Margin Of Error Confidence Level P And Q Unknown
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Use The Given Data To Find The Minimum Sample Size Required To Estimate A Population Proportion
West Emily Blunt Lady Gaga Randy Travis Chris Wallace Car Insurance Violett Beane Luxury SUV Star Wars 2016 Crossovers Answers Best Answer: Sample size = [ z sd / error]^2 Find z such that (from normal probability table) P( -z < X < z) =0.96 z = 2.05 error = 0.005 when p (and q) are unknown, they are assumed to be 0.5 Standard deviation = sqrt[p(1-p)] = sqrt[0.5(0.5)]=0.5 Sample size = [ (2.05)(0.5) / 0.005] ^2 = [205]^2 = 42025 Source(s): cidyah · 6 years ago 0 Thumbs up 0 Thumbs down Comment Add a comment Submit · just now Report Abuse margin of error: confidence level: 99%; p and q unknown Isiah Williams · 3 years ago 0 Thumbs up 0 Thumbs down Comment Add a comment Submit · just now Report Abuse Add your answer Statistics problem: a margin of error- 0.005, confidence level- 96%, p and q are unknown.? find the minimum sample size you should use to assure that your estimate of p will be within the required margin of error around the population p. Add your answer Source Submit Cancel Report Abuse I think this question viol
given data to find the use the given data to find the minimum sample size calculator minimum sample size required to estimate the population proportion. 13) margin of error 0.09 confidence level 90 p and q unknown ^ ^ Margin of error: 0.04; confidence level: 94%; p and q unknown Log On Ad: Mathway https://answers.yahoo.com/question/index?qid=20100803151348AArtprl solves algebra homework problems with step-by-step help! Algebra: Probability and statisticsSection SolversSolvers LessonsLessons Answers archiveAnswers Immediate math help from PAID TUTORS. (paid link) Click here to see ALL problems on Probability-and-statistics Question 695851: http://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.695851.html Use the given data to find the minimum sample size required to estimate the population proportion. 13) ^ ^ Margin of error: 0.04; confidence level: 94%; p and q unknown Answer by stanbon(72917) (Show Source): You can put this solution on YOUR website! Use the given data to find the minimum sample size required to estimate the population proportion. 13) ^ ^ Margin of error: 0.04; confidence level: 94%; p and q unknown ------ n = [z/E]^2*pq --- Note: The least-biased estimate for p and q is 1/2 ---- n = [2.0537/0.04]^2*(1/2)^2 --------------------- n = 659 =============== Cheers, Stan H. ===============
error Tweet Welcome to Talk Stats! Join the discussion today by registering your FREE account. Membership benefits: • Get your questions answered by community gurus and expert researchers. • Exchange your learning and research experience among peers and get advice and insight. Join Today! + Reply http://www.talkstats.com/showthread.php/704-margin-of-error to Thread Results 1 to 4 of 4 Thread: margin of error Thread Tools Show Printable Version Email this Page… Subscribe to this Thread… Display Linear Mode Switch to Hybrid Mode Switch to Threaded Mode 03-14-200608:42 AM #1 hannao1 View Profile View Forum Posts Give Away Points Posts 12 Thanks 0 Thanked 0 Times in 0 Posts margin of error Find the minimum sample size you should use to assure that your estimate of p^ will be within the required margin of margin of error around the population p. Margin of error: 0.001; confidence level: 92%; p^ and ^q unknown (my work) E= 0.001, conf level= 92%, x= .08, x/2= .04= .004 A= 1 - .004= .996, Z x/2= 2.65 N= (2.65)squared (.5)(.5) / (0.001) squared= 1.755625 / 0.001= 1755625 I was wondering if I done this correctly and if I should round off the answer. Thanks for any help!!! Reply With Quote 03-14-200608:47 AM #2 JohnM View Profile View Forum Posts TS Contributor Posts 1,948 margin of error Thanks 0 Thanked 5 Times in 4 Posts For Z, you should get 1.75. Don't round off anything. You should get n = 765,625 Reply With Quote 03-14-200608:53 AM #3 hannao1 View Profile View Forum Posts Posts 12 Thanks 0 Thanked 0 Times in 0 Posts Thanks again, your a great help!! Reply With Quote 03-15-200605:26 AM #4 Arora View Profile View Forum Posts Posts 4 Thanks 0 Thanked 0 Times in 0 Posts ? Where does that 0.5 come from in your equation? Reply With Quote + Reply to Thread Tweet « confidence interval | Principal Component Analysis » Similar Threads Determining margin of error By guerro in forum Statistics Replies: 3 Last Post: 04-29-2008, 02:40 AM Margin of Error By chopkins in forum Statistics Replies: 0 Last Post: 04-23-2008, 01:38 AM Help with margin of error problem By maharriso in forum Statistics Replies: 1 Last Post: 11-02-2006, 09:59 AM Margin of Error help By shelld in forum Psychology Statistics Replies: 0 Last Post: 03-31-2006, 03:00 PM confidence and margin of error By tejay in forum Statistics Replies: 2 Last Post: 03-12-2006, 07:29 PM Posting Permissions You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules Contact Us Talk Stats Forum Archive Top Advertise on Talk Stats All times are GMT -5.