A forensic psychologist conducted a study to examine whether being hypnotized du

Question

A forensic psychologist conducted a study to examine whether being hypnotized during recall improves the ability of witnesses remembering facts about an event. Eight participants watched a short film of a mock robbery, after which each participant was questioned about what he or she had seen. The four participants in the experimental group were questioned while they were hypnotized and gave 15, 22, 18, and 17 accurate responses. The four participants in the control group gave 20, 25, 24, and 23 accurate responses. Using the 0.05 significance level, do hypnotized witnesses perform better than witnesses who are not hypnotized?

a. What type of test will you use to answer this question? Why did you choose this test?
b. Use the five steps of hypothesis testing. Be sure to use the appropriate comparison distribution and

Show all calculations for your comparison distribution.
c. In two or three sentences, summarize the purpose of the forensic psychologist’s study, summarize

how the data was collected (procedure), and then explain what your results mean to a person who has never had a course in statistics.

Explanation / Answer

a. Here the two sample groups have different paticipants so we will use two sample t - test for equal variances for independent samples

We choose t - test as we dont know the population varaince..

b. Step I :

H0 : The experimentral group and control group population mean accurate responsens are same. 1 = 2

Ha : The experimentral group group mean accurate responses are better than control group population mean accurate responsens. 1  <  2

so here 1 is experiment group and 2 is control group.

Step II : Confidence level = 0.05

Step III

Critical value; rejection region

here dF = 4 + 4 -2 = 6 and alpha = 0.05 for one tailed test

so tcriiticall = t6,0.05 =1.9432

so we will reject the null if

l t l > tcritical

Step IV: Test statisti c

sample mean of experiment group x1 = 18

sample mean of control group x2 = 23

sample standard deviaiton s1 = 2.944

sample standard deviation of control group s2 = 2.160

pooled standard deviation sp = sqrt [(s1 2 + s1 2)/2 = 2.582

so Test statistic

t =( x1 - x2)/ sp * sqrt (2/n) =

t = (18 - 23)/ [2.582 * sqrt(2/3)] = -2.7386

Step IV "

p- value = 0.0169 < 0.05

l t l > tcriticala

Step V : so we shall reject the null hypothesis.

(c) Here as we reject the null hypothesis so we can conclude that hypnotized witnesses perform better than witnesses who are not hypnotized.

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