The signed-rank test, also known as the Wilcoxon signed-rank test, is a non-parametric statistical test used to compare two related or paired samples. It is an alternative to the paired t-test when the assumptions of the t-test are not met, such as when the data is not normally distributed or when outliers are present.
The signed-rank test is typically used when the data is measured on an ordinal or interval scale. It tests whether the median difference between the paired observations is significantly different from zero. The test does not assume any specific distribution of the data, making it robust against violations of normality assumptions.
Here's a step-by-step overview of how the signed-rank test is conducted:
1. Hypotheses:
- Null Hypothesis (H₀): There is no significant difference between the paired observations.
- Alternative Hypothesis (H₁): There is a significant difference between the paired observations.
2. Data Preparation:
- Take the differences between the paired observations.
- Discard any zero differences (no change) as they do not contribute to the test.
3. Rank the Absolute Differences:
- Rank the absolute differences, disregarding the sign.
- Assign the smallest absolute difference the rank of 1, the second smallest the rank of 2, and so on.
- If there are ties (two or more absolute differences are equal), assign the average rank to each tied observation.
4. Calculate the Test Statistic:
- Calculate the sum of the ranks of the positive differences (R⁺) and the sum of the ranks of the negative differences (R⁻).
- Take the smaller of R⁺ and R⁻ as the test statistic (T).
5. Calculate the Critical Value:
- Determine the critical value or p-value associated with the chosen significance level (e.g., α = 0.05) from the signed-rank table or using statistical software.
6. Compare the Test Statistic with the Critical Value:
- If the absolute value of the test statistic (|T|) is greater than or equal to the critical value, reject the null hypothesis. There is sufficient evidence to conclude that there is a significant difference between the paired observations.
- If the absolute value of the test statistic is less than the critical value, fail to reject the null hypothesis. There is not enough evidence to conclude a significant difference between the paired observations.
Reporting Results:
When reporting the results of the signed-rank test, it is common to include the test statistic (T), the p-value, and the decision (reject or fail to reject the null hypothesis) based on the chosen significance level.
Additionally, you may provide information about the direction and magnitude of the effect, along with any effect size measures used.
Remember, it's always a good practice to consult with a statistician or data analysis expert to ensure you are selecting the appropriate statistical test and correctly interpreting the results for your specific research or analysis.
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The Signed Rank Test (1).pptx
1. The signed rank test (Wilcoxon)
• In this test, we rank the differences without
consider their signs, assigning rank 1 to the
smallest difference in absolute value, rank 2 to
the second smallest difference in absolute
value and so on and rank ‘n’ to the largest
difference in absolute value.
• Zero differences are not considered
• If the absolute value of two or more
differences are same, we assign each one the
mean of the ranks which they jointly occupy.
2. T+, the sum of + ranks
T–, the sum of – ranks
T = min(T+, T–)
Depending on the alternative hypothesis, we
formulate the signed rank test on T, T+, T– with
the assumptions and the null hypothesis being
the same.
Alternative Hypothesis Reject the null hypothesis if
𝜇 ≠ 𝜇0 𝑇 < 𝑇𝛼
𝜇 > 𝜇0 𝑇
−
< 𝑇2𝛼
𝜇 < 𝜇0 𝑇
+
< 𝑇2𝛼
3. For 𝑛 ≥ 15, it is reasonable to assume that T+
is a value of a random variable having
approximately a normal distribution. To
perform the signed rank test based on the this
assumption, T+ is a value of a random variable
with the mean 𝜇 =
𝑛 𝑛+1
4
and the variance
𝜎2
=
𝑛 𝑛+1 2𝑛+1
24
. And Test statistics z =
𝑇+−𝜇
𝜎
.
4. Problems:
1) The following data constitute a random
sample of 15 measurements of the octane
rating of a certain kind of gasoline:
Use the signed rank test at 0.05 level of
significance to test whether or not the mean
octane rating of the given kind of gasoline is
98.5.
97.5 95.2 97.3 96.0 96.8 100.3 97.4 95.3
93.2 99.1 96.1 97.6 98.2 98.5 94.9
5. Solution:
Null Hypothesis: 𝜇 = 98.5
Alternative Hypothesis: 𝜇 ≠ 98.5
Level of significance: 𝛼 = 0.05
Test: Two tailed test
Criterion: Reject the null hypothesis if 𝑇 < 𝑇𝛼.
i.e. if 𝑇 < 𝑇0.05
7. From the above table
T-=4+12+6+10+7+5+11+14+9+3+1+13=95
T+=8+2=10
T = min(T+, T–)=min(10,95)=10
Here n = 14.
From the table, 𝑇0.05=21
Hence 𝑇 < 𝑇0.05.
Conclusion: The null hypothesis is rejected.
The mean octane rating of the given kind of
gasoline is not 98.5
8. Problems:
2) In a random sample taken at a public play
ground, it took 38, 43, 36, 29, 44, 28, 40, 50,
39, 47 and 33 minutes to play a set of tennis.
Use the signed-rank test at the 0.05 level of
significance to test whether the average is
greater than 35 minutes to play a set of tennis
at that public play ground.
9. Solution:
Null Hypothesis: 𝜇 = 35
Alternative Hypothesis: 𝜇 > 35
Level of significance: 𝛼 = 0.05
Test: One tailed test
Criterion: Reject the null hypothesis if 𝑇
−
<
𝑇2𝛼.
i.e. if 𝑇
−
< 𝑇0.1
11. From the above table
T-=6+7+2=15
T+=3+8+1+9+5+11+4+10=51
T = min(T+, T–)=min(51,15)=15
Here n = 11.
From the table, 𝑇0.1=14
Hence 𝑇
−
> 𝑇0.1.(15>14)
Conclusion: The null hypothesis is accepted.
The average is equal to 35 minutes to play a
set of tennis at that public play ground
12. Problems:
3) The following are the numbers of passengers
on flights 136 and 137 between Chicago and
Phoenix on 12 days.
232 and 189, 265 and 230, 249 and 236
250 and 261, 255 and 249, 236 and 218
270 and 258, 247 and 253, 249 and 251
240 and 233, 257 and 254, 239 and 249
Use the signed rank test at the 0.01 level of
significance to test the null hypothesis against 𝜇1 =
𝜇2 the alternative hypothesis 𝜇1 > 𝜇2.
13. Solution:
Null Hypothesis: 𝜇1 = 𝜇2
Alternative Hypothesis:𝜇1 > 𝜇2
Level of significance: 𝛼 = 0.01
Test: One tailed test
Criterion: Reject the null hypothesis if 𝑇
−
<
𝑇2𝛼.
i.e. if 𝑇
−
< 𝑇0.02
14. Measurements Difference Rank
232 and 189 43 12
265 and 230 35 11
249 and 236 13 9
250 and 261 -11 7
255 and 249 6 3.5
236 and 218 18 10
270 and 258 12 8
247 and 253 -6 3.5
249 and 251 -2 1
240 and 233 7 5
257 and 254 3 2
239 and 249 -10 6
15. From the above table
T-=7+3.5+1+6=17.5
T+=12+11+9+3.5+10+8+5+2=60.5
T = min(T+, T–)=min(60.5,17.5)=17.5
Here n = 12.
From the table, 𝑇0.01=10
Hence 𝑇
−
> 𝑇0.1.(17.5>10)
Conclusion: The null hypothesis is accepted.
Thus 𝜇1 = 𝜇2.
16. Problems:
4) The following are the weights in pounds,
before and after, of 16 persons who stayed on a
certain reducing diet for four weeks:
Use the signed rank test at 0.05 level of
significance to test whether the weight
reducing diet is effective or not.
Before 147.0 183.5 232.1 161.6 197.5 206.3 177.0 215.4 147.7 208.1 166.8 131.9 150.3 197.2 159.8 171.7
After 137.9 176.2 219.0 163.8 193.5 201.4 180.6 203.2 149.0 195.4 158.5 134.4 149.3 189.1 159.1 173.2
17. Solution:
Null Hypothesis: 𝜇1 = 𝜇2
Alternative Hypothesis:𝜇1 > 𝜇2
Level of significance: 𝛼 = 0.01
Test: One tailed test
Criterion: Reject the null hypothesis if z > 𝑧𝛼.
19. From the above table
T-=5+7+3+6+4=25
T+=13+10+16+8+9+14+15+12+2+11+1=111
Here n=16. So we use
𝜇 =
𝑛 𝑛 + 1
4
=
16 17
4
= 68
𝜎2 =
𝑛 𝑛+1 2𝑛+1
24
=
16 17 33
24
=374.
Test statistics z =
𝑇+−𝜇
𝜎
=
111−68
374
= 2.2235.
From table 𝑧𝛼=1.645. z > 𝑧𝛼
Conclusion: Null hypothesis is rejected.
The diet is effective.
