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Case uitwerking

BBS1003 Answers to all seminar reports

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2020/2021

These are all the answers to the seminars that are given in block 3 in the first year of Biomedical Sciences in Maastricht. I don't recommend copying them all, but it is a nice document to have to compare your own answers with.

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Geüpload op
13 december 2021
Aantal pagina's
40
Geschreven in
2020/2021
Type
Case uitwerking
Docent(en)
Sophie vanbelle
Cijfer
7-8

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Voorbeeld van de inhoud

Seminar 1.1
1. D ratio, Because you’re talking about a number and you can also have 0
children
2. D When the data are on ordinal scale level, because data that we want to be
organised we can take a sample of which is the median. Mode represents
highest frequency. Makes sense to see which one has highest frequency.
3. C because when its distributed normally its symmetrical so nothing is skewed
to the right
4. A Mean is smaller than median and the median is smaller than the mode if the
distribution is skewed to the left. Because, Median is the middle of all the
data, mean is the average of the data and the mode is the highest frequency
of the data. If it is skewed to the left the mean is lower than median dataset.
The median determines what’s left and right within a boxplot. (?) when
skewed to right it is reversed.
5. C 68% between minus one and plus one standard deviation. It is a symmetric
distribution
6. C It is 4 because the mean is 40 all together divided by 10 is 4
7. C, 1.2 (Written down, worked with formula)
8. D They are very similar to each other if the variance is low because then the
difference between the 2 extremes is really small
9. D Nominal because the mode is just about the frequency of how many of
something is in the data
10. A = nominal
B= ordinal
C=ratio
D= interval
E=ratio
F=ratio
G= nominal
11. a. mode = 1600
median = 1600
mean = 1820
b. mode = 1600
mean = 5340
median = 1600
It illustrates that if you only change one variable the median and mode barely
change. But the mean changes a lot. So each variable influences the median
12. a. 79.7%
b. 1.6%
c. 9 mmol/l
d. Yes because it shows the average amount of haemoglobin in students. You
can draw conclusions out of an average. Ratio measurement
e. histogram check? Centre of the data is around 8  highest frequency.
Range is between 7 and 12. (need help) knowing how to construct a

, histogram, range is different max. and min, 16-3=13. Frequency is height of
box.

13. a. 7.1%
b. no.
c. mode = 2.000 = aluminium put spss output in this document next time
14. a. Mean: 9.85
Median: 8.50
Mode: 8.00
b. 10%  look at 14 and lower and subtract that from 100
c. Male, negative: Mode = 13.00, median = 14.00, mean =15.00
Male, positive: Mode = 8.00, median = 8.00, mean = 8.40
Female, negative: mode = 4.00 median = 6.00, mean = 6.00
Female positive: Mode = 7.00, median: 10.00, mean = 10.00

d. Is teaching with positive or negative reinforcement more effective in
obtaining higher scores and what works better in each gender? The best
statistical technique is making histograms to compare the scores each gender
obtained when taught with positive of negative reinforcement. The SPSS
output is that the mean score of negative reinforcement in males is 15.00.
Positive reinforcement caused males to obtain a mean score of 8.40. In
females the mean score with positive reinforcement is 10.00. The mean score
with negative reinforcement is just 6.00 in females.
Males do better when they are taught the negative reinforcement way and
females do better with positive reinforcement.

,15. a. Need explanation about SPSS ( I get separate graphs, I want just all in one.
It doesn’t work)

,

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