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 76 students in the highest quartile of the distribution, the mean score was x = 178.10. Assume a population standard deviation of σ = 8.21. 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 2.1 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.)

Answer 1

Solution:

Given:

Sample mean =

Population standard deviation =

Confidence level = c = 99% = 0.99

Margin of Error = E = 2.1

We have to find sample size n.

Z_c is z critical value for c = 0.99 confidence level.

Find Area = ( 1+c)/2 = ( 1 + 0.99 ) / 2 = 1.99 /2 = 0.9950

Thus look in z table for Area = 0.9950 or its closest area and find corresponding z critical value.

From above table we can see area 0.9950 is in between 0.9949 and 0.9951 and both are at same distance from 0.9950, Hence corresponding z values are 2.57 and 2.58

Thus average of both z values is 2.575

Thus Z_c = 2.575

Thus

( Note: Sample size is always rounded up)

Thus required sample size = n = 102