Re: E as a % of a standard deviation
Algebraically, E = [z(a/2) / SQRT(n)] x SD, so it must be that the margin of error (maximum error as you called it) is a multiple of the population standard deviation. Keep in mind what these values represent. E is the margin of error of the estimate of mu, the population mean. SD is the population standard deviation. So, what you are asking becomes "Is the margin of error of an estimate for the population mean a fraction of the population standard deviation?" Does it make sense to even ask that question? Furthermore, if you know the population SD, you have enough information to calculate the population mean, so why estimate it? In otherwords, the "SD" you use in the formula is generally only an estimate of sigma itself. In practice, we can only estimate sigma. Using an algebraic "trick" to eliminate the need for getting a good estimate for sigma doesn't make sense to me. It sounds like your suggestion (although technically correct) would only cloud the real issue of either 1) estimating the population mean or 2) calculating the sample size needed to find such an estimate. I am not able to see the efficacy of your suggestion. Barry Edwards "John Jackson" <[EMAIL PROTECTED]> wrote in message J%4t7.57364$[EMAIL PROTECTED]">news:J%4t7.57364$[EMAIL PROTECTED]... > Really sorry. > > My formula is a rearrangement of the confidence interval formula shown below > for ascertaining the maximum error. > > E = Z(a/2) x SD/SQRT N > > The issue is you want to solve for N, but you have no standard deviation > value. > > The formula then translates into n = (Z(a/2)*SD)/E)^2Note: ^2 stands for > squared. > > You have only the confidence interval, let's say 95% and E of 1%. Lets say > that you want to find out how many people in the US have fake drivers > licenses using these numbers. How large (N) must your sample be? > > Do you have to hypothesize a US population to solve this, ie. 300 m or can > you solve it another way. It was suggested you can express the SD as a > fraction of the E. ie. E = SD/2. > > > > "Randy Poe" <[EMAIL PROTECTED]> wrote in message > [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > > John Jackson wrote: > > > > > the forumla I was using was n = (Z?/e)^2 and attempting to express .05 > as a > > > fraction of a std dev. > > > > I think you posted that before, and it's still getting > > garbled. We see a Z followed by a question mark, and > > have no idea what was actually intended. > > > > - Randy > > = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
SRS = simple random sample. cheers Michelle "Paul Bellamy" <[EMAIL PROTECTED]> wrote in message UD2t7.56886$[EMAIL PROTECTED]">news:UD2t7.56886$[EMAIL PROTECTED]... > Thanks alot - what does "SRS" mean? > Also what does "frequentist" mean - I have also seen that word? > > "Dennis Roberts" <[EMAIL PROTECTED]> wrote in message > [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > > this is the typical margin of error formula for building a confidence > > interval were the sample mean is desired to be within a certain distance > of > > the population mean > > > > n = sample size > > z = z score from nd that will produce desired confidence level (usually > > 1.96 for 95% CI) > > e = margin of error > > > > so, typical CI for mu would be: > > > > samp mean +/- z times standard error of mean > > > > e or the margin of error here is z * stan error of the mean (let me > > symbolize se) > > > > e = z * se > > > > for 95% CI .. e = 1.96 * se > > > > e = 1.96 * (sigma / sqrt n) > > > > now, what n might it take to produce some e? we can rearrange the formula > ... > > > > sqrt n = (1.96 * sigma) / e > > > > but, we don't want sqrt n ... we WANT n! > > > > n = ((1.96 * sigma)/ e) ^2 > > > > so, what if we wanted to be within 3 points of mu with our sample mean the > > population standard deviation or sigma were 15? > > > >n = ((1.96 * 5) / 3)^2 = about 11 ... > > > > only would take a SRS of about 11 to be within 3 points of the true mu > > value in your 95% confidence interval > > > > unless i made a mistake someplace > > > > > > At 09:54 AM 9/28/01 -0400, Randy Poe wrote: > > >John Jackson wrote: > > > > > > > the forumla I was using was n = (Z?/e)^2 and attempting to express > .05 > > > as a > > > > fraction of a std dev. > > > > > >I think you posted that before, and it's still getting > > >garbled. We see a Z followed by a question mark, and > > >have no idea what was actually intended. > > > > > > - Randy > > > > > > > > >= > > >Instructions for joining and leaving this list and remarks about > > >the problem of INAPPROPRIATE MESSAGES are available at > > > http://jse.stat.ncsu.edu/ > > >= > > > > _ > > dennis roberts, educational psychology, penn state university > > 208 cedar, AC 8148632401, mailto:[EMAIL PROTECTED] > > http://roberts.ed.psu.edu/users/droberts/drober~1.htm > > > > > > > > = > > Instructions for joining and leaving this list and remarks about > > the problem of INAPPROPRIATE MESSAGES are available at > > http://jse.stat.ncsu.edu/ > > = > > = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
On Sun, 30 Sep 2001 00:34:40 GMT, "John Jackson" <[EMAIL PROTECTED]> wrote: > Here is my solution using figures which are self-explanatory: > > Sample Size Determination > > pi = 50% central area 0.99 > confid level= 99% 2 tail area 0.5 > sampling error 2% 1 tail area 0.025 > z =2.58 > n1 4,146.82 Excel function for determining central interval > NORMSINV($B$10+(1-$B$10)/2) > n 4,147 > > The algebraic formula for n was: n = ?(1-?)*(z/e)2 > > > > If you can't read the above: > > n = pi(1-pi)*(z/e)^2 > > Let me know if this makes sense. > > > > It is simply amazing to me that you can do a random sample of 4,147 people > out of 50 million and get a valid answer. What is the reason for taking > mulitple samples of the same n - to achieve more accuracy? Is there a rule > of thumb on how many repetitions of the same sample you would take? > I have not followed your steps in detail, but: I think you just took a random sample to show that the number of ballots left blank, intentionally, is 1%, plus or minus 2 points. That is using a crude, generous estimate of the variance instead of conditioning on the small p. - A three-fold estimate (over the mean) for the maximum is not good accuracy. - When the minimum estimate of p goes negative, it is time to try an estimation based on something different. If I want an accurate estimate of a rare percentage, I often find it easier to think of the number-of-instances. One percent of 4000 is 40. What is the accuracy with 40 seen in the sample? (95% CI is wider than 30 to 50, but not by a whole lot.) -- Rich Ulrich, [EMAIL PROTECTED] http://www.pitt.edu/~wpilib/index.html = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
Donald, I totally agree w/your point about the stratification of the sample. My facts were set up merely for simplicity's sake notwithstanding their clear artificiality. The only instances of multiple samples I have seen are in textbooks to prove the CLT; that w/increasing numbers of sample means, the distribution (of sample means) becomes normal even if the population isn't. Statistics is a relatively new area study for me and I never would have intuitively thought that a sample of a few thousand could reveal such meaningful results. But I understand your point completely. I suppose like you say that when you factor in stratification and clustering, it isn't such a no brainer as in my example. Thank you again for enlightening me. "Donald Burrill" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > On Sun, 30 Sep 2001, John Jackson wrote: > > > Here is my solution using figures which are self-explanatory: > > > > Sample Size Determination > > > > pi = 50% central area 0.99 > > confid level= 99% 2 tail area 0.5 > > sampling error 2% 1 tail area 0.025 > > z =2.58 > > n1 4,146.82 Excel function for determining central interval > > NORMSINV($B$10+(1-$B$10)/2) > > n 4,147 > > > > The algebraic formula for n was: > > > > n = pi(1-pi)*(z/e)^2 > > > > > > It is simply amazing to me that you can do a random sample of 4,147 > > people out of 50 million and get a valid answer. > > It is not clear what part of this you find "amazing". > (Would you otherwise expect an INvalid answer, in some sense?) > Thme hard part, of course, is taking the random sample in the first > place. The equation you used, I believe, assumes a "simple random > sample", sometimes known in the trade as a SRS; but it seems to me > VERY unlikely that any real sampling among the ballots cast in a > national election would be done that way. I'd expect it to involve > stratifying on (e.g.) states, and possibly clustering within states; > both of which would affect the precision of the estimate, and therefore > the minimum sample size desired. > As to what may be your concern, that 4,000 looks like a small > part of 50 million, the precision of an estimate depends principally > on the amount of information available -- that is, on the size of the > sample; not on the proportion that amount bears to the total amount > of information that may be of interest. Rather like a hologram, in > some respects; and very like the resolving power of an optical > instrument (e.g., a telescope), which is a function of the amount of > information the instrument can receive (the area of the primary lens > or reflector), not on how far away the object in view may be nor what > its absolute magnitude may be. > > > What is the reason for taking multiple samples of the same n - > > to achieve more accuracy? > > I, for one, don't understand the point of this question at all. > Multiple samples? Who takes them, or advocates taking them? > > < snip, the rest > > > > Donald F. Burrill [EMAIL PROTECTED] > 184 Nashua Road, Bedford, NH 03110 603-471-7128 > > > > = > Instructions for joining and leaving this list and remarks about > the problem of INAPPROPRIATE MESSAGES are available at > http://jse.stat.ncsu.edu/ > = = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
On Sun, 30 Sep 2001, John Jackson wrote: > Here is my solution using figures which are self-explanatory: > > Sample Size Determination > > pi = 50% central area 0.99 > confid level= 99% 2 tail area 0.5 > sampling error 2% 1 tail area 0.025 > z =2.58 > n1 4,146.82 Excel function for determining central interval > NORMSINV($B$10+(1-$B$10)/2) > n 4,147 > > The algebraic formula for n was: > > n = pi(1-pi)*(z/e)^2 > > > It is simply amazing to me that you can do a random sample of 4,147 > people out of 50 million and get a valid answer. It is not clear what part of this you find "amazing". (Would you otherwise expect an INvalid answer, in some sense?) Thme hard part, of course, is taking the random sample in the first place. The equation you used, I believe, assumes a "simple random sample", sometimes known in the trade as a SRS; but it seems to me VERY unlikely that any real sampling among the ballots cast in a national election would be done that way. I'd expect it to involve stratifying on (e.g.) states, and possibly clustering within states; both of which would affect the precision of the estimate, and therefore the minimum sample size desired. As to what may be your concern, that 4,000 looks like a small part of 50 million, the precision of an estimate depends principally on the amount of information available -- that is, on the size of the sample; not on the proportion that amount bears to the total amount of information that may be of interest. Rather like a hologram, in some respects; and very like the resolving power of an optical instrument (e.g., a telescope), which is a function of the amount of information the instrument can receive (the area of the primary lens or reflector), not on how far away the object in view may be nor what its absolute magnitude may be. > What is the reason for taking multiple samples of the same n - > to achieve more accuracy? I, for one, don't understand the point of this question at all. Multiple samples? Who takes them, or advocates taking them? < snip, the rest > Donald F. Burrill [EMAIL PROTECTED] 184 Nashua Road, Bedford, NH 03110 603-471-7128 = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
Here is my solution using figures which are self-explanatory: Sample Size Determination pi = 50% central area 0.99 confid level= 99% 2 tail area 0.5 sampling error 2% 1 tail area 0.025 z =2.58 n1 4,146.82 Excel function for determining central interval NORMSINV($B$10+(1-$B$10)/2) n 4,147 The algebraic formula for n was: n = ?(1-?)*(z/e)2 If you can't read the above: n = pi(1-pi)*(z/e)^2 Let me know if this makes sense. It is simply amazing to me that you can do a random sample of 4,147 people out of 50 million and get a valid answer. What is the reason for taking mulitple samples of the same n - to achieve more accuracy? Is there a rule of thumb on how many repetitions of the same sample you would take? "John Jackson" <[EMAIL PROTECTED]> wrote in message s1ot7.61225$[EMAIL PROTECTED]">news:s1ot7.61225$[EMAIL PROTECTED]... > Donald - Thank you for your cogent explanation of a concept that is a bit > hard to grasp. > After researching it more, I determined that there is a gaping hole in my > knowldege relating to the area of inferences on a population proportion so I > am somethat admittedly in the dark and have to study up a bit. > > Having said that, here are some answers to ?s you posed and some additional > comments. > > Instead of a warehouse full of CDs, lets work w/a much larger population. > > Revised fact pattern: > > Suppose you want to estimate the % of voters who acutally voted in the 2000 > U.S. Presidential election who failed to make a choice for any candidate > (blank ballot). Assume (forgetting about politics) that this was simply a > matter of inadvertance, error on the part of the voter, that all voting > machines worked properly, and that the problem manifested itself the same > way all over the country. You want to estimate how many ballots were blank > and be 98% confident that the error of estimate is 2% or less. So you have a > universe of 50m voters or however many went to the polls. Assume you don't > really know if its is 50m or 75m or 100m. You just know its in the tens of > millions. > > So you want to estimate the proportion of blank ballots, knowing that a huge > number of people went to the polls. You mention and I see it stated in some > books that when you don't know the SD and don't know the exact population > size, other than that is in the millions, the safest choice is p = .5 - that > apparently is a sort of worse case scenario it seems. I have to > reread my material and also revisit the binomial distribution area which I > have studied extensively. However that knowledge has been pushed out of the > way by this complex area of sampling. > > Anyway, if you have some further thoughts given my clarification, I would > welcome your insights. > > > "Donald Burrill" <[EMAIL PROTECTED]> wrote in message > [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > > On Fri, 28 Sep 2001, John Jackson wrote in part: > > > > > My formula is a rearrangement of the confidence interval formula shown > > > below for ascertaining the maximum error. > > E = Z(a/2) x SD/SQRT N > > > The issue is you want to solve for N, but you have no standard > > > deviation value. > > Oh, but you do. In the problem you formulated, unless I > > misunderstood egregiously, you are seeking to estimate the proportion of > > defective (or pirated, or whatever) CDs in a universe of 10,000 CDs. > > There is then a maximum value for the SD of a proportion: > > SD = SQRT[p(1-p)/n] > > where p is the proportion in question, n is the sample size. > > This value is maximized for p = 0.5 (and it doesn't change much > > between p = 0.3 and p = 0.7 ). If you have a guess as to the value > > of p, you can get a smaller value of SD, but using p = 0.5 will > > give you a conservative estimate. > > You then have to figure out what that "5% error" means: it might > > mean "+/- 0.05 on the estimated proportion p" (but this is probably not a > > useful error bound if, say, p = 0.03), or it might mean "5% of the > > estimated proportion" (which would mean +/- 0.0015 if p = 0.03). > > (In the latter case, E is a function of p, so the formula for n > > can be solved without using a guesstimated value for p until the last > > step.) > > Notice that throughout this analysis, you're using the normal > > distribution as an approximation to the binomial b(n,p;k) distribution > > that presumably "really" applies. That's probably reasonable; but the > > approximation may be quite lousy if p is very close to 0 (or 1). > > Thbe thing is, of course, that if there is NO pirating of the CDs, p=0, > > and this is a desirable state of affairs from your clients' perspective. > > So you might want to be in the business of expressing the minimum p > > that you could expect to detect with, say, 80% probability, using the > > sample size eventually chosen
Re: E as a % of a standard deviation
Donald - Thank you for your cogent explanation of a concept that is a bit hard to grasp. After researching it more, I determined that there is a gaping hole in my knowldege relating to the area of inferences on a population proportion so I am somethat admittedly in the dark and have to study up a bit. Having said that, here are some answers to ?s you posed and some additional comments. Instead of a warehouse full of CDs, lets work w/a much larger population. Revised fact pattern: Suppose you want to estimate the % of voters who acutally voted in the 2000 U.S. Presidential election who failed to make a choice for any candidate (blank ballot). Assume (forgetting about politics) that this was simply a matter of inadvertance, error on the part of the voter, that all voting machines worked properly, and that the problem manifested itself the same way all over the country. You want to estimate how many ballots were blank and be 98% confident that the error of estimate is 2% or less. So you have a universe of 50m voters or however many went to the polls. Assume you don't really know if its is 50m or 75m or 100m. You just know its in the tens of millions. So you want to estimate the proportion of blank ballots, knowing that a huge number of people went to the polls. You mention and I see it stated in some books that when you don't know the SD and don't know the exact population size, other than that is in the millions, the safest choice is p = .5 - that apparently is a sort of worse case scenario it seems. I have to reread my material and also revisit the binomial distribution area which I have studied extensively. However that knowledge has been pushed out of the way by this complex area of sampling. Anyway, if you have some further thoughts given my clarification, I would welcome your insights. "Donald Burrill" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > On Fri, 28 Sep 2001, John Jackson wrote in part: > > > My formula is a rearrangement of the confidence interval formula shown > > below for ascertaining the maximum error. > E = Z(a/2) x SD/SQRT N > > The issue is you want to solve for N, but you have no standard > > deviation value. > Oh, but you do. In the problem you formulated, unless I > misunderstood egregiously, you are seeking to estimate the proportion of > defective (or pirated, or whatever) CDs in a universe of 10,000 CDs. > There is then a maximum value for the SD of a proportion: > SD = SQRT[p(1-p)/n] > where p is the proportion in question, n is the sample size. > This value is maximized for p = 0.5 (and it doesn't change much > between p = 0.3 and p = 0.7 ). If you have a guess as to the value > of p, you can get a smaller value of SD, but using p = 0.5 will > give you a conservative estimate. > You then have to figure out what that "5% error" means: it might > mean "+/- 0.05 on the estimated proportion p" (but this is probably not a > useful error bound if, say, p = 0.03), or it might mean "5% of the > estimated proportion" (which would mean +/- 0.0015 if p = 0.03). > (In the latter case, E is a function of p, so the formula for n > can be solved without using a guesstimated value for p until the last > step.) > Notice that throughout this analysis, you're using the normal > distribution as an approximation to the binomial b(n,p;k) distribution > that presumably "really" applies. That's probably reasonable; but the > approximation may be quite lousy if p is very close to 0 (or 1). > Thbe thing is, of course, that if there is NO pirating of the CDs, p=0, > and this is a desirable state of affairs from your clients' perspective. > So you might want to be in the business of expressing the minimum p > that you could expect to detect with, say, 80% probability, using the > sample size eventually chosen: that is, to report a power analysis. > > > The formula then translates into n = (Z(a/2)*SD)/E)^2 > > Note: ^2 stands for squared. > > > > You have only the confidence interval, let's say 95% and E of 1%. > > Let's say that you want to find out how many people in the US have > > fake driver's licenses using these numbers. How large (N) must your > > sample be? > > Again, you're essentially trying to estimate a proportion. (If it is > the number of instances that is of interest, the distribution is still > inherently binomial, but instead of p you're estimating np, with > SD = SQRT[np(1-p)] > and you still have to decide whether that 1% means "+/- 0.01 on the > proportion p" or "1% of the value of np". > -- DFB. > > Donald F. Burrill [EMAIL PROTECTED] > 184 Nashua Road, Bedford, NH 03110 603-471-7128 > > > > = > Instructions for joining and leaving this list and remarks about > the problem of INAPPROPRIATE MESSAGES are av
Re: E as a % of a standard deviation
On Fri, 28 Sep 2001, John Jackson wrote in part: > My formula is a rearrangement of the confidence interval formula shown > below for ascertaining the maximum error. E = Z(a/2) x SD/SQRT N > The issue is you want to solve for N, but you have no standard > deviation value. Oh, but you do. In the problem you formulated, unless I misunderstood egregiously, you are seeking to estimate the proportion of defective (or pirated, or whatever) CDs in a universe of 10,000 CDs. There is then a maximum value for the SD of a proportion: SD = SQRT[p(1-p)/n] where p is the proportion in question, n is the sample size. This value is maximized for p = 0.5 (and it doesn't change much between p = 0.3 and p = 0.7 ). If you have a guess as to the value of p, you can get a smaller value of SD, but using p = 0.5 will give you a conservative estimate. You then have to figure out what that "5% error" means: it might mean "+/- 0.05 on the estimated proportion p" (but this is probably not a useful error bound if, say, p = 0.03), or it might mean "5% of the estimated proportion" (which would mean +/- 0.0015 if p = 0.03). (In the latter case, E is a function of p, so the formula for n can be solved without using a guesstimated value for p until the last step.) Notice that throughout this analysis, you're using the normal distribution as an approximation to the binomial b(n,p;k) distribution that presumably "really" applies. That's probably reasonable; but the approximation may be quite lousy if p is very close to 0 (or 1). Thbe thing is, of course, that if there is NO pirating of the CDs, p=0, and this is a desirable state of affairs from your clients' perspective. So you might want to be in the business of expressing the minimum p that you could expect to detect with, say, 80% probability, using the sample size eventually chosen: that is, to report a power analysis. > The formula then translates into n = (Z(a/2)*SD)/E)^2 > Note: ^2 stands for squared. > > You have only the confidence interval, let's say 95% and E of 1%. > Let's say that you want to find out how many people in the US have > fake driver's licenses using these numbers. How large (N) must your > sample be? Again, you're essentially trying to estimate a proportion. (If it is the number of instances that is of interest, the distribution is still inherently binomial, but instead of p you're estimating np, with SD = SQRT[np(1-p)] and you still have to decide whether that 1% means "+/- 0.01 on the proportion p" or "1% of the value of np". -- DFB. Donald F. Burrill [EMAIL PROTECTED] 184 Nashua Road, Bedford, NH 03110 603-471-7128 = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
your formula is right on the money, but suppose your problem supplies no SD - see my recent message in this thread. "Dennis Roberts" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > this is the typical margin of error formula for building a confidence > interval were the sample mean is desired to be within a certain distance of > the population mean > > n = sample size > z = z score from nd that will produce desired confidence level (usually > 1.96 for 95% CI) > e = margin of error > > so, typical CI for mu would be: > > samp mean +/- z times standard error of mean > > e or the margin of error here is z * stan error of the mean (let me > symbolize se) > > e = z * se > > for 95% CI .. e = 1.96 * se > > e = 1.96 * (sigma / sqrt n) > > now, what n might it take to produce some e? we can rearrange the formula ... > > sqrt n = (1.96 * sigma) / e > > but, we don't want sqrt n ... we WANT n! > > n = ((1.96 * sigma)/ e) ^2 > > so, what if we wanted to be within 3 points of mu with our sample mean the > population standard deviation or sigma were 15? > >n = ((1.96 * 5) / 3)^2 = about 11 ... > > only would take a SRS of about 11 to be within 3 points of the true mu > value in your 95% confidence interval > > unless i made a mistake someplace > > > At 09:54 AM 9/28/01 -0400, Randy Poe wrote: > >John Jackson wrote: > > > > > the forumla I was using was n = (Z?/e)^2 and attempting to express .05 > > as a > > > fraction of a std dev. > > > >I think you posted that before, and it's still getting > >garbled. We see a Z followed by a question mark, and > >have no idea what was actually intended. > > > > - Randy > > > > > >= > >Instructions for joining and leaving this list and remarks about > >the problem of INAPPROPRIATE MESSAGES are available at > > http://jse.stat.ncsu.edu/ > >= > > _ > dennis roberts, educational psychology, penn state university > 208 cedar, AC 8148632401, mailto:[EMAIL PROTECTED] > http://roberts.ed.psu.edu/users/droberts/drober~1.htm > > > > = > Instructions for joining and leaving this list and remarks about > the problem of INAPPROPRIATE MESSAGES are available at > http://jse.stat.ncsu.edu/ > = = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
Really sorry. My formula is a rearrangement of the confidence interval formula shown below for ascertaining the maximum error. E = Z(a/2) x SD/SQRT N The issue is you want to solve for N, but you have no standard deviation value. The formula then translates into n = (Z(a/2)*SD)/E)^2Note: ^2 stands for squared. You have only the confidence interval, let's say 95% and E of 1%. Lets say that you want to find out how many people in the US have fake drivers licenses using these numbers. How large (N) must your sample be? Do you have to hypothesize a US population to solve this, ie. 300 m or can you solve it another way. It was suggested you can express the SD as a fraction of the E. ie. E = SD/2. "Randy Poe" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > John Jackson wrote: > > > the forumla I was using was n = (Z?/e)^2 and attempting to express .05 as a > > fraction of a std dev. > > I think you posted that before, and it's still getting > garbled. We see a Z followed by a question mark, and > have no idea what was actually intended. > > - Randy = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
Thanks alot - what does "SRS" mean? Also what does "frequentist" mean - I have also seen that word? "Dennis Roberts" <[EMAIL PROTECTED]> wrote in message [EMAIL PROTECTED]">news:[EMAIL PROTECTED]... > this is the typical margin of error formula for building a confidence > interval were the sample mean is desired to be within a certain distance of > the population mean > > n = sample size > z = z score from nd that will produce desired confidence level (usually > 1.96 for 95% CI) > e = margin of error > > so, typical CI for mu would be: > > samp mean +/- z times standard error of mean > > e or the margin of error here is z * stan error of the mean (let me > symbolize se) > > e = z * se > > for 95% CI .. e = 1.96 * se > > e = 1.96 * (sigma / sqrt n) > > now, what n might it take to produce some e? we can rearrange the formula ... > > sqrt n = (1.96 * sigma) / e > > but, we don't want sqrt n ... we WANT n! > > n = ((1.96 * sigma)/ e) ^2 > > so, what if we wanted to be within 3 points of mu with our sample mean the > population standard deviation or sigma were 15? > >n = ((1.96 * 5) / 3)^2 = about 11 ... > > only would take a SRS of about 11 to be within 3 points of the true mu > value in your 95% confidence interval > > unless i made a mistake someplace > > > At 09:54 AM 9/28/01 -0400, Randy Poe wrote: > >John Jackson wrote: > > > > > the forumla I was using was n = (Z?/e)^2 and attempting to express .05 > > as a > > > fraction of a std dev. > > > >I think you posted that before, and it's still getting > >garbled. We see a Z followed by a question mark, and > >have no idea what was actually intended. > > > > - Randy > > > > > >= > >Instructions for joining and leaving this list and remarks about > >the problem of INAPPROPRIATE MESSAGES are available at > > http://jse.stat.ncsu.edu/ > >= > > _ > dennis roberts, educational psychology, penn state university > 208 cedar, AC 8148632401, mailto:[EMAIL PROTECTED] > http://roberts.ed.psu.edu/users/droberts/drober~1.htm > > > > = > Instructions for joining and leaving this list and remarks about > the problem of INAPPROPRIATE MESSAGES are available at > http://jse.stat.ncsu.edu/ > = = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
this is the typical margin of error formula for building a confidence interval were the sample mean is desired to be within a certain distance of the population mean n = sample size z = z score from nd that will produce desired confidence level (usually 1.96 for 95% CI) e = margin of error so, typical CI for mu would be: samp mean +/- z times standard error of mean e or the margin of error here is z * stan error of the mean (let me symbolize se) e = z * se for 95% CI .. e = 1.96 * se e = 1.96 * (sigma / sqrt n) now, what n might it take to produce some e? we can rearrange the formula ... sqrt n = (1.96 * sigma) / e but, we don't want sqrt n ... we WANT n! n = ((1.96 * sigma)/ e) ^2 so, what if we wanted to be within 3 points of mu with our sample mean the population standard deviation or sigma were 15? n = ((1.96 * 5) / 3)^2 = about 11 ... only would take a SRS of about 11 to be within 3 points of the true mu value in your 95% confidence interval unless i made a mistake someplace At 09:54 AM 9/28/01 -0400, Randy Poe wrote: >John Jackson wrote: > > > the forumla I was using was n = (Z?/e)^2 and attempting to express .05 > as a > > fraction of a std dev. > >I think you posted that before, and it's still getting >garbled. We see a Z followed by a question mark, and >have no idea what was actually intended. > > - Randy > > >= >Instructions for joining and leaving this list and remarks about >the problem of INAPPROPRIATE MESSAGES are available at > http://jse.stat.ncsu.edu/ >= _ dennis roberts, educational psychology, penn state university 208 cedar, AC 8148632401, mailto:[EMAIL PROTECTED] http://roberts.ed.psu.edu/users/droberts/drober~1.htm = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
John Jackson wrote: > the forumla I was using was n = (Z?/e)^2 and attempting to express .05 as a > fraction of a std dev. I think you posted that before, and it's still getting garbled. We see a Z followed by a question mark, and have no idea what was actually intended. - Randy = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
There is a specific problem associated w/this formula: "I need to do a spot check of our inventory of CDs to ascertain which ones are genuine CDs from the factory and which ones are counterfeit CDs "burned" by a forger. I have recommeded that we take a random sample by SKU numbers and I was told it was OK as long as whatever estimate I produced, my error of estimate would be no more than 5% with a 99% confidence level. I am wondering how I should go about determining how many CDs, of which there are 10,000, to randomly look at." the forumla I was using was n = (Z?/e)^2 and attempting to express .05 as a fraction of a std dev. "Glen Barnett" <[EMAIL PROTECTED]> wrote in message 9oug3c$su1$[EMAIL PROTECTED]">news:9oug3c$su1$[EMAIL PROTECTED]... > > John Jackson <[EMAIL PROTECTED]> wrote in message > MGns7.49824$[EMAIL PROTECTED]">news:MGns7.49824$[EMAIL PROTECTED]... > > re: the formula: > > > > n = (Z?/e)2 > > This formula hasn't come over at all well. Please note that newsgroups > work in ascii. What's it supposed to look like? What's it a formula for? > > > could you express E as a % of a standard deviation . > > What's E? The above formula doesn't have a (capital) E. > > What is Z? n? e? > > > In other words does a .02 error translate into .02/1 standard deviations, > > assuming you are dealing w/a normal distribution? > > ? How does this relate to the formula above? > > Glen > = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
John Jackson <[EMAIL PROTECTED]> wrote in message MGns7.49824$[EMAIL PROTECTED]">news:MGns7.49824$[EMAIL PROTECTED]... > re: the formula: > > n = (Z?/e)2 This formula hasn't come over at all well. Please note that newsgroups work in ascii. What's it supposed to look like? What's it a formula for? > could you express E as a % of a standard deviation . What's E? The above formula doesn't have a (capital) E. What is Z? n? e? > In other words does a .02 error translate into .02/1 standard deviations, > assuming you are dealing w/a normal distribution? ? How does this relate to the formula above? Glen = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
Re: E as a % of a standard deviation
At 04:49 PM 9/26/01 +, John Jackson wrote: >re: the formula: > > n = (Z?/e)2 > > >could you express E as a % of a standard deviation . > >In other words does a .02 error translate into .02/1 standard deviations, >assuming you are dealing w/a normal distribution? well, let's see ... e is the margin of error ... using the formula for a CI for a population mean .. X bar +/- z * stan error of the mean so, the margin of error or e ... is z * standard error of the mean now, let's assume that we stick to 95% CIs ... so the z will be about 2 ... that leaves us with the standard error of the mean ... or, sigma / sqrt n let's say that we were estimating SAT M scores and assumed a sigma of about 100 and were taking a sample size of n=100 (to make my figuring simple) ... this would give us a standard error of 100/10 = 10 so, the margin of error would be: e = 2 * 10 or about 20 so, 20/100 = .2 ... that is, the e or margin of error is about .2 of the population sd if we had used a sample size of 400 ... then the standard error would have been: 100/20 = 5 and our e or margin of error would be 2 * 5 = 10 so, the margin of error is now 10/100 or .1 of a sigma unit OR 1/2 the size it was before but, i don't see what you have accomplished by doing this ... rather than just reporting the margin of error ... 10 versus 20 ... which is also 1/2 the size since z * stan error is really score UNITS ... and, the way you done it ... .2 or .1 would represent fractions of sigma ... which still amounts to score UNITS ... i don't think anything new has been done ... certainly, no new information has been created >= >Instructions for joining and leaving this list and remarks about >the problem of INAPPROPRIATE MESSAGES are available at > http://jse.stat.ncsu.edu/ >= _ dennis roberts, educational psychology, penn state university 208 cedar, AC 8148632401, mailto:[EMAIL PROTECTED] http://roberts.ed.psu.edu/users/droberts/drober~1.htm = Instructions for joining and leaving this list and remarks about the problem of INAPPROPRIATE MESSAGES are available at http://jse.stat.ncsu.edu/ =
