Chapter 14
nonparametric methods and chi-square tests
14-1. n = 14 T =
Critical pts. at =2(.0176) = .035 are 3 and n - 3 = 12.
Cannot reject H0 (p-value = 0.18)
Sign TestData
1 / +
2 / +
3 / +
4 / -
5 / -
6 / +
7 / -
8 / +
9 / +
10 / +
11 / +
12 / -
13 / +
14 / +
n = / 14
Test Statistic
T = / 10 / <- number of + signs
At an a of
Null Hypothesis / p-value / 5%
H0: p = / 0.5 / 0.1796
H0: p >= / 0.5 / 0.9713
H0: p <= / 0.5 / 0.0898
14-2. n = 42 T = z = (54 – 42)/ = 1.85
At =.05, cannot reject H0 (no preference).
14-3. / - / - / - / - / - / - / - / + / + / -- / + / + / + / + / - / - / - / + / -
- / + / + / + / - / - / - / - / + / -
T = 19 n = 30 z = (2T – n)/ = 8/ = 1.46
We cannot reject H0 that median is $78.50.
14-4. n = 15 T = 10 p = .18
Accept H0. The machine’s identification rate is 50%.
14-5. n = 18 T = 9 p = .593
Accept H0. Median Age is 41.
Sign TestData
1 / -
2 / -
3 / -
4 / +
5 / +
6 / -
7 / +
8 / +
9 / -
10 / +
11 / -
12 / -
13 / -
14 / +
15 / +
16 / -
17 / +
18 / +
n = / 18
Test Statistic
T = / 9 / <- number of + signs
At an a of
Null Hypothesis / p-value / 5%
H0: p = / 0.5 / 1.1855
H0: p >= / 0.5 / 0.5927
H0: p <= / 0.5 / 0.5927
14-6. n 1 = 43 n 2 = 17 R = 10 + 1 = 25.367
z = = = -4.947
Reject H0. Not random.
14-7. n = 72 Runs = 46 z = 2.145 p-value = 0.032. Reject H0. The sequence is probably not random.
14-8. z = = -4.42
Reject H0. The regression errors are probably not random.
Runs Test
n1 / 12n2 / 18
Test Statistic
R / 4 / <- Number of runs in the data
In case of large samples (n1 or n2 > 10)
Test Statistic
z / -4.4195 / E(R) / 15.4
s(R) / 2.57949
At an a of
Null Hypothesis / p-value / 5%
Data is random / 0.0000 / Reject
14-9. n = 17 Runs = 2 z = -3.756 The p-value is very small. If assignments are truly random, such an occurrence would happen less than once in 2,500 days. If messengers work 5 days a week, this should occur about once in ten years! Discrimination is likely.
14-10. n 1 = 30 n 2 = 12 R = 10
z = = -3.13
Reject H0. Foreign firms can claim discrimination.
Runs Test
n1 / 12n2 / 30
Test Statistic
R / 10 / <- Number of runs in the data
In case of large samples (n1 or n2 > 10)
Test Statistic
z / -3.1343 / E(R) / 18.1429
s(R) / 2.598
At an a of
Null Hypothesis / p-value / 5%
Data is random / 0.0017 / Reject
14-11. n = 17 Runs = 2 z = –3.756. Reject H0. There is evidence that ad 1 is preferred to ad 2.
14-12. R = 6 n 1 = 7 n 2 = 8 Cannot reject H0. (The sample size is probably too small.) p-value = 2(.149) = 0.298.
14-13. U = 100 + (110/2) - 151.5 = 3.5. Reject H0; p-value < 0.0002. Nautical Design seems to be preferred.
14-14. R 1 = 228 U = 18(16) + 18(19)/2 – 228 = 231 z = (231 – 144)/28.98 = 3.00
Reject H0. The concentration in Antarctica is probably lower.
Mann-Whitney
Counts1 / 18
2 / 16
Test Statistic / Rank Sums
U / 231 / 1 / 228
2 / 367
In case of large samples (n1 or n2 > 10)
Test Statistic / E[U] / 144
z / 3.0018 / s(U) / 28.9828
At an a of
Null Hypothesis / p-value / 5%
H0:m1 = m2 / 0.0027 / Reject
H0:m1 >= m2 / 0.0013 / Reject
H0:m1 <= m2 / 0.9987
14-15. Both the Wald-Wolfowitz and the Mann-Whitney tests are nonparametric and do not require the assumption of normality required for the t test. Mann-Whitney requires that the data be on an ordinal scale, and in such cases the test should be used as it is a powerful alternative to the t test, more powerful than Wald-Wolfowitz.
14-16. R 1 = = 160 U = (12)(11) + 12(13)/2 – 160 = 50
E(U) = 12(11)/2 = 66
z = (50 – 66)/16.248 = -0.984 Do not reject H0.
14-17. U = 8(8) + 8(9)/2 - 88 = 12 Reject H0; p-value = 2(0.019) = 0.038. There is some evidence that the Chicago investment may be better.
Mann-Whitney
Counts1 / 8
2 / 8
Test Statistic / Rank Sums
U / 12 / 1 / 88 / Look up U tables for p-values
2 / 48
14-18. U = 16(17) + 17(18)/2 – 229.5 = 195.5 z = = 2.14
Reject H0. The black-market commissions are probably cheaper.
14-19. The Wilcoxon signed-rank test is a useful alternative to the paired-difference t test when it may not be assumed that the differences between the paired observations are normally distributed.
14-20. T = 6. We cannot reject H0 at = 0.05. p-value = 0.10.
14-21. T = 10. Do not reject H0; no evidence of differences in management style at = 0.05.
Wilcoxon
TestStatistic
S(+) / 10 / Null Hypothesis / T
S(-) / 45 / H0: m1 = m2 / 10
H0: m1 >= m2 / 10
n / 10 / H0: m1 <= m2 / 45 / Look up T tables for p-values
14-22. T = 56. Reject H0 at = 0.05.
14-23. The sign test is apporpraite since the data are binary (0 or 1 depending on the sound preference).
14-24. Use the parametric t test (actually use z since the sample is large). This is the approporiate test since commissions can be assumed to be normally distributed.
14-25. Use the Wilcoxon signed-rank test for the mean of a single population.
14-26. T = 18 n = 14 Reject H0 at = 0.025. The main limitation of the analysis is that we have time-series data, which are likely correlated. This is not a random sample. (Another limitation is that the operation is just beginning and may not have yet reached a stable level.)
14-27. n = 12 T = 27 > 17, so we cannot reject H0 at = 0.10.
Wilcoxon
TestStatistic
S(+) / 27 / Null Hypothesis / T
S(-) / 51 / H0: m1 = m2 / 27
H0: m1 >= m2 / 27
n / 12 / H0: m1 <= m2 / 51 / Look up T tables for p-values
14-28. n = 10 T = 4.5 < 5. Reject H0 at = 0.01.
14-29. p-value < 0.001. There is evidence that the airline’s delays have increased after the deregulation.
14-30. Using SYSTAT: p-value = 0.468, so we cannot reject H0.
TestStatistic
S(+) / 320.5 / Null Hypothesis / T
S(-) / 420.5 / H0: m1 = m2 / 320.5
H0: m1 >= m2 / 320.5
n / 38 / H0: m1 <= m2 / 420.5
For Large Samples (n > 25)
Test Statistic
z / 0.725114 / E[T] / 370.5
s(T) / 68.9547
At an a of
Null Hypothesis / p-value / 5%
H0: m1 = m2 / 0.4684
H0: m1 >= m2 / 0.2342
H0: m1 <= m2 / 0.7658
14-31. H = 12.5 p-value = 0.002. Reject H0. There is evidence that the frequency of use is not the same in all three industry groups.
Group / R / n / Avg. R / Test Statistic
1 / 66.5 / 10 / 6.65 / H / 12.491
2 / 142 / 8 / 17.75
3 / 116.5 / 7 / 16.643 / Null Hypothesis
H0: All the Groups have the same distribution
At an a of
p-value / 5%
0.0019 / Reject
3 / Total / 25
Further Analysis
1
2 / Sig / 2
3 / Sig / 3
14-32. R 1 = 75 R 2 = 116 R 3 = 55 R 4 = 160
H = (75 2/7 + 116 2/7 + 55 2/7 +160 2/7) – 3(29) = 13.716
Compare with 2.005(3) = 12.84. Reject H0; p-value < .005
14-33. H = 29.60 p-value < 0.001. Reject H0. Not all three songs are equally liked. Now continue with further analysis:
D 1 = || = |(366/12) – (214.5/12)| = 12.625
D 2 = || = |(366/12) – (85.5/12)| = 23.375
D 3 = || = |(214.5/12) – (85.5/12)| = 10.75
Using = 0.05, we find = 10.53.
This number is greater than D 3 but smaller than both D 1 and D 2. All the values of D1 are greater than 10.53. Hence, there is proof that "Revolution" is preferred over alternatives A and B, and that, 'alternative A' is preferred over 'alternative B.'
14-34. R 1 = 79 R 2 = 103 R 3 = 28
H =
Reject H0.
3Group / R / n / Avg. R / Test Statistic
1 / 79 / 8 / 9.875 / H / 13.542
2 / 103 / 6 / 17.167
3 / 28 / 6 / 4.6667 / Null Hypothesis
H0: All the Groups have the same distribution
At an a of
p-value / 5%
0.0011 / Reject
3 / Total / 20
Further Analysis
1
2 / 2
3 / Sig / 3
14-35. H = 12.985 p-value = 0.001. Reject H0. Returns on investment in the three areas are probably not all equal.
14-36. R 1 = 228 R 2 = 192 R 3 = 135 R 4 = 62 R 5 = 49
H =
Reject H0. Further analysis shows that, at = 0.005, the only significant differences are between New York and Orlando (NY >Orlando), and New York and Tampa (NY>Tampa).
14-37. We assume that the samples are random and independent of each other, the variables under study are continuous, and the measurement scale is ordinal. We do not assume a normal distribution of the populations under study. If we test for differences among population means, we must assume that when differences exist they are differences in the means.
14-38. All four fragrances are not equally liked.
Friedman Test
FragrancesR / 14 / 32 / 42 / 32
1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / RowSum
1 / 1 / 2 / 4 / 3 / 10 / n / 12
2 / 2 / 1 / 3 / 4 / 10 / k / 4
3 / 1 / 3 / 4 / 2 / 10
4 / 1 / 2 / 3 / 4 / 10 / X2 / 20.4
5 / 1 / 3 / 4 / 2 / 10
6 / 1 / 4 / 3 / 2 / 10 / p-value / 0.0001
7 / 1 / 3 / 4 / 2 / 10
8 / 2 / 1 / 4 / 3 / 10
9 / 1 / 3 / 4 / 2 / 10
10 / 1 / 3 / 2 / 4 / 10
11 / 1 / 4 / 3 / 2 / 10
12 / 1 / 3 / 4 / 2 / 10
14-39. The three managers are not equally effective.
Friedman Test / EffectivenessR / 29 / 17 / 14
1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / RowSum
1 / 3 / 2 / 1 / 6 / n / 10
2 / 3 / 2 / 1 / 6 / k / 3
3 / 3 / 1 / 2 / 6
4 / 3 / 2 / 1 / 6 / X2 / 12.6
5 / 2 / 3 / 1 / 6
6 / 3 / 1 / 2 / 6 / p-value / 0.0018
7 / 3 / 2 / 1 / 6
8 / 3 / 2 / 1 / 6
9 / 3 / 1 / 2 / 6
10 / 3 / 1 / 2 / 6
14-40. The drugs are not equally effective.
R / 16 / 24 / 14
1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / RowSum
1 / 2 / 3 / 1 / 6 / n / 9
2 / 2 / 3 / 1 / 6 / k / 3
3 / 2 / 3 / 1 / 6
4 / 2 / 3 / 1 / 6 / X2 / 6.22222
5 / 1 / 3 / 2 / 6
6 / 2 / 3 / 1 / 6 / p-value / 0.0446
7 / 2 / 1 / 3 / 6
8 / 2 / 3 / 1 / 6
9 / 1 / 2 / 3 / 6
14-41.
Friedman Test / Baking ProcessesR / 32 / 19 / 17 / 12
1 / 2 / 3 / 4 / 5 / 6 / 7 / 8 / 9 / 10 / RowSum
1 / 4 / 2 / 3 / 1 / 10 / n / 8
2 / 4 / 3 / 1 / 2 / 10 / k / 4
3 / 4 / 3 / 1 / 2 / 10
4 / 4 / 2 / 3 / 1 / 10 / X2 / 16.35
5 / 4 / 1 / 3 / 2 / 10
6 / 4 / 2 / 3 / 1 / 10 / p-value / 0.0010
7 / 4 / 3 / 2 / 1 / 10
8 / 4 / 3 / 1 / 2 / 10
14-42. H0: H1: > 0
r s = 1 – 6(Sd 2)/n(n 2 –1) = 1 – 6(406)/15(224) = 0.275
From Table 11 of Appendix C, for n = 15, the critical point for = 0.05 is 0.441. Do not reject H0 of no positive correlation.
Spearman Rank Correlation Coefficientn / 15
rs / 0.275 / Look up the tables for p-value.
14-43. n = 11 r s = 0.8793 From Table 11 of Appendix C, for n = 11, the critical points for = 0.05 in a two-tailed test are 0.623. Reject H0. There is evidence of a positive correlation.
Spearman Rank Correlation Coefficient