Researchers know that the average level of test anxiety is 70 (=10.05). Sally’s

Question

Researchers know that the average level of test anxiety is 70 (=10.05). Sally’s level of test anxiety is 50. Sally wants to know if she has a “normal” amount of test anxiety.

List the following information:

A. Independent Variable:_____________________                                    B. Scaling:________________________

C. Dependent Variable:______________________                         D. Scaling:________________________

E. Can random selection be used? Explain

F. Can random assignment be used? Explain

G. What test should the shelter workers use to examine their hypothesis?_______________________

H. Why that test?

I. What are the assumptions of this test?

J. What is the null and alternative hypotheses?

H0:

H1:

K. Draw the distribution and the critical values (cut-off values/rejection regions):

L. Calculate the test value:

M. Make the decision:          Retain/Fail to Reject H0                                  Reject H0

N. What type of error could the shelter workers have made?                       Type I              Type II

O. Why that type of error?

P. If the shelter workers found an effect size of .37, is this small, medium, or large?

Q. What can the researchers do to influence power?

R. Interpret the results

Explanation / Answer

Given that mean = 70 and sd = 10.05

What is the null and alternative hypotheses?

H0: Sally's anxiety level is not under the normal range

H1: Sally anxiety level falls under the normal range

K. Draw the distribution and the critical values (cut-off values/rejection regions):

as the sample size is not given in the question , but the mean and sd values are given so we can go for a z test

The z stat is calculated as (50-70)/10.05 = -1.99

assume alpha=0.05 , calculate the z critical as -1.64 from the z tables

as the z stat > z critical , hence we reject the null hypothesis in favor of alternate hypothesis

L. Calculate the test value:

-

The z stat is calculated as (50-70)/10.05 = -1.99 , using Z = (X-Mean)/SD

M. Make the decision:          Retain/Fail to Reject H0                                  Reject H0

as stated above , the t stat > t critical value , hence we reject ho in favor of h1

Please note that we can answer only 4 subparts of the quesiton at a time , hence have answered the most relevant ones

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