Find the confidence interval for estimating the population proportion for parts (a) through (c).a. 92.5% confidence level; n = 550; p=0.10b. 95% confidence level; n=220; p=0.02c. a 0.03; n =325; p=0.50Click the icon to view the standard normal table of the cumulative distribution function.a. The confidence interval is P(Round to four decimal places as needed.)CAD

The formula of the confidence interval for population proportion is,where, denotes the sample…

Answered: Find the confidence interval for…

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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Scores on an English test from 30 students in Class One has a mean of 33 and a standard deviation of 7.2. Scores on an English test from 40 students in Class Two has a mean of 39 and a standard deviation of 5.5. Assuming the two classes are independent, construct a 90%confidence interval for difference between the mean scores for Class One and Class Two. Round your answer to 3 decimal places. Base on the confidence interval, what conclusion can you draw? Group of answer choices (-8.530,-3.470) Group 2 is bigger (-8.530,-3.470) Group 1 is bigger (-8.639,-3.361) No conclusion (-8.639,-3.361) Group 2 is bigger (-6.769,-5.231) Group 2 is bigger
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Calculate the margin of error and construct the confidence interval for the population mean using the Student's t-distribution (you may assume the population data is normally distributed). T-Distribution Table a. x 88.2, n = 69, s 17.9, 98% confidence = E Round to two decimal places if necessary EAPA SAMSUNG SUBMIT QUESTION SAVE PROGRESS SUBT
USE 3 DECIMALS A researcher believes that reading habit of newspaper is decreasing every year. In order to see the true population proportion of people who are reading newspaper s/he collect a randomly selected people on the yearly basis. The following table gives the number of people who do not read newspaper with respect to the years. Construct a 90 % confidence interval for the true population proportion difference of people who are reading newspaper in 2005 and 2010. (P2005 - P2015) 2010 2005 In 1200 1000 Number of people who do not read newspaper 300 400 KOnceki
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Use the normal distribution to find a confidence interval for a difference in proportions p₁ Assume the results come from random samples. A 90% confidence interval for p₁ - p₂ given that p = 0.75 with n = 440 and p₂ = 0.65 with #₂ = 360 Give the best estimate for p₁ - p₂, the margin of error, and the confidence interval. Round your answer for the best estimate to two decimal places and round your answers for the margin of error and the confidence interval to three decimal places. Best estimate: i Margin of error: i Confidence interval: - P₂ given the relevant sample results. to
please help
USE 3 DECIMAL A researcher believes that reading habit of newspaper is decreasing every year. In order to see the true population proportion of people who are reading newspaper s/he collect a randomly selected people on the yearly basis. The following table gives the number of people who do not read newspaper with respect to the years. Construct a 90 % confidence interval for the true population proportion difference of people who are reading newspaper in 2005 and 2010. 20102005 n 12001000 Number of people who do not read newspaper 300 400
USE 3 DECIMAL A researcher believes that reading habit of newspaper is decreasing every year. In order to see the true population proportion of people who are reading newspaper s/he collect a randomly selected people on the yearly basis. The following table gives the number of people who do not read newspaper with respect to the years. Construct a 90 % confidence interval for the true population proportion difference of people who are reading newspaper in 2005 and 2010. 20102005 12001000 Number of people who do not read newspaper 900 700

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