Question

In an experiment to learn whether Substance M can help restore memory, the brains of $20$rats were treated to damage their memories. The rats were trained to run a maze. After a day, $10$rats (determined at random) were given M and $7$of them succeeded in the maze. Only $2$of the $10$control rats were successful. The two-sample z test for “no difference” against “a significantly higher proportion of the M group succeeds”

(a) gives $z=2.25,P<0.02$

(b) gives $z=2.60,P<0.005$

(c) gives $z=2.25,P<0.04$but not $<0.02$

(d) should not be used because the Random condition is violated

(e) should not be used because the Normal condition is violated.

Short answer

The two-sample z test for “no difference” against “a significantly higher proportion of the M group succeeds” option (e) should not be used because the Normal condition is violated.

Step-by-step solution

Given information

The treated rats to damage their memories $=20$

Number of rats given M $=10$

Succeeded rats in maze$=7$

successful rats$=2$

Explanation

To achieve the Normal requirement, the number of failures in the number of successes in both samples must be greater than $10$.

The Normal criterion is not met because there are $7$successes in the first sample and $2$successes in the second sample.

As a result, it shouldn't be used because the Normal condition has been broken.