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Psychology 10

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Psychology 10

Analysis of Psychological Data

April 2, 2014

- Another example of the two-sample t test.
- Effect sizes related to the two-sample test.
- Introducing the idea of confidence intervals.

- Stereogram fusion experiment.
- Source: the Data and Story Library (http://lib.stat.cmu.edu/DASL/)
- Members of one group were given no information about the embedded image, or were told about it in words.
- Members of the other group were shown a picture of the embedded image.
- Interest was in how long it takes to fuse the stereogram.

- Letâ€™s call the groups Group 1 (not shown object) and Group 2 (shown object).
- What are our research and null hypotheses?
- m1â€“m2 = 0.
- m1â€“m2â‰ 0.

- a = .05.

- For Group 1, n = 43, SX = 368.09998, and SX2 = 5896.80986.
- For Group 2, n = 35, SX = 194.30001, and SX2 = 1862.57036.

- SS = SX2 â€“ (SX)2 / N.
- For Group 1, this is 5896.80986 â€“ 368.099982 / 43 = 2745.702993.
- For Group 2, we have 1862.57036 â€“ 194.300012 / 35 = 783.9276775.
- The pooled variance is (SS1 + SS2) / (n1 + n2â€“ 2).
- Here, that is (2745.702993 + 783.9276775) / 76 = 46.44250882.

- The standard error of the difference between means is
- Here, that is

- The means of the two groups are 368.09998 / 43 = 8.560464651, and 194.30001 / 35 = 5.551428857.
- Recall that
- Here, that is

- We have 43 + 35 â€“ 2 = 76 degrees of freedom, so weâ€™ll have to use 60 df in the table.
- From the table, the critical value of t is 2.000.
- (The real critical value for 76 df is 1.992.)
- Our t had the value 1.939, so we fail to reject the null hypothesis.

- Because we failed to reject the null hypothesis, we have been unable to show that there is a difference in mean fusion times between the population that was shown an image of the object and the population that was not shown the object.
- Note that a full interpretation includes discussion of the ideas that were being tested.

- Independence between groups:
- We were not told there was random assignment. In fact we know nothing about how the groups were created. So we really canâ€™t evaluate this.

- Independence within groups:
- We are told that in the first group, people either received no information or they received only verbal information.
- This could create clusters of scores that are similar to each other. We should be concerned.

- Equal variances in the two populations:
- We should compare the standard deviations.
- SS1 / (n1-1) = 2745.702993 / 42 = 65.37388079.
- The square root is 8.085.
- SS2 / (n2-1) = 783.9276775 / 34 = 23.0566964.
- The square root is 4.802.

- We should compare the standard deviations.

- Those look pretty different. The first (8.1) is about 1.7 times the size of the second (4.8).
- On the other hand, itâ€™s not so hard to believe that they could both be estimating a population standard deviation around 6.5.
- The evidence is right on the border of where we might worry.
- (Brief rant about the F-max test.)

- What about normality?
No Image Image

444333322222222 | 0 | 1111222222233333444

9998888766655 | 0 | 55666667999

33220000 | 1 | 0

75 | 1 | 566

200 | 2 | 0

| 2 |

| 3 |

| 3 |

| 4 |

7 | 4 |

- There appears to be a problem with the assumption that both populations are normal.
- Moreover, we have one observation that appears to be an extreme value.
- We have at least three concerns about our assumptions, then. We really shouldnâ€™t put much credence in this test (even if we had rejected the null hypothesis).

- The equivalent of Cohenâ€™s d for the two-sample situation (sometimes called Hedgesâ€™ g) is:
- In our example, this is (8.560464651 â€“ 5.551428857) / 6.81487409 = 0.44.
- Note, however, that the time is measured in seconds. We understand seconds.
- A better effect size, then, would be 8.560 â€“ 5.551 = 3.01 seconds.

- Your book also mentions r2, defined as t2/ (t2 + df).
- In our example, this would be 1.9392 / (1.9392 + 76) = .047.
- Note, however, that this measure of effect size is rarely used in practice.

- Sometimes, instead of testing a hypothesis about a parameter, we are interested in identifying a range of reasonable values for the parameter.
- A logical way to do that is to figure out what values of the parameter would not lead to rejection if we used them in a null hypothesis.

- Example: the sample mean.
- If we were conducting a one-sample t test, we would reject the null whenever
- That will happen for null values between

- We call such a range of values a 100 times (1-alpha) percent confidence interval.
- For example, if alpha = .05, the corresponding confidence interval is a 95% confidence interval.
- For alpha = .01, the related confidence interval is a 99% confidence interval.

- In the stereogram fusion example, the first group had a mean of 8.5605, N = 43, and SS = 2745.702993.
- If we were conducting a one-sample t test using that group, the variance would be 2745.702993 / 42 = 65.37388079, s = 8.085411603,and the standard error would be that value over âˆš43, or 1.233.

- If we were conducting a t test, our df would be 42, so the critical value of t would be approximately 2.021 (using 40 df).
- A 95% confidence interval would be given by 8.5605 Â± 2.021(1.233) = (6.07, 11.05).
- (The actual critical value is 2.018, but the CI would round to the same bounds.)

- A very common error is to say that there is a .95 probability that mis within the bounds of a particular interval.
- Thatâ€™s wrong. Once an interval has been calculated, mis either in it or not in it (with probability 1.0). We just donâ€™t know which is true.

- What we can say is that the interval was calculated in such a way that, if we were to repeat the process of sampling and calculating an interval, 95% of the time the resulting interval would contain mu.
- Therefore it makes sense to behave as though this particular interval contains mu.
- More on interpretation next time.

- We can also do a confidence interval for the difference between means.
- In the dot fusion example, the estimated difference between means was 8.560464651-5.551428857 = 3.009036.
- The standard error of the difference was 1.551446798.
- The critical value of t was 1.992.

- The 95% confidence interval, then, is given by 3.009036 Â± 1.992 Ã— 1.551446798 = (-0.08, 6.10).
- Interpretation?

- Confidence intervals, continued.
- The t test for repeated measures (also called the t test for related samples).

- Confidence interval using male runner data from last class.