Duplicate Question
The question on this page has been marked as a duplicate question.
Original Question
Twenty students randomly assigned to an experimental group receive an instructional program: 30 in a control group do not. Afte...Asked by Anonymous
Twenty students randomly assigned to an experimental group receive an instructional program; 30 in a control group do not. After 6 months, both groups are tested on their knowledge. The experimental group has a mean of 38 on the test (with an estimated population standard deviation of 3); the control group has a mean of 35 (with an estimated population standard deviation of 5). Using the .05 level, what should the experimenter conclude?
Answers
Answered by
PsyDAG
Z = (mean1 - mean2)/standard error (SE) of difference between means
SEdiff = โ(SEmean1^2 + SEmean2^2)
SEm = SD/โ(n-1)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to your Z score.
SEdiff = โ(SEmean1^2 + SEmean2^2)
SEm = SD/โ(n-1)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to your Z score.
Answered by
Mrs. Nation
(a) and (c)
Ho: u(ex) - u(ct) = 0
Ha:u(ex) โ u(ct) is not equal to 0
Test stat: t = (38-35) / sqrt[9/20 + 25/30] = 2.6482
p-value = 2*P(t > 2.6482 with df = 48) = 0.0109
Conclusion:
At the 5% significance level, reject Ho because the p-value is less than 5%
Ho: u(ex) - u(ct) = 0
Ha:u(ex) โ u(ct) is not equal to 0
Test stat: t = (38-35) / sqrt[9/20 + 25/30] = 2.6482
p-value = 2*P(t > 2.6482 with df = 48) = 0.0109
Conclusion:
At the 5% significance level, reject Ho because the p-value is less than 5%
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.