Masters in biostatics coarse

1. Statistical Methods for Non parametric
Continuous Variables
Yilma ch, ass.t prof bio HI
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2. Objective
At the end of the presentation you will able to:
 list non parametric statistical tests
 describe sign test
 test hypothess using sign test
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3. • Wilcoxon Sign test
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4. Introduction
• When your data do not satisfy the distributional
assumptions required by parametric procedures,
other statistical methods are needed that is Non
parametric statistics.
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5. • The distributional assumptions required for non- parametric
procedures are usually less specific than those required for
parametric procedures.
• Many non-parametric tests have less power than the
corresponding parametric tests.
• Because power should never be given up unless absolutely
necessary, non-parametric methods should not be used when
parametric methods are appropriate.
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6. Cont...
• The sign test is an example of one of these non parametric
tests.
• Do not rush to use the NPT
• If your outcome variable is not normal, try to normalize using
log or ln.
• If it is still not normalized, go to non parametric tests
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7. What is the Sign Test?
 The sign test compares the sizes of two groups.
 It is a non-parametric or “distribution-free” test, which means
the test doesn’t assume the data comes from a particular
distribution, like the normal distribution.
 The sign test is an alternative to a one-sample t-test.
 It can also be used for ordered (ranked) categorical data.
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8. Cont...
 The sign test is used to test the null hypothesis that the median
of a distribution is equal to some hypothetical ( standard) value.
It can be used;
 in place of a one-sample t-test
 in place of a paired t-test or
 for ordered categorical data where a numerical scale is
inappropriate but where it is possible to rank the observations.
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9. Assumptions of sign test
1. data is non normally distributed
2. a random sample of independent measurement for a
population with unknown median
3. the variable of interest is continuous or ranked
ordinal scale of measurement
4. the one sample test handle non symmetric data set
(skewed either to right or left)
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10. procedure
• Let A and B represent two materials or treatments to be
compared.
• Let x and y represent measurements made on A and B.
• Let the number of pairs of observations be n.
• The n pairs of observations and their differences may be
denoted by:
(X1, Y1), (X2, y2), .....,(Xn, Yn) and
X1 - Y1, X2 - Y2 .............. Xn - Yn.
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11. Cont...
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12. Cont...
• The sign test is based on the signs of these differences.
• The letter Bs will be used to denote the number of times the
maximum sign has occurred.
• If some of the differences are zero, we can cancel the
observation.
• BS= Max｛N+,N-｝
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13. • As an example of the type of data for which the sign test is
appropriate, we may consider the following yields of two
hybrid lines of corn obtained from several different
experiments.
• In this example N =28
N+ = 7
N- = 21
BS = Max｛N+,N-｝
BS= 21
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14. • we can find the critical values and p-values of Bs from the sign
test table.
• if the p-value is less than significance level or alpha value,
reject the null hypothesis
P<α → reject the null hypothesis which states no median
difference.
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16. Example Hypothesis testing using sign test
N
o
Driver
injury (x)
passenger
injury (y)
sign (+,-)
(x-y)
Driver
injury (x)
passenger
injury (y)
sign (+,-)
(x-y)
1. 42 35 + 36 37 -
2. 42 35 + 36 37 -
3. 34 45 - 43 58 -
4 34 45 - 40 42 -
5. 45 45 0 43 58 -
6. 40 42 - 37 41 -
7. 42 46 - 37 41 -
8. 43 58 - 44 57 -
9. 45 43 + 42 42 0
Test at 95% confidence interval that the driver injury is equal to
passengers injury.
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17. Step 1. Ho : driver injury = passenger injury
HA : driver injury ≠ passenger injury
Step 2. N=18-2 =16 ( two observations canceled)
N+ = 3
N- = 13
BS= Max｛N+,N-｝
BS= 13
Step 3. appropriate test is sign test α=5%
step 4. find p value from the table
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18. p-value = 0.021
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19. Step 5. Interpretation
 0.021< 0.05 ( 0.021 is obtained from sign test table at n=16, BS
13 and alpha 0.05, which is 0.021
 P<α → reject the null hypothesis
 There for the drivers injury is not equal to the passengers injury.
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20. Exercise
The table below shows the hours of relief provided by two
analgesic drugs in 12 patients suffering from arthritis. Is there
any evidence that one drug provides longer relief than the other?
Test at 95% confidence interval.
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21. Solution
 In this case our null hypothesis is that the median difference is
zero.
 Our actual differences (Drug B - Drug A) are: +1.5, +2.1,
+0.3,−0.2, +2.6,−0.1, +1.8,−0.6, +1.5, +2.0, +2.3, +12.4
 Our actual median difference is 1.65 hours.
 N+ = 9, N− = 3, n = 12,
 Bs = max(N−, N+) = 9
 Our p-value at n=12, Bs= 9 and alpha =0.05
(from tables) is p = 0.146
 We would conclude that there is no evidence for a difference
between the two treatments on relief of pain.
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22. Sign test with large sample size
large sample size = N > 30 we use Z test.
Z= (X ± 0.5) -N/2
0.5 *√N
where : X= no of fewer sign
N = total pair of sample
we can use Z= (X + 0.5) -N/2 if N/2 >X
0.5* √N
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23. find the p value of Z from the table then:
P <α → reject the null hypothesis
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24. Example
 A researcher has taken 50 pair of students for the study and
obtained the data.
 can you conclude from the data by using sign test that the
training of the two groups differ significantly?
 Given no of
+ve sign = 37
 -ve sign =12
 of 0 =1
 solution
 HO= the training of one group=training of the other.
 N= 37+12 = 49
 N/2= 49/2=24.5
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25. x= 12
As 24.5 > 12 we use the formula
Z= (X + 0.5) -N/2
1/2 √N
= (12 +0.5)- 24.5
1/2 √49 Z= -3.43
3.43 > 1.96 or -3.43 < -1.96
• p-value of Z= -3.43 = 0.0003.
• Since the hypothesis is two sided, multiply 0.0003*2= 0.0006
0.0006 < 0.05 and 0.01
so, reject the null hypothesis at 5% as well as at 1%.
• Hence, training of two groups differ significantly.
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