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1. You should recognize this as a binomial problem. You want to test the hypothesis that the sample of murderers was drawn from a population where the probability of being XYY is .005. Our null and alternative hypotheses are:
H0: p = .005
H1: p > .005
You then calculate the probability of observing 2 out of 7 men with an XYY chromosome combination, under the assumption of the null hypothesis that the population probability is .005. If this probability is less than our chosen alpha of .001, we will reject the null hypothesis. In this problem, the population probability is .005 and q = (1 - .005) or .995. That probability is equal to:
If the population probability was .005, the probability that we would observe 2 out of 7 men with an XYY chromosome set is .0005. This is less than .001, so we reject the null hypothesis. The probability of having an XYY chromosome set for those convicted of murder is greater than .005.
2. Our hypotheses are:
H0: The population of white and non-white citizens are no different in their level of satisfaction with the police services they receive.
H1: White citizens are more satisfied with their police services than black citizens.
The test statistic is the Mann-Whitney U test. In order to find the critical value of U, we look at the Mann-Whitney U table. We have a one-tailed alpha of .05 and samples sizes of n1 = 10 and n2 = 8. The critical value of U is 20. Remember that values equal to or lower than the critical value are necessary to reject the null hypothesis. Our decision rule, then, will be to reject the null hypothesis if Uobt £ 20.
The test statistic is either U or U_, which ever is the smaller of the two. In order to determine these values, we need to assign ranks to the raw scores. We ignore the fact that there are two samples, and combine the data, assigning the rank of 1 to the lowest score, the rank of 2 to the next lowest score, until the highest score is assigned the highest rank. We show you the raw scores and assigned ranks below. At the bottom of the table is the sum of the ranks for each group.
White Citizens Non-White Citizens
Scores (Rank) Scores (Rank)
10 (8) 20 (18)
6 (4) 8 (6)
5 (3) 15 (13)
18 (16) 14 (12)
11 (9) 19 (17)
12 (10) 7 (5)
4 (2) 3 (1)
17 (15) 16 (14)
13 (11)
9 (7)
S = 85 S = 86
Letting the sample of non-whites be the second sample, and using our computational formulas, we can determine the values of U and U_ to be:
The smaller of these two values is U, so our value of Uobt is 30. Since this is not less than the critical value of U, we do not reject the null hypothesis. White and non-white citizens are not different in their satisfaction with police services.
3. Our hypotheses are:
H0: The two samples were drawn from the same population. In other words, criminals and non-criminals are not different in terms of their impulsivity.
H1: The two samples come from different populations. Criminals are more impulsive than non-criminals.
The test statistic is the Wald-Wolfowitz Runs Test. The test statistic is r which is the number of runs in the sample data. From the table of r values, the critical value of r with an alpha level of .025, n1 = n2 = 10 is equal to 6. Since small values of r will lead us to reject H0, our decision rule is to reject the null hypothesis if robt £ 6. In order to determine robt, we have to count the number of runs in the data. To find this, we first ignore the fact that we have two samples, combine the data, and assign ranks with the lowest score getting the rank of 1, and so on. Then the number of runs can be determined. We show you the raw scores and ranks for the data below.
Criminals Non-Criminals
Raw Score (Rank) Raw Score (Rank)
16 (4) 29 (12)
24 (9) 21 (6)
42 (19) 14 (2)
15 (3) 34 (15)
10 (1) 47 (20)
28 (11) 36 (16)
31 (13) 38 (17)
40 (18) 23 (8)
25 (10) 32 (14)
22 (7) 19 (5)
To find the number of runs, do the following. Find the sample to which the lowest rank belongs, notice that it is the criminal sample (C). Find the sample to which the next lowest rank belongs, it is the non-criminal sample (N). Do this for each rank and display the list or runs of letters corresponding to the sample to which a rank belongs. In the data above, the list of runs would be:
1 2 3 4 5 6 7 8 9 10 11 12 C N CC NN C N CCC N C NNNN CC N
As you can see, there are 12 runs in the data. A run of one C, a run of one N, and run of two C's, and run of two N's ....a run of one N. The obtained value of the Wald-Wolfowitz test statistic, r then, is equal to 12: robt = 12. Since robt > rcrit (12 > 6) we decide not to reject the null hypothesis. We cannot reject the assumption that criminals and non-criminals have the same level or degree of impulsivity.
4. The sign test is comparable to the matched pairs or dependent samples t test. In this problem you want to determine if children of single-parent families are more of a bully than children of a matched sample of two-parent families. As we noted, this test is comparable to a test for the difference between two population medians. The hypotheses are:
H0: There is no difference between the children of single and two-parent families in terms of their tendency to bully.
H1: Children from single parent families are more of bully than children from two-parent families.
The test statistic is the number of less frequently occurring signs, designated by the letter m. To obtain the number of signs, for each pair of matched scores in the sample, compare the second score with the first score. If the second score is greater than the first, record a + sign, if the second score is less than the first, record a - sign. If the scores are equal, record a 0, and exclude this score from further analysis. Then count the number of + and - signs. The test statistic m is the number of signs that occurs less frequently. The critical value of m is found by going to the table of m values. We have a one-tailed alpha of .01 and a sample of size 12 (we have 12 pairs of scores). From the table, we can see that the critical value of m is 0. Our decision rule, therefore, is to reject the null hypothesis if mobt is less than or equal to 0.
The table below shows the raw scores, and the number of + and - signs:
Children's Scores on Bullying Scale
Single-Parent Families Two-Parent Families Sign
15 11 -
18 13 -
25 10 -
32 36 +
19 12 -
27 15 -
21 24 +
14 19 +
26 20 -
39 14 -
27 20 -
23 16 -
There are 9 - signs and 3 + signs. The less frequently occurring sign, therefore, is the + sign. Since there are three of them, mobt is equal to 3. This is greater than the critical value of m (mcrit=0), so we fail to reject the null hypothesis. Children in single-parent families are not more of a bully than those in two-parent families.
5. Our hypotheses are the same. The statistical test for the Wilcoxon Signed Ranks Test is T. With 12 pairs of observations, and a one-tailed alpha of .01, we can go to the Wilcoxon T table and find that the critical value of T is 10. Tcrit, then, is 10, and our decision rule is to reject the null hypothesis if Tobt is less than or equal to 10.
T is fairly simple to calculate. We will go through the steps. First, subtract the first score from the second score, and record both the sign and the magnitude of this difference. Second, ignore the sign and rank order the difference scores assigning the rank of 1 to the lowest score. Third, after assigning a rank to each score, go back and find out if there was a - or + sign with that difference score. Replace the sign that was associated with that difference score. After recording the correct sign, sum the ranks of both the positive and negative signs. The Wilcoxon Signed Rank test Tobt is the smaller of the two sums. With the data from problem 5 above, we show you below the difference scores, the absolute value of the differences, the ranks, and the signed ranks.
Children's Scores on Bullying Scale
Single Parent Two Parent
Families Families D ½D½ Rank Signed Rank
15 11 -4 4 2.5 -2.5
18 13 -5 5 4.5 -4.5
25 10 -15 15 11 -11
32 36 +4 4 2.5 +2.5
19 12 -7 7 8 -8
27 15 -12 12 10 -10
21 24 +3 3 1 +1
14 19 +5 5 4.5 +4.5
26 20 -6 6 6 -6
39 14 -25 25 12 -12
27 20 -7 7 8 -8
23 16 -7 7 8 -8
The sum of the negative ranks is 70, while the sum of the positive ranks is equal to 8. The value of Tobt is the lesser of these two sums, so Tobt = 8. Since this is less than the critical value of 10, our decision is to reject the null hypothesis. Children in single-parent families are more of a bully than children in two-parent families. The reason why this conclusion differs from that with the sign test is that the Wilcoxon Signed Ranks test takes into account the magnitude of the difference between the matched scores, while the sign test only takes into account the direction of the difference.
6. The hypothesis test for this problem is that all three samples have been drawn from the same population. It is, then, comparable to the parametric ANOVA. To test the hypothesis, we will use the analysis of variance for ranks test called the Kruskal-Wallis test. The hypotheses are:
H0: All three samples were selected from the same population.
H1: The samples have been drawn from different populations.
An implication of the alternative hypothesis is that the populations have different medians.
As noted, the test statistic is the Kruskal-Wallis H test. The sampling distribution of H is approximated by a chi-square distribution with k - 1 degrees of freedom, where k is the number of groups or samples. In the problem here where our alpha is .01 and we have 3-1 or 2 degrees of freedom, the critical value of chi-square is 9.210. Our decision rule, then, is to reject the null hypothesis if Hobt is greater than or equal to 9.210.
To calculate the H statistic, we first pool all of the data into one group, ignoring the fact that there are different samples. Then we rank order the scores, assigning a rank of 1 to the lowest score until we rank the highest score. After assigning ranks we sum the ranks for each group separately. That is, we add the ranks for the first sample and obtain a sum, add the ranks for the second sample and obtain a sum, and then add the ranks for the third sample and obtain a sum. We show you below the raw scores, the ranks, and the sum of the ranks for each of the three samples.
Score on Strength of the Evidence Measure
Convicted: Convicted:
Acquitted Lesser Charge Original Charge
Score (Rank) Score (Rank) Score (Rank)
19 (8) 36 (21) 31 (17)
25 (12) 28 (14) 46 (27)
21 (10) 40 (24) 50 (30)
10 (1) 33 (19) 27 (13)
32 (18) 45 (26) 39 (23)
14 (4) 18 (7) 47 (28)
12 (2) 43 (25) 34 (20)
17 (6) 15 (5) 30 (16)
20 (9) 37 (22) 22 (11)
29 (15) 33 (3) 49 (29)
S = 85 S = 166 S = 214
Now that we have the sum of the ranks for each sample, we can plug the values into our formula and find Hobt:
Our value of Hobt is 26.6, while that for Hcrit is 9.210. Since our obtained value is greater than the critical value, we decide to reject the null hypothesis. Cases that result in acquittal, conviction on reduced charges, and conviction on original charges do differ in terms of the strength of the evidence.
7. Since both exams are presumed to be measured at the ordinal level, we will use Spearman's rank order correlation coefficient rs. This correlation coefficient will tell us if scores on the two exams are correlated. Hopefully, there will be a high positive correlation between the two exams. In order to calculate rs we first have to rank order the two sets of scores. Unlike the case with the Mann-Whitney, the Wilcoxon Signed Rank Test, or Kruskal-Wallis Test, we do not first pool the data. Rather, in calculating rs we rank order the data within each variable separately. That is, we assign ranks to the civil service scores by giving a rank of 1 to the lowest score up to a rank of n for the highest score. Then we assign ranks to the police academy exam scores the same way. Once we have assigned ranks to the two variables, we find the difference between each pair of scores D, by subtracting the second score from the first. Finally, we square this D score. We show the calculations below:
Civil Service Police Academy
Recruit Raw Score (Rank) Raw Score (Rank) D D2 1 65 (1) 4 (1) 0 0
2 88 (10) 15 (9) 1 1
3 97 (13) 19 (12) 1 1
4 75 (3) 5 (2) 1 1
5 82 (8) 16 (10) -2 4
6 77 (4) 8 (5) -1 1
7 80 (6) 10 (7) -1 1
8 78 (5) 9 (6) -1 1
9 69 (2) 6 (3) -1 1
10 92 (11) 12 (8) 3 9
11 81 (7) 7 (4) 3 9
12 96 (12) 20 (13) -1 1
13 83 (9) 17 (11) -2 4
S = 34
The sum of the squared difference scores (S D2), then, is 34. We can now calculate our correlation coefficient with the formula for rs:
There is a very high positive correlation between the two variables. High ranks on the civil service exam are correlated with high ranks on the police academy exam.
To determine if our sample correlation coefficient rs is significantly different from zero, we can test the hypotheses that:
H0: r = 0
H1: r > 0
Since our n is less than 30, we can find the critical value of rs by going to the exact sampling distribution. With an alpha of .01 (one-tailed) and a sample size of 13, the critical value of rs is .673. Our decision rule, then, is to reject the null hypothesis if robt ³ .673. Since our obtained value of rs is .91, we decide to reject the null hypothesis. Scores on the civil service exam and police academy exam are positively correlated in the population.
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Copyright ©2001 The McGraw-Hill Companies.
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McGraw-Hill Higher Education is one of the many fine businesses of the
The McGraw-Hill Companies.