SOLUTION: Find the minimum sample size n needed to estimate muμ for the given values of​ c=0.95, sigmaσ​ 6.9, and E1. I got that it starts with 1/2(1-.95) which equal

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Question 1083928: Find the minimum sample size n needed to estimate muμ for the given values of​ c=0.95, sigmaσ​ 6.9, and E1.
I got that it starts with 1/2(1-.95) which equals 0.025 now I am not sure how to get the next answer of 1.96




Answer by Theo(13342)   (Show Source): You can put this solution on YOUR website!
here's a good reference you might want to look at.

https://brownmath.com/stat/sampsiz.htm

yours would be the case 0 where you know the standard deviation of the population.

since you know the population standard deviation, you would use the z-score.

your margin of error formula is:

E = cz * sd / sqrt(n)

E is the margin of error.
cz is the critical z-score
sd is the standard deviation of the population
n is the sample size

ca is the critical alpha
ci is the confidence interval

these are my abbreviations, not the usually ones that you will see.
i do this because sometimes the letters are used for different things and it can become confusing.

i believe you are given that E = 1.
you are also given that ci = .95 of the normal distribution curve.
you are also given that sd = 6.9

how do you find cz?

to find cz, you need to find the critical alpha.

since you are looking for a two-tailed confidence interval (half the alpha is on the left and half the alpha is on the right), you would find the critical alpha by using the formula of ca = (1 - ci) / 2

that gets you ci = .05/2 = .025

if you look for the area under the distribution curve that is equal to .025, you will find that the z-score is -1.96.

that would be on the left side of the distribution curve.

if you look for the area under the distribution curve that is equal to 1 - .025 = .975, you will find that the z-score is 1.96.

your confidence interval would be between a z-score of -1.96 and 1.96.

if you use a z-score calculator like the ti-84 plus and look for the inverse norm of (.025), the calculator will tell you that the z-score is -1.959963... which you can round to -1.96.

if you look for the inverse norm of (.975), the calculator will tell you that the z-score is 1.959963... which you can round to 1.96.

if you look in the z-score table for an area of .025, you will find that the table will tell you that the z-score is -1.96, and if you look for an area of.975, the table will tell you that the z-score is 1.96.

the areas don't always fall exactly on the z-score, although they did in this case.

if that happens, you need to interpolate or guess at about where the z-score would be unless picking the closest z-score from the table is acceptable.

at any rate, you found the z-score and, since you know the standard deviation, you can use the formula to determine sample size needed.

the formula to find the sample size needed starts with:

E = cz * sd / sqrt(ss)

E is the margin of error which is given as 1 as far as i can tell.
if E is not 1, then just use the same formula with the correct value of E.
cz is the critical z-score which we determined to be 1.96.
sd is given as 6.9
ss is the sample size.

with E = 1, the formula becomes 1 = 1.96 * 6.9 / sqrt(ss)

solve for sqrt(ss) to get sqrt(ss) = 1.96 * 6.9 / 1

this becomes sqrt(ss) = 13.524

solve for ss to get ss = 13.524^2 = 182.89 which you would round up to 183.

you would need a sample size of 183 to get a sample that would be equal to the mean plus or minus 1 at 95% confidence level.

to see if that holds water, we use the z-score formula of:

cz = (x-m) / se

cz is the critical z-score
x is the raw score
m is the mean
se is the standard error which is equal to sd/sqrt(n).

with a sample size of 183, se = 6.9/sqrt(183) = .510062677
cz is equal to 1.96
x-m is equal to 1.

the formula becomes 1.96 = 1 / .510062677 = 1.960543371 which you round off to 1.96.

you will get a margin of error of 1, regardless of what the mean is.

if you want your margin of error to be 10 rather than 1, you would use the formula of E = cz * sd / sqrt(ss) and replace E with 10 instead of 1.

since E = 10 and cz = 1.96 and sd = 6.9, the formula would become:

10 = 1.96 * 6.9 / sqrt(ss)

you would solve for sqrt(ss) to get sqrt(ss) = 1.96 * 6.9 / 10.

this gives you sqrt(ss) = 1.3524

solve for ss to get ss = 1.3524^2 = 1.82898576 = 2.

with an ss of 2, you would always get a margin of error that will be within 10 units from the mean.

for example:

assume the mean is 1000.

with an ss of 2, the se would be equal to 6.9/sqrt(2) = 4.87903679

to solve for x, you would use the z-score formula of cz = (x-m)/se which becomes 1.96 = (x-1000) / 4.87903679

solve for x to get x = 1.96 * 4.879303679 + 1000 = 1009.562912 which is within 10 of the mean.

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