Hi !I do not understand the difference between margin error and confidence interval.I am struggling to solve Question 3 that is circled in red. I would appreciate any help, thank you :) We want to know the true proportion of students that are ok with online courses.We take a sample of 120 students, and 72 said they are ok with online courses. We want to createa 95% confidence interval. Answer the questions below:3) What is the margin of error for this data? 1) What is the pointestimate for the populationproportion?Sample proportion >= x/nX= 72; n = 120point estimate ofthePopulation proportion is 9.602) What is a 95% confidence interval for thisdata?= 72/120= 0.60faveralde cases >> X = 72Sample size → n = 120confidence level = 95%critical value for 2 tailedatis 7 = 1.96en exel fiuction, type in = NORM. S. INV (1-0.05/2)= 0.05x = 5% =Sample population:第一张72/120 = 0.695% confidence interval isC.I. - (P + * *√√(1-P))n0.6 ± 1.96 + √0.6(1-0.6))120C.I. (0.512, 0.688)

Sample size studentsNumber of students ok with online classes Confidence level is 95% i.e., 0.95.

We want to know the true proportion of students that are ok with online courses. We take a sample of 120 students, and 72 said they are ok with online courses. We want to create a 95% confidence interval. Answer the questions below: 3) What is the margin of error for this data?

We want to know the true proportion of students that are ok with online courses. We take a sample of 120 students, and 72 said they are ok with online courses. We want to create a 95% confidence interval. Answer the questions below: 3) What is the margin of error for this data?

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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Question

100%

Hi ! I do not understand the difference between margin error and confidence interval. I am struggling to solve Question 3 that is circled in red. I would appreciate any help, thank you :)

We want to know the true proportion of students that are ok with online courses.
We take a sample of 120 students, and 72 said they are ok with online courses. We want to create
a 95% confidence interval. Answer the questions below:
3) What is the margin of error for this data?

1) What is the point
estimate for the population
proportion?
Sample proportion >= x/n
X= 72; n = 120
point estimate of
the
Population proportion is 9.60
2) What is a 95% confidence interval for this
data?
= 72/120
= 0.60
faveralde cases >> X = 72
Sample size → n = 120
confidence level = 95%
critical value for 2 tailed
at
is 7 = 1.96
en exel fiuction, type in = NORM. S. INV (1-0.05/2)
= 0.05
x = 5% =
Sample population:
第一张
72/120 = 0.6
95% confidence interval is
C.I. - (P + * *√√(1-P))
n
0.6 ± 1.96 + √0.6(1-0.6))
120
C.I. (0.512, 0.688)

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