Specific Comparisons (Correlated Observations)

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Learning Objectives

  1. Determine whether to use the formula for correlated comparisons or independent-groups comparisons
  2. Compute t for a comparison for repeated-measures data


Specific Comparisons (Correlated Observations)

  • In the Weapons and Aggression case study, subjects were asked to read words presented on a computer screen as quickly as they could
  • Some of the words were aggressive words such as injure or shatter
  • Others were control words such as relocate or consider
  • These two types of words were preceded by words that were either the names of weapons such as shot gun and grenade or non-weapon words such as rabbit or fish
  • For each subject, the mean reading time across words was computed for these four conditions
  • The four conditions are labeled as shown in Table 1. Table 2 shows the data for five subjects.
Description of Conditions
Variable Description
aw The time in milliseconds (msec) to name aggressive word following a weapon word prime.
an The time in milliseconds (msec) to name aggressive word following a non-weapon word prime.
cw The time in milliseconds (msec) to name a control word following a weapon word prime.
cn The time in milliseconds (msec) to name a control word following a non-weapon word prime.
Data from Five Subjects
Subject aw an cw cn
1 447 440 432 452
2 427 437 469 451
3 417 418 445 434
4 348 371 353 344
5 471 443 462 463


One question was whether reading times would be shorter when the preceding word was a weapon word (aw and cw conditions) than when it was a non-weapon word (an and cn conditions). In other words, is

L1 = (an + cn) - (aw + cw)

greater than 0?

This is tested for significance by computing L1 for each subject and then testing whether the mean value of L1 is significantly different from 0.

Table 3 shows L1 for the first five subjects. L1 for Subject 1 was computed by

L1 = (440 + 452) - (447 + 432) = 892 - 885 = 13
L1 for Five Subjects
Subject aw an cw cn L1
1 447 440 432 452 13
2 427 437 469 451 -8
3 417 418 445 434 -10
4 348 371 353 344 14
5 471 443 462 463 -27


  • Once L1 is computed for each subject, the significance test described in the section Testing a Single Mean can be used
  • First we compute the mean and the standard error of the mean for L1
  • There were 32 subjects in the experiment
  • Computing L1 for the 32 subjects, we find that the mean and standard error of the mean are 5.875 and 4.2646 respectively.

We then compute:

T mean.gif
M is the sample mean
μ is the hypothesized value of the population mean (0 in this case)
and sM is the estimated standard error of the mean
  • The calculations show that t = 1.378
  • Since there were 32 subjects, the degrees of freedom is 32 - 1 = 31
  • The t distribution calculator shows that the two-tailed probability is 0.1782

Priming Effect

  • A more interesting question is whether the priming effect (the difference between words preceded with a non-weapon word and words preceded by a weapon word) is different for aggressive words than it is for non-aggressive words
  • That is, do weapon words prime aggressive words more than they prime non-aggressive words?
  • The priming of aggressive words is (an - aw)
  • The priming of non-aggressive words is (cn - cw)
  • The comparison is the difference:
L2 = (an - aw) - (cn - cw)

Table 4 shows L2 for five of the 32 subjects.

L2 for Five Subjects
Subject aw an cw cn L2
1 447 440 432 452 -27
2 427 437 469 451 28
3 417 418 445 434 12
4 348 371 353 344 32
5 471 443 462 463 -29
  • The mean and standard error of the mean for all 32 subjects are 8.4375 and 3.9128 respectively
  • Therefore, t = 2.156 and p = 0.039.


Multiple Comparisons

Issues associated with doing multiple comparisons are the same for related observations as they are for multiple comparisons among independent groups.


Orthogonal Comparisons

  • The most straightforward way to assess the degree of dependence between two comparisons is to correlate them directly
  • For the weapons and aggression data, the comparisons L1 and L2 are correlated 0.24
  • Of course, this is a sample correlation and only estimates what the correlation would be if L1 and L2 were correlated in the whole population
  • Although mathematically possible, orthogonal comparisons with correlated observations are very rare.

Questions

1 Here you see taste ratings for 4 cola types: Cola A (A), generic Cola A (GA), Cola B (B), and generic Cola B (GB). L1 is the difference in taste scores between non-generic and generic Colas (nongeneric - generic). What is L1 for the fourth subject?

A	GA	B	GB
  9	  6	  8	  5
 10	  6	  9	  7
 10	  5	  8	  7
  8	  4	  7	  5
 10	  5	  8	  5
  9	  7	  7	  6

Answer >>

(A+B) - (GA + GB) specifically compares generic and nongeneric cola. It is (8 + 7) - (4 + 5).


2 Calculate the standard error for L1 (L1 = A + B - GA - GB). Keep in mind this is a repeated-measures design.

A	GA	B	GB
  9	  6	  8	  5
 10	  6	  9	  7
 10	  5	  8	  7
  8	  4	  7	  5
 10	  5	  8	  5
  9	  7	  7	  6

Answer >>

Compute L1 for each subject: (6, 6, 6, 6, 8, 3) and then compute the standard deviation (1.602) which you then divide by the square root of n (n = 6) to get 0.654.


3 Calculate t for L1 (L1 = A + B - GA - GB)

A	GA	B	GB
  9	  6	  8	  5
 10	  6	  9	  7
 10	  5	  8	  7
  8	  4	  7	  5
 10	  5	  8	  5
  9	  7	  7	  6

Answer >>

Compute L1 for each subject: (6, 6, 6, 6, 8, 3) and then compute the mean (5.833) and standard deviation (1.602) which you then divide by the square root of n (n = 6) to get 0.654 Then divide the mean by the standard error of 0.654 to get 8.919.