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STA 2023 EXAM-1

Few Solutions for the

Practice Problems

Mudunuru, Venkateswara Rao

STA 2023

Fall 2015

From Chapters 1, 2, and 3

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© Venkateswara Rao Mudunuru Fall 2015 Exam-1 PRACTICE SHEET Few Solutions 1

Practice Problem Sheet

Important vocabulary from Chapter-1:

1) For the following data given in tables A, B, C. Do the following:

a) Find 10%, 15% trimmed mean.

b) Write the five number summary. Calculate IQR.

c) Draw a box and whisker plot.

12 25 10 38 12 19 8 12

41 7 22 10 19 12 16 5

31.0 29.8 37.4 40.1 35.1 41.5 12.6 39.7 28.4

38.6 187.4 40.6 39.7 31.0 29.8 42.0 30.8 35.5

Statistics Survey Bias Stratified

Inferential Parameter Cluster Sampling Error

Systematic Subjects Convenience Blind/Single Blind

Quantitative Census Individuals Systematic

Qualitative Ordinal Simple Random Continuous

Interval Statistic Discrete Descriptive

Ratio Double Blind Nominal Multi-Stage

Sampling

A)

B)

C) 9 12 15 11 8 10 5 11 14

13 6 11 16 11 14 4 12 11 Venk

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http://www.facebook.com/venkatvnv.significantly.insignificant


2) For the interval data given in tables D, E, F, G. Calculate the following. Write your values

of (b),(c),(d),(e) in the Table format given below:

Note: Remember that the table with the following column headings with details of first two

columns and the required number of rows based on the problem will be given on the exam.

a) Class width.

b) Cumulative frequency

c) Class midpoint (or class mark) d) Draw a frequency histogram e) Class boundaries

f) Draw a ogive graph

g) Relative frequency

h) Draw a relative frequency

histogram

Interval Frequency Class

Midpoint

Class

Boundaries

Relative

Frequency

Cumulative

Frequency

D)

Weight(lb) Client

frequency

30-49 8

50-69 10

70-89 12

90-109 7

E)

Time in

seconds

Freq

1 to 10 15

11 to 20 20

21 to 30 29

31 to 40 25

41 to 50 22

51 to 60 14

Cost

in $

Sales

frequency

10-26 2

27-43 15

44-60 48

61-77 23

F)

G)

Length in

Meters Frequency

1.1 to 1.5 2

1.6 to 2.0 4

2.1 to 2.5 7

2.6 to 3.0 10

3.1 to 3.5 11

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3) Estimate the sample mean, sample standard deviation, Sample variance for the

following interval data in tables D, E, F, and G. Use the Table format given

below.

Note: Remember that the table with the following column headings with details of first

two columns and the required number of rows based on the problem will be given on the

exam.

Interval 𝑭𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚, 𝒇 𝑴𝒊𝒅𝒑𝒐𝒊𝒏𝒕, 𝒙 𝒐𝒓 𝒎 𝒙𝒇 𝒙𝟐𝒇

4) Draw Stem and Leaf graph for the following and identify the shape. (Symmetric/Right Skewed/Left Skewed)

a) Exam Scores: 35, 36, 38, 40, 42, 42, 44, 45, 45, 47, 48, 49, 50, 50, 50.

b) Final grades of 33 students:

85,77,91,85,90,79,81,92,99,86,87,88,93,88,87,90,94,82,76,72,77,72,89,

69,62,63,60,59,58,90,45,42,40.

5) Draw the Pie Chart for the following data in tables H, I, J

Note: Remember that the table with the following column headings with details of first

two columns and the required number of rows based on the problem will be given on the

exam

6) Extract the data from the stem and leaf plots given below in K, L:

Category 𝑭𝒓𝒆𝒒𝒖𝒆𝒏𝒄𝒚, 𝒇 𝑷𝒆𝒓𝒆𝒏𝒕𝒂𝒈𝒆 (%) 𝑫𝒆𝒈𝒓𝒆𝒆𝒔

H) I) J)

ICU

Type

Frequency

Medical 12

Surgery 6

Cardiac 5

Other 2

Expenses Cost in $

Housing & Food 130

Savings 25

Miscellaneous 20

Insurance 35

Major Students

Maths 28

Biology 54

Nursing 20

Business 16

Stem Leaf

2 3,4,5,6,8,9

3 5,7

4 1,4,4,6,6,8,9

5 5,8,9

K) Stem Leaf

7 0.3, 3.5, 6.7

8 2.4, 5.5, 4

9 0.0, 5.3, 7.1

L)

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REMEMBER:

Graph Type X-Axis Y-Axis Histogram or

Frequency Histogram

Class

Boundaries/Midpoints Frequency

Relative Frequency

histogram

Class

Boundaries/Midpoints Relative Frequency

Ogive Graph Class Upper

Boundaries Cumulative frequency

𝑪𝒉𝒂𝒑𝒕𝒆𝒓 − 𝟑 𝑸𝒖𝒊𝒄𝒌 𝒇𝒐𝒓𝒎𝒖𝒍𝒂𝒔 𝑳𝒊𝒔𝒕𝒆𝒅 𝑫𝒂𝒕𝒂

𝑺𝒂𝒎𝒑𝒍𝒆 𝑴𝒆𝒂𝒏: �̅� =∑ 𝒙𝒊

𝒏𝒊=𝟏

𝒏

𝑺𝒂𝒎𝒑𝒍𝒆 𝑺𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝑫𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏

𝒔 = √∑(𝒙𝒊 − �̅�)𝟐

𝒏 − 𝟏 𝑶𝑹

𝒔 =√∑ 𝒙𝒊

𝟐 −(∑ 𝒙𝒊)𝟐

𝒏𝒏 − 𝟏

𝑺𝒂𝒎𝒑𝒍𝒆 𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆, 𝒔𝟐, 𝒊𝒔 𝒐𝒃𝒕𝒂𝒊𝒏𝒆𝒅 𝒃𝒚 𝒔𝒒𝒖𝒂𝒓𝒊𝒏𝒈 𝒕𝒉𝒆

𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏, 𝒔

𝑮𝒓𝒐𝒖𝒑𝒆𝒅 𝒐𝒓 𝑰𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒅𝒂𝒕𝒂. 𝑵𝒐𝒕𝒆: 𝑰𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒅𝒂𝒕𝒂 𝒊𝒔 𝒔𝒂𝒎𝒆 𝒂𝒔 𝒈𝒓𝒐𝒖𝒑𝒆𝒅 𝒅𝒂𝒕𝒂 𝒘𝒊𝒕𝒉 𝒂𝒏 𝒂𝒅𝒅𝒊𝒕𝒊𝒐𝒏𝒂𝒍

𝒔𝒕𝒆𝒑 𝒐𝒇 𝒇𝒊𝒏𝒅𝒊𝒏𝒈 𝒕𝒉𝒆 𝒎𝒊𝒅 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒆𝒂𝒄𝒉 𝒄𝒍𝒂𝒔𝒔

𝒂𝒏𝒅 𝒕𝒓𝒆𝒂𝒕𝒊𝒏𝒈 𝒊𝒕 𝒂𝒔 𝒙𝒊 𝒐𝒇 𝒕𝒉𝒆 𝒈𝒓𝒐𝒖𝒑𝒆𝒅 𝒅𝒂𝒕𝒂

𝑺𝒂𝒎𝒑𝒍𝒆 𝑴𝒆𝒂𝒏: �̅� =∑ 𝒙𝒊𝒇𝒊

𝒌𝒊=𝟏

∑ 𝒇𝒊𝒌𝒊=𝟏

=∑ 𝒎𝒊𝒇𝒊

𝒌𝒊=𝟏



𝒌𝒊=𝟏

𝒏


𝒔 = √∑(𝒙𝒊−�̅�)𝟐𝒇𝒊

𝒏−𝟏 𝑶𝑹 𝒔 = √∑ 𝒙𝒊

𝟐𝒇𝒊−(∑ 𝒙𝒊𝒇𝒊)

𝟐

𝒏

𝒏−𝟏

𝑺𝒂𝒎𝒑𝒍𝒆 𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆, 𝒔𝟐, 𝒊𝒔 𝒐𝒃𝒕𝒂𝒊𝒏𝒆𝒅 𝒃𝒚 𝒔𝒒𝒖𝒂𝒓𝒊𝒏𝒈 𝒕𝒉𝒆 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏, 𝒔

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FEW SOLUTIONS FOR YOUR HELP. PLEASE NOTE THAT YOU MUST SHOW THE

FORMULAS USED FOR EVERY PROBLEM.

1) [FOR TABLE: A] Ordering the data in ascending form, we get

5 7 8 10 10 12 12 12

12 16 19 19 22 25 38 41

Here we have 16 data values.

a) TRIMMED MEAN

10% Trimmed Mean: We have 10% of 16 = 1.6 ~ 2 { (10

100) × 16 = 1.6 }

Hence Trim 2 data points from both the ends of the ordered data. So we have 12

remaining points.

5, 7, 8, 10, 10, 12, 12, 12, 12, 16, 19, 19, 22, 25, 38, 41

The Mean of 12 points is the 10% Trimmed mean and is obtained by summing

the remaining 12 data points and dividing by 12.

10% Trimmed mean, �̅� =∑ 𝑥𝑖

𝑛=

177

12= 14.75


100) × 16 = 2.4 }


remaining points. Which in turn gives us the same trimmed mean as we have for 10%

above.

b) Five Number Summary: 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 = 5;

𝑄1 =10 + 10

2= 10; 𝑄2 =

12 + 12

2= 12; 𝑄3 =

19 + 22

2= 20.5

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 = 41; 𝐼𝑄𝑅 = 𝑄3 – 𝑄1 = 20.5 – 10 = 10.5

c) Box Plot:

5 10 12 20.5 41

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Outlier Identification:

Any point which does not fall in the interval (𝑄1 − 1.5 ∗ 𝐼𝑄𝑅, 𝑄3 + 1.5 ∗ 𝐼𝑄𝑅) is considered as

an outlier.

i.e., ( 𝟏𝟎 – (𝟏. 𝟓 ∗ 𝟏𝟎. 𝟓), 𝟐𝟎. 𝟓 + (𝟏. 𝟓 ∗ 𝟏𝟎. 𝟓)) = ( −𝟓. 𝟕𝟓, 𝟑𝟔. 𝟐𝟓 ). Hence the potential outliers in

the data are 38 and 41.

[FOR TABLE: B] Ordering the data in ascending form, we get

12.6 28.4 29.8 29.8 30.8 31 31 35.1 35.5

37.4 38.6 39.7 39.7 40.1 40.6 41.5 42 187.4

Here we have 20 data values.

a) TRIMMED MEAN


100) × 18 = 1.8 }


remaining points.

12.6, 28.4, 29.8, 29.8, 30.8, 31.0, 31.0, 35.1, 35.5, 37.4, 38.6, 39.7, 39.7, 40.1, 40.6,

41.5, 42.0, 187.0


the remaining 14 data points and dividing by 14.


𝑛=

500.6

14= 35.757


100) × 18 = 2.7 }


remaining points.

12.6, 28.4, 29.8, 29.8, 30.8, 31.0, 31.0, 35.1, 35.5, 37.4, 38.6, 39.7, 39.7, 40.1, 40.6,

41.5, 42.0, 187.0


these remaining 12 data points and dividing by 12.


𝑛=

429.3

12= 35.775

b) Five Number Summary: 𝑀𝑖𝑛𝑖𝑚𝑢𝑚 = 12.6;

𝑄1 = 30.8; 𝑄2 =35.5 + 37.4

2= 36.45; 𝑄3 = 40.1

𝑀𝑎𝑥𝑖𝑚𝑢𝑚 = 187.0; 𝐼𝑄𝑅 = 𝑄3 – 𝑄1 = 40.1 – 30.8 = 9.3

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Fre

qu

ency

Midpoints

c) Box Plot:

Outlier Identification:

Any point which does not fall in the interval (𝑄1 − 1.5 ∗ 𝐼𝑄𝑅, 𝑄3 + 1.5 ∗ 𝐼𝑄𝑅) is

considered as an outlier.

i.e., ( 𝟑𝟎. 𝟖 – (𝟏. 𝟓 ∗ 𝟗. 𝟑), 𝟒𝟎. 𝟏 + (𝟏. 𝟓 ∗ 𝟗. 𝟑)) = ( 𝟏𝟔. 𝟖𝟓, 𝟓𝟒. 𝟎𝟓 )

Hence the potential outliers in the data are 12.6 and 187.

1) [FOR TABLE: D]: 𝐶𝐿𝐴𝑆𝑆 𝑊𝐼𝐷𝑇𝐻: (49 − 30) + 1 = 20

NOTE: Using the data in columns draw the requested graphs. Check slides for more detail.

Frequency Histogram:

Weight(lb) Client

frequency

Class

Mid-Point

Class

Boundaries

Relative

Frequency

(%)

Cumulative

Frequency

30-49 8 39.5 29.5 49.5 21.62 8

50-69 10 59.5 49.5 69.5 27.03 18

70-89 12 79.5 69.5 89.5 32.43 30

90-109 7 99.5 89.5 109.5 18.92 37

37

12.6 30.8 36.45 40.1 187

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Rel

ativ

e F

req

uen

cy

Midpoints

Cu

mu

lati

ve

Fre

qu

ency

Class Upper Boundaries

Relative Frequency Histogram

Ogive Graph:

[FOR TABLE: G] CLASS WIDTH: (1.5-1.1) + 0.1 = 0.5

Length in

Meters Frequency

Class

Midpoint

Class

Boundaries

Relative

Frequency

(%)

Cumulative

Frequency

1.1 to 1.5 2 1.3 1.05 1.55 5.88 2

1.6 to 2.0 4 1.8 1.55 2.05 11.76 6

2.1 to 2.5 7 2.3 2.05 2.55 20.59 13

2.6 to 3.0 10 2.8 2.55 3.05 29.41 23

3.1 to 3.5 11 3.3 3.05 3.55 32.35 34

34

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Fre

qu

ency

Midpoints Midpoints

Rel

ativ

e F

req

uen

cy

Class Upper Boundaries

Cu

mu

lati

ve

Fre

qu

ency

3) [TABLE: D] For Sample Mean, Sample Standard Deviation and Sample

Variance. Follow these steps:

Weight(lb) Client

frequency Midpoint, m 𝒇. 𝒎 𝒇. 𝒎𝟐

30-49 8 39.5 316 12482

50-69 10 59.5 595 35402.5

70-89 12 79.5 954 75843

90-109 7 99.5 696.5 69301.75

37 2561.5 193029.3

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You MUST mention the formulas and steps [given in green color below] in your exam)

Sample mean:

𝑁𝑂𝑇𝐸: �̅� =∑ 𝑓𝑖𝑚𝑖

𝑘𝑖=1

∑ 𝑓𝑖𝑘𝑖=1

=∑ 𝑓𝑖𝑚𝑖

𝑘𝑖=1

𝑛 𝑜𝑟 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑠𝑎𝑚𝑒 𝑎𝑠

∑ 𝑥𝑖𝑓𝑖𝑘𝑖=1


�̅� =∑ 𝑓𝑖𝑚𝑖

𝑘𝑖=1

𝑛=

2561.5

37= 69.23

Sample Standard Deviation,

𝑠 =√∑ 𝑥𝑖

2𝑓𝑖 −(∑ 𝑥𝑖𝑓𝑖)2

𝑛𝑛 − 1

=√193029.3 −

(2561.5)2

3737 − 1

= 20.88

Sample Variance: 𝑠2 = (15.531)2 = 436.04

[TABLE: G] For Sample Mean, Sample Standard Deviation and Sample Variance.

Follow these steps:

Length in

Meters Frequency Midpoint,

m 𝒇. 𝒎 𝒇. 𝒎𝟐

1.1 to 1.5 2 1.3 2.6 3.38

1.6 to 2.0 4 1.8 7.2 12.96

2.1 to 2.5 7 2.3 16.1 37.03

2.6 to 3.0 10 2.8 28 78.4

3.1 to 3.5 11 3.3 36.3 119.79

34 90.2 251.56

Sample mean:

𝑁𝑂𝑇𝐸: �̅� =∑ 𝑓𝑖𝑚𝑖

𝑘𝑖=1


=∑ 𝑓𝑖𝑚𝑖

𝑘𝑖=1

𝑛 𝑜𝑟 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑠𝑎𝑚𝑒 𝑎𝑠

∑ 𝑥𝑖𝑓𝑖𝑘𝑖=1


�̅� =∑ 𝑓𝑖𝑚𝑖

𝑘𝑖=1

𝑛=

90.2

34= 2.65

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Sample Standard Deviation,

𝑠 =√∑ 𝑥𝑖

2𝑓𝑖 −(∑ 𝑥𝑖𝑓𝑖)2

𝑛𝑛 − 1

=√251.56 −

(90.2)2

3434 − 1

= 0.61

Sample Variance: 𝑠2 = (15.531)2 = 0.372

5) For Table –H: You must compute the Last Two Column values.

NOTE: These steps need not be shown in exam

Percentage for Surgery: 6

25× 100 = 24%

Degrees for Surgery: 24

100× 360 = 86.4

For Table –J:

ICU Type Frequency

Medical 12

Surgery 6

Cardiac 5

Other 2

ICU Type Frequency Percentage Degree

Medical 12 48 172.8

Surgery 6 24 86.4

Cardiac 5 20 72

Other 2 8 28.8

25 100 360

Major Students Percentage Degree

Maths 28 23.73 85.43

Biology 54 45.76 164.74

Nursing 20 16.95 61.02

Business 16 13.56 48.82

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4) Draw Stem and Leaf graph for the following and identify the shape.

(Symmetric/Right Skewed/Left Skewed)

a) Exam Scores: 35, 36, 38, 40, 42, 42, 44, 45, 45, 47, 48, 49, 50, 50, 50.

Stem and Leaf Plot:

Clearly, Data is symmetric (or bell shaped)

6) [FOR TABLE-K]:

The data points from the stem-leaf plot are:

23, 24, 25, 26, 28, 29, 35, 37, 41, 44, 44, 46, 46, 48, 49, 55, 58, 59

[FOR TABLE-L]:

The data points from the stem-leaf plot are:

70.3, 73.5, 76.7, 82.4, 85.5, 84, 90, 95.3, 97.1

Stem Leaf

3 5,6,8

4 0,2,2,4,5,5,7,8,9

5 0,0,0

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Note: Formulas will not be provided nor any cheat sheets are allowed on the exams.

𝑪𝒉𝒂𝒑𝒕𝒆𝒓 − 𝟑 𝑸𝒖𝒊𝒄𝒌 𝒇𝒐𝒓𝒎𝒖𝒍𝒂𝒔

𝑳𝒊𝒔𝒕𝒆𝒅 𝑫𝒂𝒕𝒂

𝑺𝒂𝒎𝒑𝒍𝒆 𝑴𝒆𝒂𝒏: �̅� =∑ 𝒙𝒊

𝒏𝒊=𝟏

𝒏


𝒔 = √∑(𝒙𝒊 − �̅�)𝟐

𝒏 − 𝟏 𝑶𝑹

𝒔 =√∑ 𝒙𝒊

𝟐 −(∑ 𝒙𝒊)𝟐

𝒏𝒏 − 𝟏

𝑺𝒂𝒎𝒑𝒍𝒆 𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆, 𝒔𝟐, 𝒊𝒔 𝒐𝒃𝒕𝒂𝒊𝒏𝒆𝒅 𝒃𝒚 𝒔𝒒𝒖𝒂𝒓𝒊𝒏𝒈 𝒕𝒉𝒆

𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏, 𝒔

𝑮𝒓𝒐𝒖𝒑𝒆𝒅 𝒐𝒓 𝑰𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒅𝒂𝒕𝒂.

𝑵𝒐𝒕𝒆: 𝑰𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒅𝒂𝒕𝒂 𝒊𝒔 𝒔𝒂𝒎𝒆 𝒂𝒔 𝒈𝒓𝒐𝒖𝒑𝒆𝒅 𝒅𝒂𝒕𝒂 𝒘𝒊𝒕𝒉 𝒂𝒏 𝒂𝒅𝒅𝒊𝒕𝒊𝒐𝒏𝒂𝒍 𝒔𝒕𝒆𝒑 𝒐𝒇 𝒇𝒊𝒏𝒅𝒊𝒏𝒈 𝒕𝒉𝒆 𝒎𝒊𝒅 𝒑𝒐𝒊𝒏𝒕 𝒐𝒇 𝒕𝒉𝒆 𝒊𝒏𝒕𝒆𝒓𝒗𝒂𝒍 𝒐𝒇 𝒆𝒂𝒄𝒉 𝒄𝒍𝒂𝒔𝒔

𝒂𝒏𝒅 𝒕𝒓𝒆𝒂𝒕𝒊𝒏𝒈 𝒊𝒕 𝒂𝒔 𝒙𝒊 𝒐𝒇 𝒕𝒉𝒆 𝒈𝒓𝒐𝒖𝒑𝒆𝒅 𝒅𝒂𝒕𝒂

𝑺𝒂𝒎𝒑𝒍𝒆 𝑴𝒆𝒂𝒏: �̅� =∑ 𝒙𝒊𝒇𝒊

𝒌𝒊=𝟏



𝒌𝒊=𝟏



𝒌𝒊=𝟏

𝒏


𝒔 = √∑(𝒙𝒊 − �̅�)𝟐𝒇𝒊

𝒏 − 𝟏 𝑶𝑹

𝒔 =√∑ 𝒙𝒊

𝟐𝒇𝒊 −(∑ 𝒙𝒊𝒇𝒊)𝟐

𝒏𝒏 − 𝟏

𝑺𝒂𝒎𝒑𝒍𝒆 𝑽𝒂𝒓𝒊𝒂𝒏𝒄𝒆, 𝒔𝟐, 𝒊𝒔 𝒐𝒃𝒕𝒂𝒊𝒏𝒆𝒅 𝒃𝒚 𝒔𝒒𝒖𝒂𝒓𝒊𝒏𝒈 𝒕𝒉𝒆 𝒔𝒂𝒎𝒑𝒍𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏, 𝒔

Venk

at Vn

V


sta 2023 exam-1 few solutions for the practice …vmudunur.myweb.usf.edu/fall 2015 exam-1 review...

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