REVIEW & THEORETICAL EXAM BY
EXAMPLE
Sorana D. Bolboacă
OUTLINE
OUTLINE
 About theoretical exam
 About the final mark
 About the second theoretical assignment
 Topics on theoretical exam
 Example of theoretical exam
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©2017 - Sorana D. BOLBOACĂ
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ABOUT THEORETICAL EXAM
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©2017 - Sorana D. BOLBOACĂ
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ABOUT THEORETICAL EXAM
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©2017 - Sorana D. BOLBOACĂ
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ABOUT THEORETICAL EXAM
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ABOUT THEORETICAL EXAM
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EXAM SCHEDULING
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3rd February 2017: 9.00 a.m.
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Be at the location in time (8.30 a.m.)
Have with you the student card and an ID card /
passport
A hand calculator will be useful (you will not be
allow to use your phone)
At the end of the exam each student let the
answer sheet to the commission and sign
for attendance
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
Maximum of 30% of the questions could be with
one correct answer (10 out of 35).
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Marked with and asterisk: “*”
There are no questions with 5 correct answers
(could be 2 / 3 / 4)
The correct answer is given as filling the circle
(blackening) with black pen / pencils
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©2017 - Sorana D. BOLBOACĂ
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
Correct answers: will be published at the
discipline at 2 p.m.
To pass the exam:
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The mark for theoretical exam ≥ 5
The mark for practical exam ≥ 5
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©2017 - Sorana D. BOLBOACĂ
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THEORETICAL MARK
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Mark for theoretical exam (written mark): (the sum of
points)/(no. of possible points)*9+1
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©2017 - Sorana D. BOLBOACĂ
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Theoretical Exam
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1. question with one correct answer:
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2. questions with 2 correct answers:
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5 concordances = 1 point
< 5 conc. = 0 points
5 concordances = 1 point
4 concordances = 0.8 point
< 4 conc. = 0 points
3. questions ≥ 3 correct answers:
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


5 concordances = 1 point
4 concordances = 0.8 point
3 concordances = 0.3 point
< 3 concordances = 0 points
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©2017 - Sorana D. BOLBOACĂ
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FINAL MARK
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Theoretical mark: = 1 + 9/35 * (sum of points)
 pass if is > 5 without rounding
Practical mark = 1 + 9  pass if is > 5 without
rounding
Final mark = 0.3 * practical mark +
0.7*theoretical mark + bonus(max 3*0.2) +
rounding if necessary
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©2017 - Sorana D. BOLBOACĂ
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LECTURES AND PRACTICAL ACTIVITY EVALUATION
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University: https://eval.umfcluj.ro/

When? Until January 22, 2017
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Information: storage – upload – download & operation (binary: memories,
optical disc, etc. vs. decimal: HDD, memory stick, etc.)
Descriptive statistics: statistical series & graphical representation &
calculation of centrality parameters, coefficient of variation (+ interpretation)
& recognition of normal distribution
Probabilities: operations and probabilities & calculation of conditional
probabilities & conditional probabilities for diagnosis (Se, Sp, PPV, NPV) and
risk factors (RR, OR + interpretation) & construction of contingency table
based on a given scenario
Probability distributions: problems
Confidence intervals: calculation for mean, difference between two means,
frequency and difference between frequencies + interpretation
Statistical tests: statistical hypotheses & computation of the test statistic &
computation of degrees of freedom & interpretation of Z-test, t-test for
independent samples (independent samples choose based on the results of
variance tests – Levene or Barlett), paired t-test, statistical hypotheses &
interpretation for test of normality, ANOVA, Mann-Whitney, Wilcoxon,
Kruskal-Wallis
Correlation and regression analysis: choosing the proper correlation
coefficient (Pearson vs. Spearman), interpretation of correlation coefficient
(Colton rules and statistical test of significance)
©2017 - Sorana D. BOLBOACĂ
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Second Theoretical Assignment
Min = 6
Max = 35
Participation
140%
130%
125%
120%
120%
100%
120%
110%
100%
100%
90%
Participation
83%
82% 82%
80%
80%
82%
80% 80%
70%
64%
60%
50%
40%
36%
40%
20%
0%
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Group no
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Second Theoretical Assignment
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REVIEW & SUBJECTS BY EXAMPLE
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21% correct answers

A study was conducted to measure the effect of mother alcohol
consumption upon the birth weight of a baby. The following variables
were measured for each woman included in the study: alcohol
consumption status (yes/no), baby birth weight (g), and APGAR score.
The scales of these variables are:
A.
nominal, interval, ordinal
B.
nominal, ratio, ratio
C.
ordinal, ratio, interval
D.
nominal, ratio, nominal
E.
nominal, ratio, ordinal
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21% CORRECT ANSWERS

The arithmetic mean of total of decayed, missing and filled teeth on a
sample of children who attended to community dental surgery was of
1.71 with a standard deviation of 1.64 (sample size of 34). The 95%
confidence interval (Zα = 1.96) associated to mean is:
a) 0.2813
b) 1.4287 - 1.9913
c) 1.1587 - 2.2613
d) 0.0700 - 3.3500
e) Could not be calculated based on provided data
mean = 1.71
stdev = 1.64
n=34
1.71±1.96*1.64/sqrt(34) = 1.71±0.5513
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A physician wants to known if the number of male oesophageal
cancer patients diagnosed with multiple primary tumours
differs from the proportion of female oesophageal cancer
patients with the same diagnosis. A random sample of 60 male
and 40 female oesophageal cancer patients was selected and the
numbers with multiple primary tumours in each sample were
recorded. Forty men and ten women with multiple primary
tumours were identified. The calculated value of test statistic is:
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Χ2
16.67
0.65
0.72
14.04
None is correct
(40-30)2/30
Obs
MPT+
MPT-
Total
M
40
20
60
F
10
30
40
Total
50
50
100
(20-30)2/30
=
+
+
(10-20)2/20 + (30-20)2/20 = 16.67
11% CORRECT
ANSWERS
Exp
MPT+
MPT-
Total
M
30
30
60
F
20
20
40
Total
50
50
100
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57% correct answers
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Which of the following is a conclusion of carrying out a given statistical
test of hypothesis at a 5% significance level?
A.
H0 is more likely to be rejected than if the investigator had chosen
a level of significance of 5%
B.
The probability of accepting H0 when it is false is 0.01
C.
The probability of accepting H0 when it is false is 0.05
D.
The probability of rejecting H0 when it is true is 0.01
E.
The probability of observing a statistically significant result is 0.1
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©2017 - Sorana D. BOLBOACĂ
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REVIEW & SUBJECTS BY EXAMPLE
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REVIEW & SUBJECTS BY EXAMPLE
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REVIEW & SUBJECTS BY EXAMPLE
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REVIEW & SUBJECTS BY EXAMPLE
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©2017 - Sorana D. BOLBOACĂ
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REVIEW & SUBJECTS BY EXAMPLE
135±2.579*20/sqrt(169) = 135±3.97 = [131-139]
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©2017 - Sorana D. BOLBOACĂ
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REVIEW & SUBJECTS BY EXAMPLE
Reoccurrence
Non-Reoccurrence Total
A
=160-106=54
106
160
B
=120-104 = 16
104
120
Total
280
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©2017 - Sorana D. BOLBOACĂ
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REVIEW & SUBJECTS BY EXAMPLE
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REVIEW & SUBJECTS BY EXAMPLE
OC+
M+
45
M-
30
Total
75
OC-
Total
120
195
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©2017 - Sorana D. BOLBOACĂ
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REVIEW & SUBJECTS BY EXAMPLE
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©2017 - Sorana D. BOLBOACĂ
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REVIEW & SUBJECTS BY EXAMPLE
Reoccurrence
Non-Reoccurrence Total
A
=160-106=54
106
160
B
=120-104 = 16
104
120
Total
=54/160=0.3375
©2017 - Sorana D. BOLBOACĂ
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REVIEW & SUBJECTS BY EXAMPLE
Group 1: mean =140, stdev=36
Group 2: mean = 168, stdev = 36, n=81
Z = (168-140)/(36/sqrt(81)) = 7
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©2017 - Sorana D. BOLBOACĂ
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‘These things will become clear to you,’ said the old man gently,
‘at least,’ he added with slight doubt in his voice, ‘clearer than
they are at the moment.’
Douglas Adams
‘Today’s truths become errors tomorrow.’
Ursula K. Le Guin
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©2017 - Sorana D. BOLBOACĂ
16-Jan-2017