Question

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 83 students in the highest quartile of the distribution, the mean score was x = 177.30. Assume a population standard deviation of σ = 7.69. These students were all classified as high on their need for closure. Assume that the 83 students represent a random sample of all students who are classified as high on their need for closure.

Find a 95% confidence interval for the population mean score μ on the "need for closure scale" for all students with a high need for closure. (Round your answers to two decimal places.)

lower limit'

upper limit

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 177.30

Population standard deviation =    = 7.69

Sample size n =83

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E = Z/2 * ( /n)

=1.96 * (7.69 /   83)

= 1.65
At 95% confidence interval estimate of the population mean
is,

- E < < + E

177.30 - 1.65 <   <177.30 + 1.65

175.65 <   < 178.95

( lower limit=175.65, upper limit= 178.95)