝑥 ) = 0

6. Sum of Squares (SS): The Squared of the deviation (d2) or Sum of Squares (SS) =

526.4(from the above Excel Worksheet) (OR) SS can be calculated using the formula:

𝛴 (𝑥 − 𝑥 )2 = 𝟓𝟐𝟔. 𝟒

7. Variance (S2) = SS


= 526.4

10−1 = 58.488

8. Standard Deviation(S/SD):√𝑆𝑆


= √526.4

10−1

= √58.488 = 7.647

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Means that the 4th batch B.Sc. agriculture heights are expected to be within the range of 161.6 ±

7.647, i.e. 153.95 to 169.247 of sample mean

9. Standard Error (SE) = σ 𝑥 =σ

√𝑛. Where σ =7.648 and n=10

= 7.648

√10 =

7.648

3.162 = 2.418

This means that the “mean” of the height is estimated to be within the sample mean of 161.6 ±

2.418


CI 0.95 = x̅ ± t (0.05, df ) x SE





Where 𝑥 =161.6, Z=2.262 and SE=2.418

= 161.6 + (2.262x2.418)

= 161.6+5.46 =167.06


Where 𝑥 =161.6, Z=2.262 and SE=2.41

= 161.6 – (2.262x2.418)

= 161.6-5.46= 156.14

So, 156.14< 𝒙 ̅ < 167.06

Therefore, at 95% CI for mean, the lower limit lies at 156.6 and upper limit at 167.06. So, it is

95% confident that the BSc Agriculture heights are between 156.14 and 167.06 cm.

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Fig 5.Box Plot for the heights of 4th cohort BSc Agriculture students

The Box Plot shows that the average height is 161.6cm. The minimum height is 150cm and the

maximum is 175cm. There is no outlier as their heights between 150 cm and 175cm.



students.


Mean 1.6160E2 2.41845


Mean

Lower Bound 1.5613E2


Median 1.6000E2

Variance 58.489


Fig 5. SPSS descriptive statistics for age, height and weight.

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8. Manual calculations of central tendency and dispersion of B.Sc. AG Student’s (Weight)

1. Mean (𝑋) = ∑𝑥

𝑛 where ∑𝑥 =646 and n=10

= 646

10=64.6


= n+1

2=

10+1

2=

11


44 59 60 60 65 65 68 68 75 82

65(Median): 65+65

2=

130

2= 𝟔𝟓

3. Mode: The mode is the value that occurs the most frequently in a data set.

Rearranging the data in ascending order:

44 59 60 60 65 65 68 68 75 82

In this data set there is existence of multimodal, so smallest is shown here as the mode i,e 60



x d=(x-x̄) d²=(x-x̄)²

59 -5.6 31.36

68 3.4 11.56

60 -4.6 21.16

65 0.4 0.16

82 17.4 302.76

44 -20.6 424.36

68 3.4 11.56

60 -4.6 21.16

75 10.4 108.16

65 0.4 0.16

sum Σx=646 Σd=0 Σd²=932.4

mean 64.6

mode 60

median 65


of 56

5. Deviation (d) = (𝑥 − 𝑥 ) = 0

6. Sum of Squares (SS) = 𝛴 (𝑥 − 𝑥 )2 = 𝟗𝟑𝟐. 𝟒

7. Variance = (S2) = SS

𝑛−1 =

932.4

10−1

=103.6


𝑛−1 = √

932.4

10−1

= √103.6 = 10.178

Therefore, mean is expected to be within the range of 64.6 ± 10.178, i.e. 54.422 to 74.778 of

sample mean.

9. Standard Error (SE) = σ 𝑥 = σ

√𝑛

= 10.178

√10 =

10.178

3.162 = 3.2188

This means that the 4th batch B.Sc. agriculture weight is estimated to be within the sample mean

of 64.6 ± 3.2188


CI 0.95 = x̅ ± t (0.05, df ) x SE





Where 𝑥 =64.6, Z=2.262 and SE=3.2188

= 64.6 + (2.262x3.2188)

= 64.6+7.28 =71.88


of 56

Where 𝑥 =64.6, Z=2.262 and SE=3.2188

= 64.6 – (2.262x3.2188

= 64.6-7.28 = 57.32

So, 57.32< 𝒙 ̅ < 71.88

It is 95% confident that the BSc Agriculture average weights are between 57.32and 71.88 kg

because the lower limit lies at 57.32kg and upper limit at 71.88kg.

Fig 6. Box Plot for the weight of 4th cohort Agriculture

The average weight is 64.6 kg. The minimum weight in the class is 44kg and the maximum is

82kg. There are two students whose weights are 44kg and 82 kg showing outliers in the Box Plot

among the 4th cohort Agriculture students.

of 56



students.


Weight Mean 64.6000 3.21870


Mean

Lower Bound 57.3188

Upper Bound 71.8812

Median 65.0000

Variance 103.600


Fig 6 S PSS descriptive statistics for age, height and weight.

9. Age, Height and Weight data of B.Sc. Forestry (Worked out in excel).

Age (x)

Height

(cm)

Weight

(Kg)

34.00 173.00 58.00

33.00 167.00 57.00

38.00 177.00 75.00

33.00 170.69 76.00

32.00 158.00 68.00

30.00 167.00 68.00

35.00 167.64 48.00

36.00 150.00 54.00

34.00 170.69 70.00

35.00 164.00 65.00

Sum Σx=340.00 Σx=1665.02 Σx=639.00

Mean

(x̄) 34.00 166.50 63.90

mode 33 167 68

median 34.00 167.32 66.50

of 56

Manual calculations of central tendency and dispersion of B.Sc. Forestry Student’s (Age)

1. Mean (𝑿) = ∑𝒙

𝒏 =

340

10= 34


n+1

2=

10+1

2=

11

2= 𝟓.5(Position)

30 32 33 33 34 34 35 35 36 38

34(Median): 𝟑𝟒+𝟑𝟒

𝟐=

𝟔𝟖

𝟐= 𝟑𝟒

3. Mode: Rearranging the data in ascending order

30 32 33 33 34 34 35 35 36 38

The mode is the value that occurs the most frequently in the data set. So, there is existence of

multimodal in this data set so, smallest is shown here. Therefore mode is 33.



Age (x) d=(x-x̄) d² =(x-x̄)²

34.00 0.00 0.00

33.00 -1.00 1.00

38.00 4.00 16.00

33.00 -1.00 1.00

32.00 -2.00 4.00

30.00 -4.00 16.00

35.00 1.00 1.00

36.00 2.00 4.00

34.00 0.00 0.00

35.00 1.00 1.00

Sum Σx=340.00 Σd=0 Σd²=44.00

Mean 34.00

mode 33

median 34.00

of 56

5. Deviation (d) = (𝑥 − 𝑥 ) =0

6. Sum of Squares (SS): The Squared of the deviation (d2) or Sum of Squares (SS) = 44

(from the above Excel Worksheet) (OR) SS can be calculated using the formula:

SS= 𝛴 (𝑥 − 𝑥 )2 = 𝟒𝟒


𝑛−1 =

44

10−1 = 4.89

8. Standard Deviation = (S/SD)

√𝑆𝑆

𝑛−1 = √

44

10−1 = √4.89=2.211

Therefore, about 99% of the 4th batch B.Sc. Forestry age is expected to be within the range of 34

± 2.211, i.e. 31.789 to 36.211 of sample mean

8. Standard Error (SE) = σ 𝑥 = 𝛔

√𝒏

2.211

√𝟏𝟎 =

𝟐.𝟐𝟏𝟏

𝟑.𝟏𝟔𝟐 = 0.699

This means that the mean is estimated to be within the sample mean of 34 ± 0.699.


CI 0.95 = x̅ ± t (0.05, df ) x SE





Where 𝑥 = 34, Z=2.262 and SE=0.699

= 34 + (2.262x0.699)

= 34+1.58 = 35.58


Where 𝑥 = 34, Z=2.262 and SE=0.699

of 56

= 34 - (2.262x0.699)

= 34-1.58 = 32.42

Therefore, 32.42< 𝒙 ̅ < 35.58 .So, at 95% CI for mean, the lower limit lies at 32.42 and upper

limit at 35.58 Therefore, BSc Forestry age is between 32.42and 35.58 years.

Fig 7.Whisker Box Plot for the age of 4th cohort BSc Forestry student

The Box Plot shows that the average age is 34. The minimum age is 30 and the maximum is

38years.Their age doesn’t show outlier in the SPSS generated Box Plot.


Standard deviation for the age of B.Sc. FO students.


Age Mean 34.0000 .69921


Mean

Lower Bound 32.4183

Upper Bound 35.5817

Median 34.0000

Variance 4.889


Fig 7. SPSS generated result for age of FO students

10. Manual calculations of central tendency and dispersion of B.Sc. FO Student’s (Height)

1. Mean = (𝑋) = ∑𝑥

𝑛

of 56

Where ∑𝑥 = 1665.02, and n=10

= 1665.02

10= 166.50

2. Median = n+1

2=

10+1

2=

11

2= 𝟓.5(Position)

150 158 164 167 167 167.64 170.69 170.69 173 177

167.32(Median):167+167.64

2=

334.64

2= 𝟏𝟔𝟕. 𝟑𝟐

3. Mode: Rearranging the data in ascending order

150 158 164 167 167 167 170 170 173 177

The mode is the value that occurs the most frequently in the data set. So, mode is 167



Height (cm) d=(x-x̄) d²=( x-x̄)²

173.00 6.50 42.23

167.00 0.50 0.25

177.00 10.50 110.22

170.69 4.19 17.53

158.00 -8.50 72.28

167.00 0.50 0.25

167.64 1.14 1.30

150.00 16.50 272.30

170.69 4.19 17.53

164.00 -2.50 6.26

Sum Σx=1665.02 Σd=0 Σd²=540.13

Mean 166.50

mode 167

median 167.32

5. Deviation (d) = (𝑥 − 𝑥 ) = 0

of 56

6. Sum of Squares (SS) = 𝛴 (𝑥 − 𝑥 )2 = 𝟓𝟒𝟎. 𝟏𝟑


𝑛−1 =

540.13

10−1 = 60.01


𝑛−1 = √

540.13

10−1

= √60.01 = 7.747

Means that the 4th batch B.Sc. Forestry heights are expected to be within the range of 166.50 ±

7.747, i.e. 158.753 to 174.247 of sample mean

9. Standard Error (SE) =σ 𝑥 = σ

√𝑛

= 7.747

√10 =

7.747

3.162 = 2.450

Therefore, the mean of the height is estimated to be within166.50 ± 2.450


CI 0.95 = x̅ ± t (0.05, df ) x SE





Where x̅=166.50, z=2.262 and SE= 2.450

= 166.50 + (2.262x2.450)

= 166.50+5.54

=172.04


Where x̅=166.50, z=2.262 and SE= 2.450

= 166.50 – (2.262x2.450)

= 166.50-5.54 = 160.96

So, 160.96< 𝒙 ̅ < 172.04

of 56

At 95% CI for mean, the lower boundary lies at 160.96cm and upper boundary at 172.04cm. So,

the BSc Forestry height is between 160.96 and 172.04 cm.

Fig 8 Whisker Box Plot for the height of 4th cohort BSc Forestry

In the 4th cohort B.Sc. Forestry, their average age lies in 166cm. The minimum height is 150 cm

and the maximum is 177cm. There is one student whose height is 150 cm showing outlier in the

SPSS generated Box Plot among the height of 4th cohort B.Sc. Forestry.


Standard deviation for the Height of B.Sc. FO students


Height Mean 1.6650E2 2.44986


Mean



Median 1.6732E2

Variance 60.018


Fig 8. SPSS descriptive results of the Height of B.Sc. Forestry

of 56

11. Manual calculations of central tendency and dispersion of B.Sc. FO Student’s (Weight)

1) Mean = (𝑋) = ∑𝑥

𝑛 =

639

10= 63.9

2) Median = n+1

2=

10+1

2 =

11


48 54 57 58 65 68 68 70 75 76

66.5(Median): 65+68

2=

133

2= 𝟔𝟔. 𝟓

3. Mode: The mode is the value that occurs the most frequently in the data set.

Arranging the data in ascending order:

48 54 57 58 65 68 68 70 75 76

In this data set the mode is 68 as it occurs frequently than other data.

4) To find the Deviation and Sum of Squares (using MS Excel Worksheet)


Weight

(Kg) d=(x-x̄) d²=(x-x̄) ²

58.00 -5.90 34.81

57.00 -6.90 47.61

75.00 11.10 123.21

76.00 12.10 146.41

68.00 4.10 16.81

68.00 4.10 16.81

48.00 15.90 252.81

54.00 -9.90 98.01

70.00 6.10 37.21

65.00 1.10 1.21

Sum Σx=639.00 Σd=0 Σd²=774.90

Mean 63.90

mode 68

median 66.50

of 56

5) Deviation (d) = (𝑥 − 𝑥 ) = 0

6) Sum of Squares (SS) = 𝛴 (𝑥 −)2 = 𝟗𝟑𝟐. 𝟒


𝑛−1 =

932.4

10−1

=103.6

8. Standard Deviation (S/SD):√𝑆𝑆

𝑛−1 = √

774.90

10−1

= √86.1 = 9.279

This means that the weights are expected to be within the range of 63.9 ± 9.279, i.e. 54.621 to

73.179 of sample mean.

9) Standard Error (SE) = σ 𝑥 = σ

√𝑛

= 9.279

√10 =

9.279

3.162 = 2.934

Therefore, 4th batch B.Sc. Forestry weight is estimated to be within the sample mean of 63.9 ±

2.934


CI 0.95 = x̅ ± t (0.05, df ) x SE





Where x̅=63.9, z=2.262 and SE=2.934

= 63.9 + (2.262x2.934)

= 63.9+6.64=70.54


Where x̅=63.9, z=2.262 and SE=2.934

of 56

= 63.9 – (2.262x2.934)

= 63.9-6.64 = 57.26

So, 57.26< 𝒙 ̅ < 70.54

Therefore, at 95% CI for mean, the lower limit lies at 57.26 and upper limit at 70.54. So, it is

95% confident that the BSc Forestry weights are between 57.26and 70.54 kg.

Fig 9.SPSS generated Box Plot for the height of BSc 4th cohort Forestry


Mean, median, variance and Standard deviation for the Weight of B.Sc. FO

students


Weight Mean 63.9000 2.93428


Mean

Lower Bound 57.2622

Upper Bound 70.5378

Median 66.5000

Variance 86.100


Fig 9. SPSS generated results for the weight of B.Sc. FO

The average weight of the 4th cohort BSc

Forestry is 63.9kg. The minimum weight

in the Forestry class is 48kg and the

maximum measures to 76kg as it shows

clearly in the SPSS generated Box Plot.

There is no outlier because weight of the

class ranges between 48kg and 76kg.

of 56

12. Age, Height and Weight data of B.Sc. SD student (Worked out in excel).


x d=(x-x̄) d²=(x-x̄)² x d=(x-x̄) d²=(x-x̄)² x d=(x-x̄) d²=(x-x̄)²

22 -2.2 4.84 160 -3 9 52 -6.7 44.89

22 -2.2 4.84 156 -7 49 59 0.3 0.09

23 -1.2 1.44 155 -8 64 46 -12.7 161.29

21 -3.2 10.24 158 -5 25 56 -2.7 7.29

33 8.8 77.44 164 1 1 65 6.3 39.69

34 9.8 96.04 180 17 289 90 31.3 979.69

22 -2.2 4.84 169 6 36 55 -3.7 13.69

22 -2.2 4.84 164 1 1 59 0.3 0.09

21 -3.2 10.24 154 -9 81 45 -13.7 187.69

22 -2.2 4.84 170 7 49 60 1.3 1.69

Sum 242 0 219.6 1630 0 604 587 0 1436.1

Mean 24

163

58.7

mode 22

164

59

median 22

162

57.5

Table 4: Computation of central tendencies for age, height and weight of B.Sc. SD students

13. Manual calculations of central tendency and dispersion of B.Sc. SD Student’s (Age)

1) Mean(𝑿) = ∑𝒙

𝒏 where ∑𝑥 = 242 𝑎𝑛𝑑 𝑛 = 10

=242

10=24.2

2) Median: The Median is the 'middle value' in the data set.

n+1

2=

10+1

2=

11

2= 𝟓.5

(Position)

21 21 22 22 22 22 22 23 23 34

22(Median): 22+22

2=

44

2= 𝟐𝟐


Rearranging the data in ascending order:

of 56

21 21 22 22 22 22 22 23 23 34

So, in this data set mode is 22


Table 1.7: Deviation and Sum of Squares

Age

x d=(x-x̄) d²=(x-x̄)²

22 -2.2 4.84

22 -2.2 4.84

23 -1.2 1.44

21 -3.2 10.24

33 8.8 77.44

34 9.8 96.04

22 -2.2 4.84

22 -2.2 4.84

21 -3.2 10.24

22 -2.2 4.84

Sum 242 0 219.6

Mean 24

mode 22

median 22

5. Deviation (d) = (𝒙 − 𝒙 ) =0


219.6(from the above Excel Worksheet) (OR) SS can be calculated using the formula:

𝛴 (𝑥 − 𝑥 )2 = 𝟐𝟏𝟗. 𝟔


𝑛−1 =

219.6

10−1 = 24.4

8. Standard Deviation = (S/SD)

of 56

√𝑺𝑺

𝒏−𝟏 = √

𝟐𝟏𝟗.𝟔

𝟏𝟎−𝟏 = √24.4=4.94

This means that the age is expected to be within the range of 24.2 ± 4.94, i.e. 19.26 to 29.14 of

sample mean.

9. Standard Error = SE = σ 𝑥 = 𝛔

√𝒏

= 4.94

√𝟏𝟎 =

4.94

3.162 = 1.562

This means that the Sustainable Development students age is estimated to be within the sample

mean of 24.2 ± 1.562


CI 0.95 = x̅ ± t (0.05, df ) x SE





Where x̅=24.2, z=2.262 and SE=1.562

= 24.2 + (2.262x1.562)

= 24.2+3.53 = 27.73


Where x̅=24.2, z=2.262 and SE=1.562

= 24.2 - (2.262x1.562)

= 24.2-3.53 = 20.67

This shows that 57.26< 𝒙 ̅ < 70.54

Therefore, at 95% CI for mean, the lower limit lies at 20.67 years and upper limit at 27.73 years.

So, it is 95% sure that the Sustainable Development students’ age is between 20.67and 27.73

years.

of 56

Fig10. Box Plot for the age of BSc Sustainable Development students

The average age in the BSc Sustainable Development class is 24.2 years. The lowest age is 21

and the highest is 34 years. There are two students whose ages are 33 and 34 showing outlier in

the SPSS generated Box plot for the age of the SD class.


Mean, median, variance and Standard deviation for the age of B.Sc.SD

students


Age Mean 24.2000 1.56205

95% Confidence Interval

for Mean

Lower Bound 20.6664

Upper Bound 27.7336

Median 22.0000

Variance 24.400


Fig 10. SPSS results for the age of SD students

of 56

13. Manual calculations of central tendency and dispersion of B.Sc. SD Student’s (Height)

1. Mean (𝑋) = ∑𝑥

𝑛 =

1630

10= 163


= n+1

2=

10+1

2=

11


154 155 156 158 160 164 164 179 170 180

162(Median): 160+164

2=

324

2= 𝟏𝟔𝟐



154 155 156 158 160 164 164 179 170 180

So, in this data set mode is 164



x d=(x-x̄) d²=(x-x̄)²

160 -3 9

156 -7 49

155 -8 64

158 -5 25

164 1 1

180 17 289

169 6 36

164 1 1

154 -9 81

170 7 49

Sum 1630 0 604

Mean 163

mode 164

median 162

of 56

5. Deviation (d) = (𝑥 − 𝑥 ) = 0


604(from the above Excel Worksheet) (OR) SS can be calculated using the formula:

𝛴 (𝑥 − 𝑥 )2 = 𝟔𝟎𝟒


𝑛−1 =

604

10−1 = 67.11

8. Standard Deviation(S/SD): √𝑆𝑆

𝑛−1 = √

604

10−1

= √67.11 = 8.192

This shows that the Sustainable Development students heights are expected to be within the

range of 163 ± 8.192, i.e. 154.808 to 171.192 of sample mean.

9. Standard Error (SE) = σ 𝑥 = σ

√𝑛

= 8.192

√10 =

8.192

3.162 = 2.591

This means that the Sustainable Development students’ height is estimated to be within the

sample mean of 163± 2.591i.e 160.409 to 165.591cm.


CI 0.95 = x̅ ± t (0.05, df ) x SE





Where x̅=163, z=2.262 and SE=2.591

= 163 + (2.262x2.591)

= 163+5.86 =168.86


of 56

Where x̅=163, z=2.262 and SE=2.591

= 163 – (2.262x2.591)

= 163-5.86 = 157.14

Shows that 157.14< 𝒙 ̅ < 168.86

Therefore, it is confident that at 95% CI for mean, the lower limit lies at 157.14 and upper limit

at 168.86. So, the BSc Sustainable Development heights are between 157.14and 168.86cm.

Fig11.SPSS generated Box Plot for height of SD students

The average height of the BSc Sustainable Development class is 163cm. The minimum height is 154 and

the maximum is 180cm. There is no outlier in the height of the SD class because the age ranges between

154 and 180 cm.

of 56


Mean, median, variance and Standard deviation for the Height of B.Sc.SD

students


Height Mean 1.6300E2 2.59058


for Mean



Median 1.6200E2

Variance 67.111


Fig 11.SPSS results for the height of SD students

14. Manual calculations of central tendency and dispersion of B.Sc. SD Student’s (Weight)

1. Mean (𝑋) = ∑𝑥

𝑛

Where ∑𝑥 =587 and n=10. So, 587

10=58.7


n+1

2=

10+1

2=

11


45 46 52 55 56 59 59 60 65 90

57.5(Median): 56+59

2=

115

2= 𝟓𝟕. 𝟓



45 46 52 55 56 59 59 60 65 90

So, in this rearranged data set mode is 59

4. To find the Deviation and Sum of Squares (using MS Excel Worksheet).

of 56


x d=(x-x̄) d²=(x-x̄)²

52 -6.7 44.89

59 0.3 0.09

46 -12.7 161.29

56 -2.7 7.29

65 6.3 39.69

90 31.3 979.69

55 -3.7 13.69

59 0.3 0.09

45 -13.7 187.69

60 1.3 1.69

Sum 587 0 1436.1

Mean 58.7

mode 59

median 57.5

5. Deviation (d) = (𝑥 − 𝑥 ) = 0

6. Sum of Squares (SS): Sum of Squares (SS): The Squared of the deviation (d2) or Sum of

Squares (SS) = 1436.1(from the above Excel Worksheet)

(OR) SS can be calculated using the formula:

𝛴 (𝑥 − 𝑥 )2 = 𝟏𝟒𝟑𝟔. 𝟏


𝑛−1

= 1436.1

10−1 =159.57


𝑛−1 = √

1436.1

10−1

= √159.57 = 12.632

of 56

Therefore, the Sustainable Development students weights are expected to be within the range of

58.7 ± 12.632, i.e. 46.068 to 71.332of sample mean.

9. Standard Error (SE) = σ 𝑥 =σ

√𝑛. Where σ= 12.632 and n=10

= 12.632

√10 =

12.632

3.162 = 3.994

This means that the Sustainable Development students weight is estimated to be within the

sample mean of 58.7± 3.994


CI 0.95 = x̅ ± t (0.05, df ) x SE





Where 𝑥 = 58.7, Z = 2.262 and SE = 3.994

= 58.7 + (2.262x3.994)

= 58.7+9.034=67.734


Where 𝑥 = 58.7, Z = 2.262 and SE = 3.994

= 58.7 – (2.262x3.994)

= 58.7-9.034= 49.67

The average weight is 58.7 of the

BSc Sustainable Development

students. Their minimum weight is

45 and the maximum weights

90kg. There is one student whose

age shows outlier in the SPSS

generated Box Plot for the weight

of Sustainable Development class.

Fig12. SPSS Generated Box Plot showing

the weight of Sustainable Development

students

of 56

SPSS generated result showing mean, 95% Confidence Interval for Mean, median, variance

and Standard deviation for the Height of B.Sc.SD students


Weight Mean 58.7000 3.99458


for Mean

Lower Bound 49.6636

Upper Bound 67.7364

Median 57.5000

Variance 159.567


Fig 12.SPSS results for the weight of SD students

of 56

Question 2: Using a sample of 30 students for the ‘age’ data of your statistics class

(considering the population) decide an appropriate class interval; construct a frequency

distribution table and draw appropriate graph(s). Using SPSS construct an error bar with one

SEM. Report outlier if any. (10 marks)

Solution:

1. Age of 30 students from the statistics class (randomly selected)

Fig 1 Age of 30 students from statistics class Fig 2. Age of 30 students rearranged in ascending order

Age data of statistics

class(30 students)

23 33

33 31

21 31

22 32

21 29

24 35

20 35

22 38

20 34

22 34

32 32

31 34

32 33

31 36

30 32

Age data rearranged in

ascending order (30 students)

20 32

20 32

21 32

21 32

22 32

22 33

22 33

23 33

24 34

29 34

30 34

31 35

31 35

31 36

31 38

of 56

2. Age interval of 30 students

Total 30

Fig 1.1 Frequency distribution of age data of statistic class

Fig. 1.2 Frequency graph for the age data of statistic class

Age interval Frequency

20-24 9

25-29 1

30-34 16

35-39 4

0

2

4

6

8

10

12

14

16

18

20-24 25-29 30-34 35-39

Fre

qu

ency

Age interval

Frequency distribution of age data

of 56

Fig. 1.3 Error bar with one standard error of mean (SEM)

3. Report outliers if any

Upper Limit Lower limit

Inter quartile range (IQR): Q3-Q1

Add 1.5xIQR to Q3

Q3 = 33.25

Q1 = 22.75

IQR= Q3-Q1

=33.25-22.75=10.5

1.5xIQR+Q3=1.5x10.5+33.25

=15.75+33.25=49

Subtract1.5xIQRfrom Q1

Q1 = 22.75

IQR= 33.25-22.75

=10.5

Q1-1.5xIQR = 22.75-1.5x10.5

=22.75-15.75=7

Fig.1.4 Table showing Upper

Limit and Lower Limit including

Q1, Q2, Q3 and IQR

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The Interquartile Range (IQR) = Q3 – Q1 = 33.25 – 22.75 = 10.5

Therefore, the outliers can be calculated by: 1.5 x IQR = 1.5 x 10.5 = 15.75

So the boundaries of the main body of the data, beyond which lie outliers are:

Upper boundary = Q3 + 1.5 IQR = 33.25 + 15.75 = 49

Lower Boundary = Q1 – 1.5 IQR = 22.75 – 15.75 =7

Therefore, in this data set, there are no outliers since all data points are within 7 and 49.

Home work

Question3. Consider the heights of your B. Sc class strength and compute the proportion (Z

score) of students taller than your own height. Also determine the percent of students who are

less than 165 cm tall.

Height (x) d= (x-x̄) d²=(x-x̄)²

165.00 2.37 5.60

157.00 -5.63 31.74

158.00 -4.63 21.47

160.00 -2.63 6.94

165.00 2.37 5.60

162.00 -0.63 0.40

171.00 8.37 70.00

154.00 -8.63 74.54

155.00 -7.63 58.27

160.00 -2.63 6.94

150.00 -12.63 159.60

170.00 7.37 54.27

168.00 5.37 28.80

149.00 -13.63 185.87

172.00 9.37 87.73

160.00 -2.63 6.94

170.00 7.37 54.27

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175.00 12.37 152.93

150.00 -12.63 159.60

152.00 -10.63 113.07

155.00 -7.63 58.27

155.00 -7.63 58.27

177.00 14.37 206.40

176.00 13.37 178.66

165.00 2.37 5.60

170.00 7.37 54.27

152.00 -10.63 113.07

155.00 -7.63 58.27

168.00 5.37 28.80

169.00 6.37 40.53

155.00 -7.63 58.27

147.00 -15.63 244.40

167.00 4.37 19.07

162.00 -0.63 0.40

150.00 -12.63 159.60

156.00 -6.63 44.00

165.00 2.37 5.60

162.00 -0.63 0.40

154.00 -8.63 74.54

156.00 -6.63 44.00

155.00 -7.63 58.27

162.00 -0.63 0.40

162.00 -0.63 0.40

159.00 -3.63 13.20

158.00 -4.63 21.47

165.00 2.37 5.60

170.00 7.37 54.27

162.00 -0.63 0.40

170.69 8.06 64.91

165.00 2.37 5.60

164.00 1.37 1.87

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167.64 5.01 25.07

150.00 -12.63 159.60

150.00 -12.63 159.60

173.00 10.37 107.47

167.00 4.37 19.07

158.00 -4.63 21.47

165.00 2.37 5.60

177.00 14.37 206.40

155.00 -7.63 58.27

192.02 29.39 863.57

173.74 11.11 123.36

177.00 14.37 206.40

155.00 -7.63 58.27

177.00 14.37 206.40

156.00 -6.63 44.00

181.00 18.37 337.33

170.00 7.37 54.27

161.54 -1.09 1.20

170.69 8.06 64.91

173.74 11.11 123.36

158.00 -4.63 21.47

167.00 4.37 19.07

160.00 -2.63 6.94

155.00 -7.63 58.27

167.64 5.01 25.07

161.54 -1.09 1.20

173.74 11.11 123.36

167.64 5.01 25.07

150.00 -12.63 159.60

170.69 8.06 64.91

155.00 -7.63 58.27

164.00 1.37 1.87

170.00 7.37 54.27

170.69 8.06 64.91

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167.64 5.01 25.07

162.00 -0.63 0.40

165.00 2.37 5.60

161.54 -1.09 1.20

156.00 -6.63 44.00

164.00 1.37 1.87

150.00 -12.63 159.60

160.00 -2.63 6.94

176.00 13.37 178.66

165.00 2.37 5.60

165.00 2.37 5.60

148.00 -14.63 214.14

156.00 -6.63 44.00

157.00 -5.63 31.74

155.00 -7.63 58.27

165.00 2.37 5.60

153.00 -9.63 92.80

160.00 -2.63 6.94

158.00 -4.63 21.47

154.00 -8.63 74.54

156.00 -6.63 44.00

158.00 -4.63 21.47

163.00 0.37 0.13

174.00 11.37 129.20

164.00 1.37 1.87

162.00 -0.63 0.40

167.00 4.37 19.07

180.00 17.37 301.60

150.00 -12.63 159.60

152.00 -10.63 113.07

172.00 9.37 87.73

169.00 6.37 40.53

179.00 16.37 267.86

164.00 1.37 1.87

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Fig 1 Summary data: Heights (cm) of B.Sc. students

The arithmetic mean 𝒙 = Σx/n. where ∑x=20329.18 and n=125 (Total BSc students)

=20329.18/125=19.20

The Sum of squares (SS) = 9032.465 (from the above Excel Worksheet)

Standard deviation (𝜎) =√𝑆𝑆

𝑛−1

Where SS/d²=9032.465

n=125

=√9032.465

125−1 = 8.53

The deviant of the normal distribution is calculated using z score, the formula is:

𝒛 =𝒙−𝝁

𝝈

Z score= x− 𝜇

𝜎 where x=169 cm (my own height)

𝜇 =162.63

𝜎 =8.53

=169−162.63

8.53= 6.37

8.53 =0.75

159.00 -3.63 13.20

155.00 -7.63 58.27

154.00 -8.63 74.54

150.00 -12.63 159.60

170.00 7.37 54.27

160.00 -2.63 6.94

∑x=20329.18 ∑d=0 ∑x²= 9032.465

𝜇= 162.63

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Now, referring to the areas of the normal distribution (z score table) at 0.75, the value derived is 0.2734

or 27.34% (0.2734x100), which means that 27.34% of the height is between 162.63 (𝝁) and 169cm(my

height)

So, 50%- 27.34% =22.66%

= 22.66% x 125(Total students)

=28.32≈28

This means that 28 persons have height taller than my height i.e. 169cm among the total students

of 125 (all BSc students).

Persons who are shorter than 165cm

50%+27.34%=77.34%

=77.34%x125 where 125 is the total population

=96.67≈97

Therefore, 76 persons have height shorter than 165cm.

Module: Statistics (STS 201) BSc Year 2, Semester 4 Royal University of Bhutan COLLEGE OF NATURAL...

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