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 76
students in the highest quartile of the distribution, the mean
score was x = 175.30. Assume a population standard deviation of σ =
7.53. These students were all classified as high on their need for
closure. Assume that the 76 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 1.7 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 = = 7.53
Margin of error = E = 1.7
At 99% confidence level the z is ,
= 1 - 99% = 1 - 0.99 = 0.01
/ 2 = 0.01 / 2 = 0.005
Z/2 = Z0.005 = 2.576
sample size = n = (Z/2* / E) 2
n = (2.576 * 7.53/ 1.7)2
n = 130.19
n = 130
Sample size = 130
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