Question

Dr. Stambaugh hypothesizes that a novel anxiety drug will affect memory. To examine this, the drug is administered to a random sample of 31 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 51 seconds. Normal rats have an average completion time of 45 seconds with a standard deviation of 12 seconds. What can Dr. Stambaugh conclude with α = 0.01?

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

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

c) Compute the appropriate test statistic(s) 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--- Reject H0 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--- na trivial effect small effect medium effect large effect
r² =  ;   ---Select--- na trivial effect small effect medium effect large effect

f) Make an interpretation based on the results.

Rats on the anxiety drug were significantly slower completing the maze

.Rats on the anxiety drug were significantly quicker completing the maze.   

The anxiety drug did not affect rats in completing the maze.

Answer 1

(f). Rats on the anxiety drug were significantly quicker completing the maze.

Ang. na 31, = 51, 4:45. , S = 12, 270.01, df=n=30 (ar: Appropriate test = One sample t test b) Penu Population Normal rats Sample - Trained rats. .161 Hypothesis Ho! u = 45 Hii Ni>. 45 Critical value tai ind=t0.01,30 - 2.46( from t table) Tent stotistio 51 - 45 2794 12 113 T+= 2:764 Decision Since t=2.784> tegint = 2.46 Hence, we reject Ho at 6=0.01 (etto 99% confidence interval for H I N=001 thizin-1 = +0.005,30 Lower limit = X-E = 51- 5.93 = 2.75 TL = 45.07 Margin of Error (E) Upper Limit= x + E - 51+5.93 E = tolent TUL = 56.93] = 2.754 1995(1 = ( 45.07, 56.93) TE = 5.93 (el Effects size (d), Jh - = 0.500 (Medium) 13 = 9 2 = t² (2.7 842 - 0.205 (amall ). 42447 (2.764)2 + 30