SOLUTION: With a sample size of 900, the standard error is 3. What should be the sample size so that we could be 95% confident that the population mean is within 4 of the sample mean? Answe

Algebra ->  Statistics  -> Confidence-intervals -> SOLUTION: With a sample size of 900, the standard error is 3. What should be the sample size so that we could be 95% confident that the population mean is within 4 of the sample mean? Answe      Log On


   



Question 1170608: With a sample size of 900, the standard error is 3. What should be the sample size so that we could be 95% confident that the population mean is within 4 of the sample mean?
Answer=1945

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the standard error is equal to the standard deviation divided by the square root of the sample size.

with a sample size of 900, the standard error is 3.

this gets you the formula se = sd / sqrt(ss).
se = standard error
sd = standard deviation
ss = sample size.

in your problem, the formula becomes 3 = sd / sqrt(900)
solve for sd to get:
sd = 3 * sqrt(900) = 3 * 30 = 90.

at 95% confidence level, the two tailed alpha will be .05/2 = .025.

the critical z-score becomes plus or minus 1.96

the z-score formula becomes plur or minus 1.96 = (x - m) / se

since se = sd / sqrt(ss), and since sd = 90, the formula becomes se = 90 / sqrt(ss).

the formula becomes plus or minus 1.96 = (x - m) / (90 / sqrt(ss)).

since you want the margin of error to be plus minus 4, then (x - m) must be equal to plus or minus 4, and the formula becomes:

plus or minus 1.96 = (plus or minus 4) / (90 / sqrt(ss))

this is the same as:

plus or minus 1.96 = (plus or minus 4) * sqrt(ss) / 90

solve for sqrt(ss) to get:

sqrt(ss) = 90 * (plus or minus 1.96) / (plus or minus 4)

just working on the plus sides of this equation, you get:

sqrt(ss) = 90 * 1.96 / 4.

that gets you sqrt(ss) = 44.1

solve for ss to get:

ss = 1944.81.

that's the answer your are looking for.

the other options will lead to the following results.

sqrt(ss) = 90 * -1.96 / -4 = 44.1, giving you the same answer.

sqrt(ss) = 90 * 1.96 / -4 = -44.1 which is not possible because it winds up being the square root of a negative number which is not real.

sqrt(ss) = 90 * -1.96 / 4 = -44.1 which is not real for the same reason.

the only real answer is 1944.81.

your solution is that the sample size is 1944.81.

to confirm, replace ss with 1944.81 in the problem to get:

plus or minus 1.96 = (x - m) / (90/sqrt(1944.81).

solve for (x - m) to get:

(x - m) = 1.96 * 90 / sqrt(1944.81) = 4 when 1.96 is positive.

(x - m) = -1.96 * 90 / sqrt(1944.81) = -4 when 1.96 is negative.

your margin of error is plus or minus 4.

this works with any mean as long as the standard error is equal to 90 / sqrt(1944.81) = 2.040800586.

for example, if the mean is 7500, then you get:

z = (7504 - 7500) / 2.040800586 = 1.96

z = (7496 - 7500) / 2.040800586 = -1.96

the confidence interval is from 7496 to 7504.

the margin of error is plus or minus 4.

normally, you would round the answer to the next highest integer which would make it equal to a sample size of 1945.

that makes your margin of error slightly less than 4.

for example:

1.96 = (x - m) / (90/sqrt(1945)

solve for (x - m) to get:

(x - m) = 3.999804622.

that's slightly less than 4, but if you round it to the 3 decimal places, it will be equal to 4.