Question

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 19 trials, with a standard deviation of 5.(The number of trials to learn this maze is normally distributed.) The researcher now tries an ordinary rat in the maze, but with sound. The rat takes 35 trials to learn the maze. Complete parts a and b below.

(a) Using a 0.05 level, what should the researcher conclude? First, determine the null hypothesis. Choose the correct answer below.

Determine the research hypothesis. Choose the correct answer below.

Determine the comparison distribution. Choose the correct answer below.

Determine the cutoff sample score(s).

Determine the sample score on the comparison distribution.

Decided whether or not to reject the null hypothesis.

Label the cutoff points on the distribution below.

Explain your findings to somebody who is not in a statistics class but is familiar with mean, standard deviation, and z-score.

Answer 1

Given:

Sample mean, =35; Standard deviation, =5 and Sample size, n=Number of different attempts(tests) on rat =1 or number of rats, n =1

Hypotheses:

Null hypothesis(H0):

The mean number of trials taken by the rat to learn the maze (with sound) is not significantly greater than 19.

Research hypothesis(H1):

The mean number of trials taken by the rat to learn the maze (with sound) is significantly greater than 19. (right-tailed test).

Test statistic:

Z = = =3.2

Critical (table) value:

Significance level,

​For right tailed test, Z_0.05 =1.645

This is the cut-off Z-score. So, the cut-off sample score, X is calculated as follows:

Z =(X - ​​​​​​)/ X =Z + =1.645(5)+19 =27.225

Labelling the cut-off sample score:

So, upto 27.225 (or 27) trials, we cannot reject the null hypothesis and claim number of trials taken by the rat is significantly > 19. But if the rat takes more than 27 trials, then we will have sufficient evidence to reject the null hypothesis and to claim the time taken by the rat is significantly > 19.

Rejection region for null hypothesis(H0):

Conclusion:

Since the test statistic of 3.2 fell in the rejection region, there is a sufficient evidence to reject the null hypothesis at 0.05 significance level and claim that the number of trials taken by the rat to learn the maze with sound is significantly greater than 19.