PLG 500 Penaakulan Statistik dalam pendidikan

PMC500

1. PLG 500 PenaakulanStatistikdalamPendidikan
Statistical Reasoning in Educational Research
DrNooraida Yakob
DrNor AsnizaIshak
DrMuzirahMusa

2. •Summarising and displaying data, describe
the distribution
–Data distribution. Constructing charts and graphs
–bar chart, histogram, box plot, stem and leaf
plot
–Describing the distribution, calculating mean,
mode and median, standard deviation and
variance, skewness of a distribution.

3. Data distribution.
Constructing charts and graphs –bar chart, histogram,
box plot, stem and leaf plot

4. 63,59,53,61,58,63, 77,65,59,63
59,59,88,58,59,64,60,64,75,66
62,61,60,70,64,63,59,62,65,76
Mathematics score
•This set of data makes sense when it is simplified
•We can report the number of students with specific scores in
a table
•This number is called frequency

7. Carta Bar
•Diagram showed frequency
•The height of each bar indicates the frequency of
each category
•The bars are the same width and separated

10. Histogram
•Showed the frequency of each interval class for
continuous variable
•The width of each bar represents the size of class
interval, the height represents the frequency of class
interval
•The bars have the same width and touch to one
nother

13. Stem & Leaf Plot
AStem and Leaf Plotis a special table where each data
value is split into a "stem" (the first digit or digits) and a
"leaf" (usually the last digit).

17. Boxplot
•A Box Plot is the visual representation of the statistical five
number summary of a given data set.
•A Five Number Summary includes:
–Minimum
–First Quartile
–Median (Second Quartile)
–Third Quartile
–Maximum

23. Box Plot
•A Five Number Summary includes:
–Minimum = 53
–First Quartile = 59
–Median (Second Quartile) = 62.5
–Third Quartile = 65
–Maximum = 88

24. Example 3-24: Solution
Step 5.
Figure 3.14
PremMann, Introductory Statistics, 7/ECopyright © 2010 John Wiley & Sons. All right reserved

26. Boxplot
To determine the outliers
Outlieris a value exceed 1.5*IQR form the nearest quartile (1.5 x 6
= 9, Q3 = 65) ; 65 + 9 = 74
Extreme outliers is a value exceed 3*IQR from the nearest quartile
-was labelled as * (3 x 6 =18, Q3 = 65); 65 + 18 = 84
Extreme outliers –ID
13
Outliers –ID 7,19,30
IQR –interquartile
range

28. Describing the distribution, calculating mean,
mode and median, standard deviation and
variance, skewness of a distribution.

29. Central Tendency Measurement
Numerical value to represent the center point of
data set
Indicate where most values in a distribution fall
Describe the whole data with one value
Three types :
Mode
Median
Mean

30. Central Tendency Measurement
Mode = score with the highest frequency
Median = score at the middle when the data is arranged
from smallest to largest
Mean = total score number of score
3, 7, 9, 4, 5, 4, 6, 9, 9
Mode =9(perhatikanmod≠3)
Median =3,4,4,5,6,7,9,9,9
Mean =(3+7+9+4+5+4+6+9+9)/9=6.22

32. Central Tendency Measurement
Advantages Weaknesses
Min Used the most as it is
stable
Take into account all
data
Suitable for continuous
data
Dipengaruhiolehnilaiekstrim
22, 25, 21, 26, 67
Min =32.20
MedianTidakdipengaruhioleh
nilaiekstrim
Hanyamengambilnilaidi tengah-
tengah
Mod The simplest
measurement
Suitable for discrete data
Some of the data did not have
mode value
11, 14, 19, 16, 25, 36

33. Measures of Dispersion
A measurement that showed how well the values in
a data set differ from one another or from the
central of the data set
Variance
Standardofdeviation

34. Varians& SisihanPiawai
Variansmerupakanstatistikyangmengukursejauhmanakahskor-skor
didalamdataberbezadenganmin.
Nilaivariansyangbesarmenunjukkandataberadajauhdaripadamin.
Inibermaksuddatatersebutlebihterserak.Sebaliknya,nilaivarians
yangkecilmenunjukkandataberadahampirdenganmin,yang
bermaksuddatatersebutlebihterkumpul.
Variansbagisatu-satudata,
2
diberikandalamrumus-rumusberikut:

2
=
σ??????−ҧ??????
2
??????
atau
2
=
σ??????
2
??????
−ҧ??????
2
,
denganҧ??????=minskorbagidata tersebut
Sisihanpiawai,
=
.  = varians .

35. Variance
4, 6, 9, 3, 5, 12, 10Min, ??????ҧ = 7
7
101253964




N
X
x x
2
4 16
6 36
9 81
3 9
5 25
12 144
10 100
x
2
= 411
x
x
(x –ҧ??????) (x –ҧ??????)
2
44 –7 = –3 (–3)
2
= 9
66 –7 = –1 (–1)
2
= 1
9 9 –7 = 2(2)
2
= 4
33 –7 = –4(–4)
2
= 16
55 –7 = –2 (–2)
2
= 4
1212 –7 = 5(5)
2
= 25
1010 –7 = 3(3)
2
= 9
(x –ҧ??????)
2
= 68

2
=
σ??????−ҧ??????
2
??????
=
68
7
=9.71

2
=
σ??????
2
??????
−ҧ??????
2
=
411
7
−7
2
=9.71

36. Variance & Standard Deviation
Advantages:
More accurate as the calculation involved all values in
the data set
Measure the dispersion of each value from the mean
value in the data set.
The calculation did not involved the square of deviation.
The unit for standard deviation and the unit for data –
same

38. Skewness Dispersion
Negative skewness –
tail on the left side
Positive skewness –
tail on the right side
Give the shape of distribution

40. Kasih
Terima