Three experiments investigating the relation between need for
cognitive closure and persuasion were performed. Part of the study
involved administering a "need for closure scale" to a group of
students enrolled in an introductory psychology course. The "need
for closure scale" has scores ranging from 101 to 201. For the 77
students in the highest quartile of the distribution, the mean
score was x = 178.30. Assume a population standard
deviation of σ = 7.99. These students were all classified
as high on their need for closure. Assume that the 77 students
represent a random sample of all students who are classified as
high on their need for closure. How large a sample is needed if we
wish to be 99% confident that the sample mean score is within 1.8
points of the population mean score for students who are high on
the need for closure? (Round your answer up to the nearest whole
number.)
students
Solution :
Given that,
Population standard deviation = = 7.99
Margin of error = E = 1.8
At 99% confidence level the z is,
= 1 - 99%
= 1 - 0.99 = 0.01
/2 = 0.005
Z/2 = Z0.005 = 2.576
sample size = n = [Z/2* / E] 2
n = [ 2.576 * 7.99 / 1.8 ]2
n = 130.75
Sample size = n = 131
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