Question

The mean amount of time that a randomly selected mouse finds its way out of a maze is 18 seconds. A researcher thought that a loud noise would make the mouse to finish the maze faster (her theory). She conducted an experiment to test her theory by placing a mouse in the maze and measuring how long it takes to finish the maze with a loud noise as stimulus. She randomly sampled 25 mice and measured their times. The average was found to be 16.8 seconds. Assume that the standard deviation of the population of mice is 4.4 seconds. Conduct a test (her theory) at a 5% significance level.

(explanation please)

(a) Check the required conditions for a hypothesis test to determine if they are satisfied.

(b) Describe, in words, what ?? would represent in this case and state the hypotheses (??0, ????) using ??.

(c) Calculate the test statistic and the p-value. (d) Determine whether we reject the null hypothesis OR fail to reject the null hypothesis at a 5% significance level.

(e) Write an appropriate conclusion in the context of the problem.

(f) If we use a 10% significance level (instead of using a 5% significance level), would we still reject the null hypothesis OR fail to reject the null hypothesis?

Answer 1

(a) Required conditions for hypothesis testing:

The sample must be randomly selected - Yes

σ must be known - Yes; Population SD is know

(b) = Represents the mean time taken by a randomly selected mouse to find its way out of the maze

H0: = 18

Ha: 18

The Null hypothesis states that the mean time taken by a randomly selected mouse to find its way out of the maze is 18 seconds even with stimulus.

Alternate hypothesis states that the mean time taken by a randomly selected mouse to find its way out of the maze is less than 18 seconds with stimulus (loud noise).

(c) Test statistics

z stat = -1.364

P( z< -1.364)

P ( Z<−1.364 ) = 1−P ( Z<1.364 ) = 1−0.9131 = 0.0869

P( z< -1.364) = 0.869

(d) The z-critical value for a left-tailed test, for a significance level of α=0.05

zc=−1.64

As the z stat (-1.364) is not in the rejection area, we fail to reject the Null hypothesis.

Also P value (0.0869) is greater than level of significane (0.05), we fail to reject the Null hypothesis.

(e) As we fail to reject the Null hypothesis, it indicates that the mean time taken by a randomly selected mouse to find its way out of the maze is 18 seconds even with stimulus. (loud noise)

(f) 10% level of significance

The z-critical value for a left-tailed test, for a significance level of α=0.10 ; zc = − 1.28

At 10% level of significance, the z stat (-1.364) falls in the rejection area. Hence we reject the Null hypothesis.

Also P value (0.0869) is less than level of significane (0.10), we reject the Null hypothesis.