Question

Dr. Stambaugh predicts that a novel antidepressant drug will effect memory. To examine this, the drug is administered to a random sample of 22 trained rats. The rats were then placed in a maze and timed on how long it took them to complete it. The average completion time was 47 seconds. Normal rats have a mean completion time of 45 seconds with a standard deviation of 11 seconds. What can be concluded with an α of 0.05?

a) What is the appropriate test statistic?
---Select one--- (na, z-test, one-sample t-test, independent-samples t-test, or related-samples t-test)

b)
Population:
---Select one--- (the maze, normal rats, time to complete maze, trained rats, memory)
Sample:
---Select one--- (the maze, normal rats ,time to complete maze, trained rats, memory)

c) Obtain/compute the appropriate values to make a decision about H₀.
(Hint: Make sure to write down the null and alternative hypotheses to help solve the problem.)
critical value =  ; test statistic =
Decision:  ---Select one--- (Reject H0 or Fail to reject H0)

d) If appropriate, compute the CI. If not appropriate, input "na" for both spaces below.
[  ,  ]

e) Compute the corresponding effect size(s) and indicate magnitude(s).
If not appropriate, input and select "na" below.
d =  ;   ---Select one--- (na, trivial effect, small effect, medium effect, or large effect)
r² =  ;   ---Select one--- (na, trivial effect, small effect, medium effect, or large effect)

f) Make an interpretation based on the results.

a.Rats on the antidepressant drug were significantly slower in completing the maze.

b. Rats on the antidepressant drug were significantly quicker in completing the maze.    

c. The antidepressant drug did not significantly impact the rats in completing the maze.

Answer 1

a)  z-test

We will use a z-test for one mean, with a known population standard deviation.

b)

Population:

normal rats

Sample:

trained rats

c)

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho:

Ha:

Test Statistics

The z-statistic is computed as follows:

The critical value for a two-tailed test is

Fail to Reject H0

d)

Confidence Interval

The 95% confidence interval is

42.403<μ<51.597

e)

small effect (mean difference is 0.2 standard deviation)

Small effect
f)

c. The antidepressant drug did not significantly impact the rats in completing the maze.

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