Summary
Summary STK 120 STUDY NOTES
- Course
- Stk 120
- Institution
- University Of Pretoria (UP)
Summary on Chapter 12-18 for STK 120
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-Chapter 18: Sign test (18. 1)
The school psychologist is currently busy studying stress during exam time. She has developed a questionaire that she
believes measures stress. A score above 80 indicates stress at a dangerous level.
A random sample of 16 senior learners revealed an average stress level of 84.25
You may assume that stress levels are normally distributed with standard deviation of 8
so
Ho :
M = So
Haims
Nonparametric tests
·
Small samples
&
No assumptions about the specific form of the populations's probability distribution
&
Categorical or Quantitative data
Sign test
~
i
median mean
Two categories: either less than median or more than median
Note: non parametric statistics in general refer to statistical methods in which the data is not required to fit a
normal distribution. It is also called ‘distribution-free’ statistics
In concluding the sign test, we compare each observation to the hypothesized value of the population median
I
Observation > hypothesized value, record a ‘+’ sign
⑧
Observation < hypothesized value, record ‘-‘ sign
A
Observation = hypothesized value, eliminate the observation and proceed with the smaller sample.
Test statistic: the number of plus signs
, Example:
Management made the decision to stock a new potato chip product
based on the manufacturer's estimate that the median sales should be
$450 per week on a per store basis
Consider the random sample of 10 stores, indicating each of their
weekly sales in the table below
Store Weekly Store Weekly Ho :
Median =
450 Ho :
p
= 0
, 5
y
E(pc) = 10 (0 5)
, number Sales $ number Sales $ Ha : Median = 450 Ha :
p + 0
, 5
= 5
Sl 485 + 63 474 +
19 562 +
39 662 + Test statistic :
7
two tailed
36 415- 84 386 -
-value =
P(x = x)
~
24 f(x)
+
(5)px(1
x
= 1 -
P(x = 6)
= -
p)b -
128 566 + 102 515 + =
2(0 1719)
,
= 0 , 3438
12 426 44 721 +
Ho
Do not
reject
:
A sales clerck from FNB would like to investigate wether the median number of savings account customers per day is
more than 64. The clerk collected the number of saving accounts per day data for 10 random days. Test at a 0,05 level
of significance
Day I 2 3 4 5 6 78 9 10
A Ho :
median = 64
bi i is z jo in is is it
t
Customers is
Ha :
median > 64
b test statistic = S
p-value = Ple,
.
c
=
1 -0
, 9805
= 0
, 0195
d Ho
Reject 0195 < 0 , 03
. ... 0
,
Normal approximation of the binomial distribution: Continuity Correction
continuous discrete
⑫ yeDaa
a
always o,
~
·
P(y = 2) =
P(y=2 -
0, 3)
·
, A physician is interested in the duration of non medicated patients' visit to the private practice. It is
believed that the median duration of visit for each patient was at least 22 minutes. Test wether the
median visit duration of patients is shorter than 22 minutes. A random sample of 25 visits in a practice
with a large Medicaid load yielded, in order. The following duration of visits:
+ + + + +
Visit . 4
9 13 4 . 15 6 . 16 . 2 16 4 . 16 8 . 18 1
.
18 3 . 18 7
. 18 8 . 18 . 9 19 1
. 19 3 .
19 5 .
20 1 .
21 6
.
21 8.
21 .
9 23 4 .
24 1 . 24 8 .
24 9 . 28 2 .
26 8 .
27/
6
Diff
12 6 8 6 4 5 8 -5 6 3 9 3 1 2 9- 2 7 3 - 1 9 - 0 4 3 2
3 7 1
-
5 2 3 3 2 -0 2 0 1 4 2 8 2 9
-
3 2 .
5
- -
2 1 4 8
. - . . - - - - . -
. . - - .
- . . . . .
. . . . . . . . .
. . .
Ho : median ? 22 Ho :
p20 , s
,
Ha : median 22 Ha :
po ,
s test statistic = >
Mea p-value = P(x < ) = P(x= , s)
P(z 5- 1]
v
= = ,
= 0
, 0228
Nonparametric test: two idependent samples
Using ranks for comparing two population distributions: Independant random samples
Ho The two identical
populations
:
are
be
Ha : The two populations are not identical ~
Always
The mann-whitney-wilcoxon test (18.3)
Example:
Exact test illustrated in textbook but only the large-sample approximation based on the normal distribution will
be used
Ho :
College and
high school students populations are identical in terms of performance Performance ratings
Ha :
College and
high school students
populations are not identical in terms of performance College student High school student
Student ID Rating Student ID Rating
*When there are
I .
Pooled 3 8 S 9 12 15 1718 20 2) 23252532
tiestea I 15 b I 18 s
Ranks 43678a
.
10 Il
I
252 3
-
209
-
2 I 2
i s 9 14
3 25 1 3 3214
P-value : Test Statistic = 45 4 8 2 . 5 4 9 4
·
m (n (n,
=
,
+
nc + 1) -
> m =
2(z)(7 + 2 + 1) =
52 , 5 5 25 12 . 5 5 25 .
12
·
:
z(n ne 1) + +
-
> 8 =
yiz(z)(z)(7 + 2+ 1) = 7, 8262
b 21 lo 6 8 2 5 .
(
48 5-52 S
7
=P
.
P(w
.
= 48) 7 8267 , 12 S
7 177
0, 3) -48 57
= 0 , 3050
Will
always
↳
use sum of ranks for
sample
2(0 3050) test statistic
, as
:
, 61
=
, 0
: Do not
reject Ho ,
do not have
enough evidence
.
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