| 95% CI for Cohen's d | |||||||
|---|---|---|---|---|---|---|---|
| t | df | p | Cohen's d | SE Cohen's d | Lower | Upper | |
| own-iq | 0.080 | 85 | .937 | 0.009 | 0.108 | -0.203 | 0.220 |
| neighbor-iq | -2.252 | 85 | .027 | -0.243 | 0.109 | -0.457 | -0.028 |
| Note. For the Student t-test, the alternative hypothesis specifies that the mean is different from 120. | |||||||
| Note. Student's t-test. | |||||||
The raincloud plots show individual data points and show that both variables have a couple of observations that might classify as "outliers"
| 95% CI for Cohen's d | |||||||||
|---|---|---|---|---|---|---|---|---|---|
| Measure 1 | Measure 2 | t | df | p | Cohen's d | SE Cohen's d | Lower | Upper | |
| own-iq | - | neighbor-iq | 1.619 | 85 | .109 | 0.175 | 0.155 | -0.039 | 0.387 |
| neighbor-iq | - | own-iq | -1.619 | 85 | .109 | -0.175 | 0.155 | -0.387 | 0.039 |
| Note. Student's t-test. | |||||||||
In the paired t-test, you can define the difference as A-B, or as B-A. This flips the sign of the t-statistic and Cohen's d, and the two-sided p-values will be identical. Exercise: how does the one-sided p-value differ between the two ways of defining the difference? Does that make sense?
The paired t-test raincloud plots show the two estimates side by side. For example, here we can see how the paired observations match with each other: some people judged their own iq as quite different from their neighbor's (steep grey lines connectin the observations), while most people had very similar estimates (i.e., flat grey lines).