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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?

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
18th Edition
ISBN:9780079039897
Author:Carter
Publisher:Carter
Chapter10: Statistics
Section10.5: Comparing Sets Of Data
Problem 1GP
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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?
Transcribed Image Text: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)
Transcribed Image Text: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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