A researcher wants to test whether a certain sound will make rats do worse on learning tasks. It is known that an ordinary rat can learn to run a particular maze correctly in 18 trials, with a standard deviation of 6. (The number of trials to learn this maze is normally distributed.) The researcher now tries an ordinary rat in the maze, but with the sound. The rat takes 38 trials to learn the maze. (a) Using the .05 level, what should the researcher conclude? Solve this problem explicitly using all five steps of hypothesis testing, and illustrate your answer with a sketch showing the comparison distribution and the rejection region.. (b) Then explain your answer to someone who has never had a course in statistics (but who is familiar with mean, standard deviation, and Z scores).

So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.

Step-by-step explanation:

In this data we have

Mean= u = 18

X= 38

Standard deviation = s= 6

1) We formulate the null and alternate hypothesis as

H0:  u = 18  against           Ha : u > 18  One tailed test .

2) The significance level alpha = ∝= 0.05 and  Z alpha has a value ± 1.645 for one tailed test.

3)The test statistics used is

Z= X- u / s

z= 38-18/6= 3.333

4) The calculated value of z = 3.33 is greater than the z∝ = 1.645

5) So we reject the null hypothesis and accept the alternate hypothesis that rats learn slower with sound.

First we set the criteria for determining the true of value of the variable. That whether the rats learn in less or more than 18 trials.

Then we find the value of z for the given significance value given and the test about to be checked.

Then the test statistic is determined and calculated.

Then both value of z and z alpha re compared. If the test statistics falls in the rejection region  reject the null hypothesis and conclude alternate hypothesis is true.

The figure shows that the calulated z value lies outside the given z values

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