Question

DQuestion 6 2 pts Researchers were interested in the relationship between processing speed, measured by rapid naming of etters (called Rapid Automatized Naming: RAN) and memory span. Hypothetical data for a group of 14 people is presented below femory Span Given that MRAN 14.36 and Mmemony span -621, calculate the covariance of the above data (round to TWO Given that SSPAN -293.21 and SSmemoy span+58.36. calculate the correlation coeficient (round to TWO decimal places 2 pts Question 7 Psychologists studied a sample of adolescents in a Midwest town, The researchers assessed a personality bleness" using a standard personality questionnaire. The questionnaire is a meas

Answer 1

there are all total n=14 values

and mean of Ram is 14.36 and mean of memory span is 6.21

then the covariance between Ram and memory span is

CV=[ $\sum$ (ram values-14.36)*(memroy span values-6.21)]/n

=[(17-14.36)*(6-6.21)+(16-14.36)*(5-6.21)+....+(14-14.36)*(8-6.21)]/14

=-8.07654=-8.08 [rounded to two decimal] [answer]

if SS of Ram is 293.21 and SS of memory is 58.36 then correlation is

(n*covariance)/sqrt[SS of Ram*SS of memory]

=14*(-8.08)/sqrt(293.21*58.36)

=-0.8647536=-0.86 [rounded to two decimal] [answer]