Question

You give your 30 students a pre-test for their knowledge of statistics on the first day of class for the semester. Their mean score was 40% correct (s = 5.00). Then on the comprehensive final exam you note that their mean score was 70% correct (s = 6.00). The standard deviation of the difference between the 2 means (40 vs. 70) is 4.00. What is the confidence interval for the mean difference in statistics knowledge before and after the course? What is the effect size in comparing knowledge students had before and after the course?

Answer 1

do you have the formula

Answer 2

Yes. Hold on.

Answer 3

CI 0.95 = XD + t 0.05 (sX ) = XD + t 0.05 (s / SR [ N ] )

This is the formula for the confidence interval.

Answer 4

what is XD

Answer 5

X Bar and I forgot what the number is called when it's little and beneath the Big number? Sorry.

Answer 6

Looks like that.

Answer 7

ok x bar d stands for the difference of the means

Answer 8

sample difference mean

Answer 9

Okay. I see.

Answer 10

and you need to find the t score for .05

Answer 11

for 29 degree of freedom
the t score is 2.0452

Answer 12

so this gives us

CI 0.95 = XD + t 0.05 (sX ) = XD + t 0.05 (s / SR [ N ] )
= (70-40) + 2.0452 * 4 / sqrt(30)

Answer 13

actually it is an interval
((70-40) - 2.0452 * 4 / sqrt(30) , (70-40) + 2.0452 * 4 / sqrt(30)

Answer 14

I was using the s= 5.00 and the s= 6.00 in my equation. haha

Answer 15

(28.506, 31.494)

Answer 16

the next question, can you double check for typo, i dont understand the english

Answer 17

What is the effect size in comparing knowledge students had before and after the course?

Answer 18

Yeah. That does sound confusing. I think it might be asking to use Cohen's equation.