A study was done to determine the effect of the MECLR (methanol extract of C lut
ID: 3356802 • Letter: A
Question
A study was done to determine the effect of the MECLR (methanol extract of C lutea) on the mating habits of rabbits.
Twenty male rabbits were randomly assigned to one of four groups:
Control: no MECLR or sildenafil citrate
40 mg/kg body weight of MECLR
80 mg/kg body weight of MECLR
0.5 mg/kg body weight of sildenafil citrate
The next day, each male was paired with a female rabbit, and the number of times the male tried to mount the female (“mounting frequency”) was recorded. The results are as follows:
Control rabbits: 2, 3, 3, 3, 3
40 mg/kg MECLR: 4, 5, 5, 5, 6
80 mg/kg MECLR: 6, 6, 7, 8, 8
0.5 mg/kg sildenafil: 8, 9, 11, 11, 11
Question:
Suppose a large number of mice will be given 60 mg/kg of MECLR. Assume that rate of “mounting frequency” for any given dosage is normally distributed (even though the actual number of mounts will of course be an integer). Based on these assumptions, what percentage of mice would be expected to have mounting frequency values of 9.5 or higher?
A. None: T = 11.5, proportion of mice with MF > 9.5 is 9.9 x 10–31
B. Very few: T = 3.5, percentage of mice with MF > 9.5 is 0.026%
C. Very few: T = 3.3, percentage of mice with MF > 9.5 is 0.045%
D. Few: T = 1.87, percentage of mice with MF > 9.5 is 3.07%
E. Some: T = 0.80, percentage of mice with MF > 9.5 is 21.0%
Explanation / Answer
Answer to the question)
from the given 20 observations of mounting frequency
We get avergae number of mounting ( M) = 6.2
Standard deviation (SD)= 2.8580
Since it is said its a large sample , we consider n to be 100
.
We need to find the probability of x > = 9.5
P(x > = 9.5) = 1 - P( x < 9.5)
P(X < 9.5) = P( t < (9.5-6.2)/(2.8580/sqrt(100)) = P(t < 11.5) = 0.99999
P(x > = 9.5) = 1 - 0.99999 = 9.9 x 10^-31
Thus answer choice A is correct
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