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We want to see if this is true for adults 35 and older. How many do we need to sample to have a margin of error of 5% at a. 90% confidence level.
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โ If we know that each member of the population has probability p of having a certain characteristic, we can use the CLT theorem to study the distribution of a sample mean. โ What if we don't know p, all we have is our data from the sample. We want to make an estimate of p, and give some margin of error. This is essentially what a confidence interval is. โ For a prescribed level of confidence (less than 100%), we want to determine a range for which we are THAT confident the true population probability โpโ is within the range.
โ Usually we want a fairly high confidence level: 75%, 95% or 99% are common, but really any percentage less than 100 is possible. The larger the confidence, the wider the interval. โ The more sure we are of the confidence interval, the less precise it is. estimate Confidence interval Margin of error Margin of error
โ For your reference, these could be useful: Confidence %
deviations (z) 50% 0. 75% 1. 90% 1. 95% 1. 97% 2. 99% 2. 99.9% 3. To calculate, use invNorm(CI + (1-CI)/2) e.g. for 75% confidence, invNorm(.75 + (1-.75)/2) =invNorm(.75+ .25/2) =invNorm(.875)
You want to give a 95% confidence interval of how many apples in a given orchard are bad this year. Of all harvested apples, you randomly test 1000 apples and find 35 of them are bad. โ p estimate is p=.035, so q=. โ SD(p)=โ(.035.965/1000)=. โ (^) The middle 95% is within 1.96 sds โ Our confidence interval is .035ยฑ1.96.0058, i.e. between and .0236 and. โ We are 95% confident that in this orchard between 2.36% and 4.64% of apples are bad.
A poll of 1654 adults asked whether they have ever bobbed for apples. 76% said โYes.โ For 93% confidence, what is the margin of error? โ To find the z-score for the central 93%, remember that 7% is in the tails, 3.5% in the upper tail and 3.5% in the lower tail. So invNorm(.965)=1.812 is our z โ (^) ME 93% = zโ(pq/n) =1.812โ(.76.24/1654) =.01903, or 1.903%
A poll of 1654 adults asked whether they have ever bobbed for apples. 76% said โYes.โ What is the margin of error for 99% confidence? โ Similarly, the z value for central 99% is invNorm(.995)=2. โ (^) ME 99% =2.576*.010501=.02705 or 2.705% โ As confidence level of the interval increases, so does the margin of error!
โ (^) For C% confidence, ME C =z C โ(pq/n) โ If we increase the sample size, the margin of error goes down, but at a rate of the square root of the change in โnโ. โ To halve ME, we need to quadruple (x4) the sample size โ To get 1/ th the ME, we need to increase sample size to be 100 times as large
โ (^) You can use the formula ME C =z C โ(pq/n) to give you the confidence level, because you can determine z C , and from that figure out the confidence level. โ Divide both sides by โ(pq/n) to give you: z C
C /โ(pq/n) โ Then Confidence level is found: normalcdf(-z C , z C
-z C z C