Question

Part 1. 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 65 students in the highest quartile of the distribution, the mean score was x = 178.30. Assume a population standard deviation of  σ = 8.29. These students were all classified as high on their need for closure. Assume that the 65 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

Part 2.

A research group conducted an extensive survey of 2938 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered through hour-long telephone interviews with a nationally representative sample. In response to the question, "What does success mean to you?" 1577 responded, "Personal satisfaction from doing a good job." Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. Find a 90% confidence interval for p. (Round your answers to three decimal places.)

Lower Limit

Upper Limit

Answer 1

Solution :

1) Given that,

Point estimate = sample mean = = 178.30

Population standard deviation =    = 8.29

Sample size = n = 65

At 95% confidence level

= 1 - 95%  

= 1 - 0.95 =0.05

/2 = 0.025

Z/2 = Z_{0.025 = 1.96}

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

= 1.96 * ( 8.29 /  65 )

= 2.02

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

  ± E

178.30 ± 2.02

( 176.28, 180.32 )  

lower limit = 176.28

upper limit = 180.32

2) Given that,

n = 2938

x = 1577

Point estimate = sample proportion = = x / n = 1577 / 2938 = 0.537

1 - = 1 - 0.537 = 0.463

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_{0.05 = 1.645}

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.537 * 0.463) /2938 )

= 0.015

A 90% confidence interval for population proportion p is ,

± E

= 0.537 ± 0.015

= ( 0.522, 0.552 )

lower limit = 0.522

upper limit = 0.552