INFERENCE: one nominal and one ordinal variable

A. Kruskal-Wallis analysis of variance (H):
1. Rank order all ordinal scores in all nominal categories.
2. Calculate the sum of ranks for each nominal category.
3. Substitute into the formula:

H = ¹²/_{[N(N + 1)]} [Rj² / nj ] - 3(N + 1)

Where: N = total in sample
nj = number in a nominal category
k = number of nominal categories
R_j = sum of ranks for a nominal category

4. Look up p-value of H in the Chi Square Significant Probability Value using the degrees of freedom formula: k - Assumptions:

• continuous distribution underlying the ordinal variables- a continuous variable is one whose unit of measure can be divided infinitely; ordinal continuous data resembles interval/ratio data in its wide possibilities of response; instead of just five or six (i.e. scores, scales, etc.)
• five or more cases in each nominal category.

Example: Data without ties on the ordinal variable

Political Interest Scores
South	North	East	West
35	36	40	41
34	31	30	28
22	23	26	27
21	20	17	14
8	9	11	13

Q: Can this be generalized to the whole population?
A: Variables: region of residence (nominal) and political interest scores (ordinal).
Check to be sure assumptions are met before proceeding. (Okay in this example)

South	Rank	North	Rank	East	Rank	West	Rank
35	17	36	18	40	19	41	20
34	16	31	15	30	14	28	13
22	9	23	10	26	11	27	12
21	8	20	7	17	6	14	5
8	1	9	2	11	3	13	4
	51=Rs		52=Rn		53=Re		54=Rw

*Note: When ranking the scores, ranking can either be from high to low or low to high.

R_S = sum of ranks for south. n_S = number in south = 5
R_N = " " " " north. n_N = " " "north = 5
R_E = " " " " east. n_E = " " " east = 5
R_W = " " " " west. n_W = " " " west = 5 and N = 20

H = ¹²/_[20(20+1)] [(51)²/5 + (52)²/5+ (53)²/5 + (54)²/5] - 3(20 + 1) =

H = .0285714 [ 2206 ] - 63**Note: Keep as much accuracy as possible for this statistic.

H = 63.028508 - 63

H =.03 (At the end, round answer to 2 decimal places.)

Look this up on Chi Square table: d.f. = k - 1 = 4 - 1 = 3

In this case, p value > .05 so one cannot generalize the difference to the whole population.

Example: Data with ties on the ordinal variable:

	Abortion Attitude Scores
	Hi	MedHi	MedLo	Lo
Experimental group #1	5	1	0	0	6
Experimental group #2	0	1	1	5	7
Control group	0	3	4	0	7
	5	5	5	5	20

Q: Can this be generalized to the whole population?
A: Put data in a table with nominal variable on left side and ordinal variable on top. Use same formula and same procedure.

Dealing with rank scores in data with ties: scores from an exam:

Score Rank

99 -----1

97 -----2.5 If these were ranked #2 and #3 it wouldn't

97 -----2.5 be fair since they have the same score, so

93 -----4 split the difference.

92 -----5

90 -----7 Same for here: Instead of numbers 6,7 and 8,

90 -----7 split the difference and gave it to each.

90 -----7

89 -----9

In the example above, take the column totals and make up a table:

	Abortion Attitude Scale
	Hi	MedHi	MedLo	Lo
Column totals	5	5	5	5
Rank range	1-5	6-10	11-15	16-20
Median range	2	8	13	18

R _{exp #1} = 5(3) + 1(8) = 23; n _{exp #1} = 6

R _{exp #2} = 1(8) + 1(13) + 5(18) = 111; n _{exp #2} = 7

R _control = 3(8) + 4(13) = 76; n _control = 7

H = ¹²/_[20(20+1)] [(23)²/6 + (111)²/7 + (76)²/7] - 3(20+1) = .0285714 [ 2673.45 ] - 63

H = 76.384209 - 63 = 13.38

Look this value up on the Chi Square table, (d.f.= 3 - 1 = 2) and find that H is greater than p < .01 so one can generalize to the entire population.

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