Experiments on learning in animals sometimes measure how long it takes for mice

Question

Experiments on learning in animals sometimes measure how long it takes for mice to find their way through a maze. The mean time is 18 seconds for one particular maze. A researcher thinks that a loud noise will cause the mice to complete the maze faster. She measures how long each of 36 randomly selected lab mice takes with noise as stimulus. Her sample data yielded a mean of 16.5 seconds and a standard deviation of 6.4 seconds Suppose the calculated test statistic is t = -2.16. Assuming all conditions are met, what is the p-value for this left-tailed test?

Explanation / Answer

For calculated test statistic t=-2.16 the P value for left tailed test is  .018852.

But the actual calculated test statistic t value is

t= 16.5-18/6.4/sqrt(36)

t= -1.5/6.4/6

t= -1.5/1.0667

t= -1.40625

So the p value for thelefttailedtestfor this t=-1.40625 is  .08427.

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