How would you figure out the lower and higher limit for these 3 problems? 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 agroup of students enrolled in an introductory psychology course. The "need for closure scale" has scores ranging from 101 to 201. For the 66 students in the highest quartile of thedistribution, the mean score was x = 178.70. Assume a population standard deviation of o = 7.65. These students were all classified as high on their need for closure. Assume that the 66students 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 u on the "need forclosure scale" for all students with a high need for closure. (Round your answers to two decimal places.)lower limitupper limit A study of parental empathy for sensitivity cues and baby temperament (higher scores mean more empathy) was performed. Let x, be a random variable that represents the score of amother on an empathy test (as regards her baby). Let x, be the empathy score of a father. A random sample of 34 mothers gave a sample mean of x1 = 67.00. Another random sampleof 35 fathers gave x, = 60.19. Assume that o, = 11.97 and o, = 10.85.(a) Let u, be the population mean of x, and let u, be the population mean of x2. Find a 95% confidence interval for u1 -(Use 2 decimal places.)lower limitupper limit(h) Examine th e confidence interval and explain what it means in the context of this problDoes the confidence interval contain all positiveall nec ative or both positive and

Answered: Three experiments investigating the…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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How would you figure out the lower and higher limit for these 3 problems?

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 66 students in the highest quartile of the
distribution, the mean score was x = 178.70. Assume a population standard deviation of o = 7.65. These students were all classified as high on their need for closure. Assume that the 66
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 u 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

A study of parental empathy for sensitivity cues and baby temperament (higher scores mean more empathy) was performed. Let x, be a random variable that represents the score of a
mother on an empathy test (as regards her baby). Let x, be the empathy score of a father. A random sample of 34 mothers gave a sample mean of x1 = 67.00. Another random sample
of 35 fathers gave x, = 60.19. Assume that o, = 11.97 and o, = 10.85.
(a) Let u, be the population mean of x, and let u, be the population mean of x2. Find a 95% confidence interval for u1 -
(Use 2 decimal places.)
lower limit
upper limit
(h) Examine th e confidence interval and explain what it means in the context of this probl
Does the confidence interval contain all positive
all nec ative or both positive and

Expert Solution

Step 1

Given,

$x = 178.70 σ = 7.65 n = 66 α = 1 - 0.95 = 0.05$

Since population standard deviation is given, therefore z-confidence interval will be obtained.

As per z-distribution table, the critical value of z at 0.05 level of significance is obtained as 1.96.

The required 95% confidence interval for the population mean score $μ$ on the "need for closure scale" for all students with high need for closure is obtained as-

$\begin{array}{rcl} C I & = & x \pm z_{c r i t} \times \frac{σ}{\sqrt{n}} \\ = & 178.70 \pm (1.96) \times \frac{7.65}{\sqrt{66}} \\ = & 178.70 \pm 1.84563 \\ = & (176.85437, 180.54563) \\ ~ (176.85, 180.55) \end{array}$

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