A researcher knows that rats learn to solve a T-maze in 15 trials on average, with a standard deviation of 8. He believes that rats treated with low doses of THC (the active ingredient in marijuana) will require more trials to learn a T-maze. He selects a random sample of 25 rats, administers a low dose of THC to each, then measures the time it takes them to learn a T-maze. The mean of these 25 scores is 4.4. Assuming that the distribution of means is normal, which of the following statements is true?
a. zobs = 1.875, p < .05
b. zobs = –1.875, p < .05
c. zobs = 1.875, p > .05
d. zobs = –1.875, p > .05
Ans: A
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