For the data provided,
x = 28
y = 36
x2 = 144
y2 = 248
xy = 187
r= [6(187) - 28(36)] / [(6(144) - 282)(6(248) - 362)].5 = .9198
b= [6(187) - 28(36)] / [6(144) - 282] = 1.425
a= 36/6 - 1.425 (28/6) = -.65
predicted y = -.65 + 1.425x
The predicted y for x=6 is 7.9
14. Procedures using Fisher's r to Z'
The question of the age-anxiety relation differing across species can be assessed with the
z-test for independent correlations. The Fisher Z' transform score for -.40 is -.424; the Z'
for +.50 is +.549
The 99% CI, in terms of Z', is given as: .549 - 2.58 (1/37.5) and .549 + 2.58
(1/37.5) falling around the population rho (again, in terms of Z').
15.
For the first predictor, .337/.259 = 1.301 (ns)
For the first step, R=.166, so...
For the second step, R=.695, so...
For the hierarchical test,
If nothing is strange, then one would expect 1/3 of the 9 people to be from each conditions.
That makes the chi-square equal to
(2-3)2/3 + (1-3)2/3 + (6-3)2/3 = 4.67
z = [.549 - -.424] / [1/40-3 + 1/20-3].5 = .973 /.29 = 3.355
Reject the null hypothesis that the two are equivalent: absolute valure of 3.355 exceeds
absolute value of 1.96
After the arithmetic, we convert back to the r metric
Given these data, one is 99% confident that .125 and .973 fall around the true population
correlation.
For the second predictor, .381/.089 = 4.28 (signif on 1,21)
[.1662/1] /[(1-.1662)/[24-1-1)] = F = .624 ns on 1,22
[.6952/2] /[(1-.6952)/[24-2-1)] = F = 9.810 on 2,21. Reject Ho
[(.6952 - .1662)/(2-1)] / [ (1-.6952)/(24-2-1) ] = F for
change in R2 on 1,21 = 18.56
Time on the computer task predicts SAT performance, above and beyond pre-existing ability.
17.Goodness of Fit
To be significant on 2 df, chi-square would need to exceed 5.991. It does not, so their is no
evidence that odd reaction times is anything other than random across conditions