Practical Connection Assignment
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Review Chi-Square
Lesson: Spearman’s rs
Review:
Chi-Square test
Practice #1
Enter the observed data
After you enter the observed values
2. Make sure the values of the sum of the expected match the sum of the observed.
The results
Look at the chi-square statistic, and determine significance.
Practice #2
Enter the observed data
After you enter the observed values
2. Make sure the values of the sum of the expected match the sum of the observed.
The results
Look at the chi-square statistic, and determine significance.
| ANSWER: There was a relationship between purchasing and newspaper advertising. The df=4, N = 101, chi-square = 9.82, | |
| and p < .05. (X2 [4, N=101], = 9.82 p < .05). The Wall Street Journal appears to produce the most purchases. |
Spearman’s rs
Email: [email protected]
Phone: 941-822-1000
Appointment: Call, text, or email
Practice your statistical analysis.
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Review
Question on p- value
p < .05 = significance
p > .05 = no significance
P-value is the probability that the data would be at least as extreme as those observed, if the null hypothesis were true.
P < .05 means results are more extreme than what you got; occur fewer than five times in 100 when the null hypothesis is true.
Chi Square Test
When do we use?
Examining frequencies (observed vs. expected)
What is the null hypothesis?
Variables are not related
If we reject the null…
Variables are related, p < .05 (Chi square statistic must be greater than value in chart)
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Spearman’s rs Rank Order Correlation
The test is ideal for comparing the strength of similarity between a set of rank-order data.
Notice that there are two tests for ranked order data, the Mann-Whitney U and Spearman’s rs. We will not be covering the Mann-Whitney U, and I am not aware of anyone in the UC EdD program using for their dissertation. I am aware of one Spearman’s rs being used, and we will talk about that in a few slides.
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Rationale and Use
This test is used to see how strongly related two sets of data are to each other. The data has to be arranged in rank-order. This, like chi square, is a non-parametric test, so it should be used when data produce at least a bimodal distribution.
Suppose you want to see if age is related to cell phone usage. This can be accomplished with a relatively simple formula.
The null hypothesis in the Spearman’s r sub s is there is zero correlation or relationship between the pairs.
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An example from a dissertation:
Because the rating patterns appeared to be similar with respect to males and females, a Spearman’s rs correlation was conducted to determine if males and females viewed the deterrent factors the same way. The results of the Spearman’s rs test showed that the rankings were significantly similar to each other, (rs[7] = .93, p < .01). This test showed that gender did not affect the way students perceived factors that contributed to decisions to drop out of the University of the Cumberlands.
Once again, you need to know about this test to read data, perform the test if necessary, and interpret the results.
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Writing & Calculating
Reporting the Spearman’s rs
(rs [N ] = correlation, p < or > .05)
(Note you report the N instead of df in your results-the dissertation example had seven pairs).
The formula:
One minus the sum of differences in rank squared times six divided by the number of pair ranks times number of pairs minus one.
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Explanation of the Formula
The formula is relatively straight forward.
Σ = sum total
d = Difference in ranks of pairs (not scores)
d2 = Difference in ranks squared
n = Number of pairs of ranks
In the next slide, we take a look at a simple use of the formula.
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Spearman Example
In this example of Spearman’s rank-order statistic, we will recreate a test Vann conducted and reported in an article for Human Resource Development Quarterly (1995). The study involved measuring deterrents to participation in adult education programs. Vann relied on an instrument that had been used in an earlier study directed by Anthony Valentine and Gordon Darkenwald (1985). Vann wanted to know how his results compared to those found by his colleagues. This test was especially interesting because the studies were conducted in two distinctly different culture areas in the USA (the Arkansas Ozarks and New England).
Replication of a study to see if findings were similar is one use of the Spearman’s as in Dr. Vann’s example on the screen.
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Avoid the Common Mistake
Before we go into that example, I wanted to point out a common mistake. For some reason, beginning researchers want to subtract the real score in each pair to get d (i.e., 4.5 -3.6 = .90). In reality you should subtract the rank order (i.e., 5th place – 3rd place = 2).
So if 4.5 is a rank of 5th place and 3.6 is the rank of 3rd place. Use the rank, not the real score.
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Comparing Two Studies: The Rank Orders
Vann Darkenwald and Valentine
TC TC
LPP LCR
LCR LPP
PP PP
LC LC
It is easy to see that only the ranks for LCR and LPP are different, but can we be confident that random chance did not cause the differences in ranks? The next slide begins our quest to answer that question.
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Calculating Differences
| Factor | Vann (Rank) | D & V (Rank) | d | D2 |
| TC | 1 | 1 | 0 | 0 |
| LPP | 2 | 3 | -1 | 1 |
| LCR | 3 | 2 | 1 | 1 |
| PP | 4 | 4 | 0 | 0 |
| LC | 5 | 5 | 0 | 0 |
| Σ 2 |
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Spearman Calculation
1 – (6x2)/5(25-1) = 1 – 12/120 = 1 -.10 = .90 = rs
Σ = sum total
d = Difference in ranks of paired scores
d2 = Difference in ranks squared
n = Number of pairs of ranks
Inverse relationship because it ends as a negative. Let’s look at the table to see if .90 is significant.
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Is .90 Significant?
df = number of pairs minus 2 (df -2), 5 – 2 = 3. Note: this table does not use df, it uses pairs.
See Spearman Table for significance levels of .01. or .05. Some tables do not require df; they simply ask for the n or number of pairs.
Note: like any significance, .0 means no correlation and 1.0 is a perfect correlation. Any number between .0 and 1.0 is the level of correlation, but it remains to be seen if the statistic is significant. The table value in the Appendix of the book tells you if the correlation is significant.
One tailed vs. two tailed. Does anyone know the difference?
One-tailed=Detects a positive difference in population means, or a negative difference, but not both.
Two-tailed= non-directional statistical test than can detect either a positive or a negative difference in population means.
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Exercise
Ten students took part in this study, the teacher was interested in knowing if the age of the participant was related to the amount of cell phone usage. In the next slide, you will find a table that ranks the student’s respective age and his or her cell phone usage.
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Table of Ranks in Age and Placement
| Name | Rank in cell phone usage | Rank in Age | D | D2 |
| Davenport | 1 | 7 | 6 | 36 |
| Switzer | 2 | 10 | 8 | 64 |
| Baker | 3 | 2 | 1 | 1 |
| Lopez | 4 | 3 | 1 | 1 |
| Williams | 5 | 9 | 4 | 16 |
| Sayles | 6 | 5 | 1 | 1 |
| Pierce | 7 | 6 | 1 | 1 |
| Martinez | 8 | 1 | 7 | 49 |
| Talbot | 9 | 4 | 5 | 25 |
| Schneider | 10 | 8 | 2 | 4 |
| Σ198 |
Σ = sum total
d = Difference in ranks of paired scores
d2 = Difference in ranks squared
n = Number of pairs of ranks
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Using formula to solve:
1- (6x 198) = 1- (1188/990) = 1-1.2
(10 x 99)
= -.2
Look on page 404 (10 pairs)
Is it significant?
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Using excel functions to solve:
First, enter ranked data in two columns. Select Correlation test.
Next, highlight ranked columns, select grouped by columns, labels, and output range.
Rank in cell phone usage
Using excel functions to solve (continued):
The Spearman’s rs calculated here is the same as the answer obtained with the formula. Which, still is: rs= [10] = .2, p > .05.
Rank in cell phone usage
Rank in cell phone usage
Rank in cell phone usage
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