Question

An educational researcher devised a wooden toy assembly project to test learning in 6-year-olds. The time in seconds to assemble the project was noted, and the toy was disassembled out of the child’s sight. Then the child was given the task to repeat. The researcher would conclude that learning occurred if the mean of the second assembly times was less than the mean of the first assembly times. Find the 99% confidence interval for the difference in means.

Child	1	2	3	4	5	6	7
Trial 1	108	140	154	115	130	111	108
Trial 2	99	118	154	96	107	102	110

Answer 1

solution:-

given that

trial 1:

mean x1 = 123.7

standard deviation s1 = 18.04

n1 = 7

trial 2:

mean x2 = 112.3

standard deviation s2 = 19.81

n2 = 7

degree of freedom df = (n1+n2)-2 = (7+7)-2 = 12

we look into t table with df and 99% confidence

critical value t = 3.055

confidence interval formula

=> (x1-x2) +/- t * sqrt(s1^2/n1 + s2^2/n2)

=> (123.7-112.3) +/- 3.055 * sqrt((18.04^2/7) + (19.81^2/7))

=> -19.5 < 42.3