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Hypothesis Tests: Two Independent SamplesPowerPoint Presentation

Hypothesis Tests: Two Independent Samples

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Hypothesis Tests: Two Independent Samples

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Hypothesis Tests: Two Independent Samples

- Bushman (1998) Violent videos and aggressive behavior
- Doing the study Bushman’s way--almost

- Bushman had two independent groups
- Violent video versus educational video
- We want to compare mean number of aggressive associations between groups

- These are Bushman’s data, though he had more dimensions.
- We still have means of 7.10 and 5.65, but both of these are sample means.
- We want to test differences between sample means.
- Not between a sample and a population mean

Cont.

- 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 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 100 subjects.
- Each group has n - 1 = 100 - 1 = 99 df
- Total df = n1- 1 + n2 - 1 = n1 + n2 - 2 100 + 100 - 2 = 198 df
- t.025(198) = +1.97 (approx.)

- Since 2.66 > 1.97, reject H0.
- Conclude that those who watch violent videos produce more aggressive associates (M = 7.10) than those who watch nonviolent videos (M = 5.65), t(198) = 2.66, p < .05 (one-tailed).

Conclusions

Since -.88 < 1.97, fail to reject H0.

Conclude that those who took the IQ pill did not score significantly higher (M = 102.4) on an IQ test than those who did not take the IQ pill (M = 99.4), t(8) = -.88, p > .10.

Larger t value for same data

- 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

- If we assume both population variances are equal, then average of sample variances would be better estimate.

Cont.

- Substitute sp2 in place of separate variances in formula for t.
- Will not change result if sample sizes equal
- Do not pool if one variance more than 4 times the other.
- Discuss

- 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.

- Extension of what we already know.
- We can often simply express the effect as the difference between means (=1.45)
- We can scale the difference by the size of the standard deviation.
- Gives d
- Which standard deviation?

Cont.

- Use either standard deviation or their pooled average.
- We will pool because neither has any claim to priority.
- Pooled st. dev. = √14.8 = 3.85

Cont.

- This difference is approximately .38 standard deviations.
- This is a medium effect.
- Elaborate

Cont.

- p = .95 that interval formed in this way includes the true value of 1 - 2.
- Not probability that 1 - 2 falls in the interval, but probability that interval includes 1 - 2.
- Comment that reference is to “interval formed in this way,” not “this interval.”