A chess team advisor wanted to examine how long it took his top two players, Fishy Bobber and Chet Mate, to complete their moves. Over the course of 50 matches each, Fishy averaged 330 seconds to move while Chet averaged 420 seconds to move. If the overall standard deviation for Fishy and Chet was 45 seconds, what was the effect size of their difference?
A) 1.0
B) 90
C) 2.0
D) Not enough information is given.

Question

Quizplus · Accepted Answer

To find the effect size of the difference between Fishy Bobber and Chet Mate, we can use Cohen's d formula:

d = (M1 - M2) / SD

where M1 is the mean of Fishy's move times, M2 is the mean of Chet's move times, and SD is the pooled standard deviation.

Pooled standard deviation, in this case, is calculated as:

SD = sqrt[(n1 - 1) x SD1^2 + (n2 - 1) x SD2^2 / (n1 + n2 - 2)]

where n1 and n2 are the sample sizes (in this case, both are 50) and SD1 and SD2 are the standard deviations of Fishy's and Chet's move times, respectively.

Plugging in the values, we get:

d = (330 - 420) / sqrt[(49 x 45^2 + 49 x 45^2) / 98]

d = -90 / 31.82

d = -2.83

Since we are interested in the absolute value of the effect size, we can take the absolute value of d, which gives us 2.83. However, since the question asks for the effect size to one decimal place only, the answer is 2.8, which corresponds to choice C.

A Chess Team Advisor Wanted to Examine How Long It