Confidence Interval of percentage of rivets
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4.) 20 rivets out 100 in a new box that we tested had break strength below the desired level. Since this is a critical part for assembly, we want to be 99% confident of how many rivets are below the desired level. Calculate the confidence interval of the percentage of rivets that are below the desired level.
5.) We expect a 1.5% response/mail rate in a particular mail campaign. We are going to test a new outer envelope with the phrase "Open me first". We would like to be accurate within 0.25% (0.0025) of the true response/mail rate for future mailings. How many mail pieces should we mail and how confident would you want to be?
6.) Write your own simple problem using the concept of Expected Value and solve.
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Solution Summary
4.) 20 rivets out 100 in a new box that we tested had break strength below the desired level. Since this is a critical part for assembly, we want to be 99% confident of how many rivets are below the desired level. Calculate the confidence interval of the percentage of rivets that are below the desired level.
5.) We expect a 1.5% response/mail rate in a particular mail campaign. We are going to test a new outer envelope with the phrase "Open me first". We would like to be accurate within 0.25% (0.0025) of the true response/mail rate for future mailings. How many mail pieces should we mail and how confident would you want to be?
6.) Write your own simple problem using the concept of Expected Value and solve.
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4) Applying the general formula for a confidence interval, the confidence interval for a proportion, p, is: p+- z*Sp
<br>where p is the proportion in the sample, z depends on the level of confidence desired, and Sp, the standard error of a proportion, is equal to: Sp=SQRT(p(1-p)/n)
<br>In this case, p=20/100=0.2, Sp=SQRT(0.2(1-0.2)/100)=0.04
<br>Since it's a two-tailed distribution, we should use the 0.995 level of confidence. (0.005 on each side), from the z-table, we can find z=2.58
<br>So upper limit is: ...
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