Question

A scientist decides she wants to determine the effect of being housed in groups or separately on the learning ability of the common mouse, Mus musculus. She sets up her experiment in the lab by having two sets of mice: two large cages with three mice each, and six smaller cages each containing only one mouse. She keeps them in the same room and gives all the mice equal access to food and water, and periodically runs the mice in a maze to assess learning ability. What is the flaw, if any, in her experimental design?

a. She should have kept the mice in solitude in a separate room.
b. She should have given the mice in solitude less food than those who lived in groups.
c. There is no problem with her experimental design.
d. She should have given the mice in solitude extra toys and a wheel in order to give them something to occupy their time.
e. She should have kept the cage size uniform for the mice in solitude and the mice in groups; thus, she inadvertently altered more than one variable.

Answer 1

E

Step-by-step explanation:

The answer is E because the size of the cage will affect the mice.

The answer cannot be A, because keeping the mice in solitude in separate rooms would have made no difference to the experiment because in either situation, the mice are in isolation.

The answer cannot be B, because in giving the one group less food, this would change the behaviour of the mice completely. If the mice don't receive enough food and water, they may become lethargic which would affect the results.

The answer cannot be C, because E is the correct answer. If she had kept the cages uniform (and hence kept the conditions the same) then this answer would be right, however in this example, her experimental design is flawed.

The answer is not D, because if she were to do this, she would, once again, be introducing factors that may affect the results of the test.