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 = 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_______________

Answer 1

Solution :

Given that,

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 = Z_0.05 = 2.576

Sample size = n = [(Z_/2 * ) / E]²

= [(2.576 * 7.53) / 1.7]²

= 130.19

Sample size = 130

Sample would be needed = 130