Question 1183927
i usually calculate it this way.


the formula to use is z = (x - m) / s


z is the z-score
x is the raw score
m is the mean
s is the standard error.


if you want the  margin of error to be equal to 1, then (x - m) must be equal to 1.


the formula becomes:


z = 1 / s


at 90% confidence level, the two tailed condidence interval will require a cr5iticql z-score of plus or minus 1.645.


i usually work with the positive z-score to find the sample size required.


the formula becomes 1.645 = 1 / s.


s is equal to the population standard deviation divided by the square root of the sample size.


you get:


s = 2.5 / sqrt(n), where n is the sample size.


the formula becomes:


1.645 = 1 / (2.5 / sqrt(n)).


since 1 / (2.5 / sqrt(n)) is equal to 1 / 2.5 * sqrt(n), the equation becomes:


1.645 = 1 / 2.5 * sqrt(n).


multiply both sides of this equaton by 2.5 to get:


1.645 * 2.5 = sqrt(n).


solve for sqrt(n) to get:


sqrt(n) = 4.1125.


solve for n to get:


n = 16.91265625.


that's your solution.


since s is equal to 2.5 / sqrt(n), then s is equal to 2.5 / 4.1125 which makes s = .607903 rounded to 6 decimal places.


this value of the standard error = s will give you the margin of error you desire.


this margin of error will be the same, regardless of what the mean is.


some samples are shown below:


<img src = "http://theo.x10hosting.com/2021/082701.jpg" >


<img src = "http://theo.x10hosting.com/2021/082702.jpg" >


<img src = "http://theo.x10hosting.com/2021/082703.jpg" >


you can see that the margin of error is plus or minus 1 regardless of what the mean is.


for your second problem, i did the same thing.


critical z was still 1.645 because the confidence level was still 90%.


the margin of eror desired is not plus or minus .5.


the population standard deviation is now 2.4, rather than 2.5


the z-score formula ia still z = (x - m) / s


z is the z-score
x is the raw score
m is the mean
s is the standard error.


if you want a margin of error of .5, then (x - m) must be equal to .5.


the formula becomes:


1.645 = .5 / s


multiply both sides of the formula by s to get:


1.645 * s = .5


since s = 2.4 / sqrt(n), the formula becomes:


1.645 * 2.4 / sqrt(n) = .5


solve for sqrt(n) to get:


sqrtn) = 1.645 * 2.4 / .5


this makes sqrt(n) = 7.896


solve for n to get:


n = 7.896^2 = 62.346816.


when sqrt(n) = 7.896, then s = 2.4 / 7.896 = .3039513 rounded to 6 decimal places.


this is the standard error that will give you a margin of error of plus or minus .5 regardless of what the mean is.


some samples are shown below.


<img src = "http://theo.x10hosting.com/2021/082704.jpg" >


<img src = "http://theo.x10hosting.com/2021/082705.jpg" >


<img src = "http://theo.x10hosting.com/2021/082706.jpg" >


i believe you can generalize the formula as shown below.


the z-score formula is z = (x - m) / s


when (x - m) is the MOE, the formula becdomes:


z = MOE / s


multiply both sides of the formula by s to get:


z * s = MOE.


since s = population standard deviation / sqrt(n), the formula becomes:


z * population standard deviation / sqrt(n) = MOE.


solve for sqrt(n) to get:


sqrt(n) = z * population standard deviation / MOE.


let's see if this works.


when z = 1.645 and MOE = 1 and population standard deviation = 2.5, the formula becomes:


sqrt(n) = 1.645 * 2.5 / 1 = 4.1125.


this agrees with what we got before.


when z = 1.645 and MOE = .5 and population standard deviation = 2.4, the formula becomes:


sqrt(n) = 1.645 * 2.4 / .5 = 7.896.


this also agrees with what we got before.


the z-score used is the critical z-score for the confidence level indicated.