Question

orrect Question 10 0/1 pts A psychologist hypothesizes that the children with type-K personality score higher than the children with type-R personality on an memory test. The estimated standard error turned out to be 0.9, while the mean score of type-K children was 18.8 and the mean score for type-R children was 21.3. What is the t score? t = (18.8-21.3)(0.3) = 8.33 t = (18.8-21.3)/0,9 = -2.78 • t = (18.8–21.3)/0.92 = -3.09 t = (18.8-21.3)/(2)(0.9) = -1.39

Answer 1

As we are testing here whether the children with type-K personality score higher than the children with type - R personality on a memory test, therefore this is a test for difference in means. The t test statistic therefore is computed here as:

$\large t^* = \frac{\mu_k - \mu_r}{SE} = \frac{18.8 - 21.3}{0.9} = -2.78$

Therefore B is the correct tets statistic computation and value here.