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Summary All Seminars & Final assignment BBS1003
- Instelling
- Maastricht University (UM)
All Seminars & Final assignment BBS1003 including the final assignment
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Voorbeeld 4 van de 72 pagina's
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Voorbeeld van de inhoud
Seminar 1.1 assignmentsMade by
1. Problem
Given. A study where the variable gender (female/non-binary/male) is studied.
Question. What is the level of measurement of the variable gender (female/non-binary/male)?
It is a categorical variable, or more specific a nominal variable.
2. Problem
Given. A median.
Question. In which situation would the median be an appropriate measure of central tendency?
(a) When the data are of nominal scale level
(b) When the average is not appropriate
(c) When the data are of ordinal scale level
(d) When there are two modes
The mean can be inappropriate because of outliers that means we can better rely on the median for a
more accurate result. NOIR = Nominal, ordinal, interval and ratio. Median can be calculated for O,I and R
and not for Nominal.
3. Problem
Given. A normal distribution.
Question. What is a characteristic of a normal distribution?
(e) Approximately 99% of the observations are located between the mean minus three times
the standard deviation and the mean plus three times the standard deviation
, (f) The distribution is skewed to the left
(g) The distribution has two modes
(h) The distribution is skewed to the right
In a normal distribution you don’t have 2 modes, and it isn’t skewed to the left or right. And the answer
has to do with the 68/95/99.7 rule with standard deviation.
4. Problem
Given. A distribution skewed to the left.
Question. What is a characteristic of a distribution skewed to the left?
(i) Median is smaller than the mean, the mean is smaller than the mode
(j) Mode, median and mean are equal
(k) Mean is smaller than the median, the median is smaller than the mode
(l) Mode is smaller than the median, the median is smaller than the mean
5. Problem
Given. A normal distribution.
Question. What is a characteristic of a normal distribution?
(m) Approximately 50% of the observations are smaller than the mean
(n) Approximately 99% of the observations are located between the mean minus two
standard deviations and the mean plus two standard deviations
(o) The mean is lower than the mode, the mode is lower than the median
(p) The distribution is not symmetric
Normal distribution is symmetric, the mean is same as median, and it is app. 50% below mean and 50%
above mean.
,6. Problem
Given. The grades of 11 students (/10) are given below
7 7 10 9 7 1 2 9 2 4 10
Question. What is the value of the mean?
(q) 5.2
(r) 7.2
(s) 6.2
(t) 11.4
(7 + 7 +10+ 9+ 7+ 1+ 2+ 9+ 2+ 4+ 10)/11 = 6.18 = 6.2
7. Problem
Given. The grades of 13 students (/10) are given below
6 8 4 1 3 6 2 7 6 10 4 6 2
Question. What is the variance? (For this question more than one answers might be correct)
(u) 6.3
(v) 6.4
(w) 6.7
(x) 6.8
Mean: (6 +8+ 4+ 1+ 3+ 6+ 2+ 7+ 6+ 10+ 4+ 6+ 2)/13 = 5
(6-5)2 + (8-5)2 + (4-5)2 + (1-5)2 + (3-5)2 + (6-5)2 + (2-5)2 + (7-5)2 + (6-5)2 + (10-5)2 + (4-5)2 + (6-5)2 + (2-5)2 /13
= 6.3
Population Variance
Sample Variance
, 8. Problem
Given. A sample with very low variance.
Question. What does a very low variance mean?
(y) The individuals in the data set are homogeneous
(z) The individuals in the data set are very different from each other
(aa) The standard deviation is large
(bb) The sample size is small
A low variance means a less disperse data set.
9. Problem
Given. One mode.
Question. What is the minimal level of measurement required to calculate a mode?
(cc) Ratio
(dd) ordinal
(ee) Interval
(ff) nominal
The mode is used almost exclusively with nominal-level data, as it is the only measure of central
tendency available for such variables.
10. Problem
Given. A variable Blood pressure (in mmHg).
Question. What is the level of measurement of the variable Blood pressure (in mmHg)?
Ratio
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