A psychologist develops a new inventory to measure depression. Using a very larg

Question

A psychologist develops a new inventory to measure depression. Using a very large standardization group of "normal" individuals, the mean score of this test is m = 54 and s = 10, and the scores are normally distributed. To determine whether the test is sensitive in detecting those individuals who are severely depressed, a random sample of patients who are described as depressed by a therapist is selected and given the test. Presumably, the higher the score on the inventory, the more depressed the patient is. The data are as follows: 53, 56, 60, 54, 56, 65, 63, 74, 62, 71, 51, 63, 81, 60, 66, and 77.
a. Calculate the mean and standard deviation of the sample.
mean= 63.25 SD=8.79

b. Calculate the SE of the distribution of sample means.
c. Using the .01 significance level (and a two-tailed test), what is the critical z-score?
d. What is the calculated z-score?

e. What should the researcher conclude (Reject null hypothesis or fail to reject null hypothesis)?
f. Write your results in APA format (1 point).

Explanation / Answer

(a) x-bar = 63.25, s = 8.79

(b) SE = /n = 10/16 = 0.625

(c) Critical z = ± 2.576

(d) = 10, = 54

Calculated z = (x-bar - )/SE = (63.25 - 54)/0.625 = 14.8

(e) Reject the null hypothesis

(f) There is sufficient evidence that the mean score is different from 54.

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