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Within subjects t tests

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- Related samples
- Difference scores
- t tests on difference scores
- Advantages and disadvantages

- The same participants give us data on two measures
- e. g. Before and After treatment
- Usability problems before training on PP and after training

- With related samples, someone high on one measure probably high on other(individual variability).

Cont.

- Correlation between before and after scores
- Causes a change in the statistic we can use

- Sometimes called matched samples or repeated measures

- Calculate difference between first and second score
- e. g. Difference = Before - After

- Base subsequent analysis on difference scores
- Ignoring Before and After data

- The training decreased the number of problems with Powerpoint
- Was this enough of a change to be significant?
- Before and After scores are not independent.
- See raw data
- r = .64

Cont.

- If no change, mean of differences should be zero
- So, test the obtained mean of difference scores against m = 0.
- Use same test as in one sample test

D and sD = mean and standard deviation of differences.

df = n - 1 = 9 - 1 = 8

Cont.

- With 8 df, t.025 = +2.306 (Table E.6)
- We calculated t = 6.85
- Since 6.85 > 2.306, reject H0
- Conclude that the mean number of problems after training was less than mean number before training

- Eliminate subject-to-subject variability
- Control for extraneous variables
- Need fewer subjects

- Order effects
- Carry-over effects
- Subjects no longer naïve
- Change may just be a function of time
- Sometimes not logically possible

- Distribution of differences between means
- Heterogeneity of Variance
- Nonnormality

- Effect of training on problems using Powerpoint
- Same study as before --almost

- Now we have two independent groups
- Trained versus untrained users
- We want to compare mean number of problems between groups

Cannot compute pairwise differences, since we cannot compare two random people

We want to test differences between the two sample means (not between a sample and population)

- How are sample means distributed if H0 is true?
- Need sampling distribution of differences between means
- Same idea as before, except statistic is (X1 - X2) (mean 1 – mean2)

- Mean of sampling distribution = m1 - m2
- Standard deviation of sampling distribution (standard error of mean differences) =

Cont.

- Distribution approaches normal as n increases.
- Later we will modify this to “pool” variances.

- Same basic formula as before, but with accommodation to 2 groups.
- Note parallels with earlier t

- Each group has 6 subjects.
- Each group has n - 1 = 9 - 1 = 8 df
- Total df = n1- 1 + n2 - 1 = n1 + n2 - 2 9 + 9 - 2 = 16 df
- t.025(16) = +2.12 (approx.)

- T = 4.13
- Critical t = 2.12
- Since 4.13 > 2.12, reject H0.
- Conclude that those who get training have less problems than those without training

- Two major assumptions
- Both groups are sampled from populations with the same variance
- “homogeneity of variance”

- Both groups are sampled from normal populations
- Assumption of normality
- Frequently violated with little harm.

- Assumption of normality

- Both groups are sampled from populations with the same variance

- Refers to case of unequal population variances.
- We don’t pool the sample variances.
- We adjust df and look t up in tables for adjusted df.
- Minimum df = smaller n - 1.
- Most software calculates optimal df.