Question 1120956: A manufacturer wishes to estimate the proportion of washing machines leaving the factory that is defective. How large a sample should she check in order to be 94 percent confident that the true proportion is estimated to within 0.015?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the formula to use for E is E = Z * S
E is the margin of error.
Z is the critical Z-score.
S is the standard error.
the formula to use for S is S = sqrt(P*Q/N)
P is the probability of success.
Q is the probability of failure which is equal to 1 - P.
N is the sample size.
the formula of E = Z * S becomes:
E = Z * sqrt(P*Q/N)
divide both sides of this formula by Z to get:
E/Z = sqrt(P*Q/N)
square both sides of this formula to get:
E^2/Z^2 = P*Q/N)
multiply both sides of this formula by N and multiply both sides of this formula by Z^2 and divide both sides of this formula by E^2 to get:
N = P * Q * Z^2 / E^2.
since you don't have a clue as to what P should normally be, then use .5 as P.
this results in .5 as Q.
your minimum sample size will be the largest it has to be when P = .5 which will ensure any P other than .5 will be covered more than enough.
your critical z-score at 96% confidence level will be Z = plus or minus 1.88079361 which results in Z^2 = 3.537384602
your desired margin of error is E = plus or minus .015 which results in E^2 = .000225
N = P * Q * Z^2 / E^2 becomes:
N = .5 * .5 * 3.537384602 / .000225
this results in N = 3930.427336
you would normally round up to the nearest integer to get N = 3931.
wheh N = 3931, your sample size will be big enough to ensure you are within .015 of the desired population percentage, regardless of what it is, with an assumed population proportion of .5 being the worst case.
for example:
the P = .5 and Q = .5 and Z = 1.88079361, the z-score formula becomes:
Z = (X - M) / S
S is equal to sqrt(P * Q / N)
N is equal to 3931.
S is equal to sqrt (.5 * .5 / 3931) = .007974758
when Z is 1.88079361 and M is .5 and S is .007974758, you solve for X to get:
X = 1.88079361 * .007974758 + .5 = .5149989074.
E is equal to that minus .5 = .0149989074 which is a smaller margin of error than .015.
when Z is -1.88079361 and M is .5 and S is .007974758, you solve for X to get:
X = -1.88079361 * .007974758 + .5 = .4850010926
E is equal to that minus .5 = -.0149989074 which is a smaller margin of error than -.015.
if your assumed population proportion is .2, then:
P = .2 and Q = .8
S becomes sqrt(P*Q/N) = sqrt(.2*.8/N) which becomes S = sqrt(.16/N)
E = Z * S becomes E = Z * sqrt(P*Q/N) which becomes E = Z * sqrt(.16/N)
we solved for N previously to get N = P * Q * Z^2 / E^2.
when P = .2 and Q = .8 and Z = plus or minus 1.88079361 and E = plus or minus .015, we solve for N to get N = 2515.473495.
you need a smaller sample size because your assumed population proportion is .2 rather than .5.
the largest sample sized you should require is when P = .5
when N = 2516, S = sqrt(.2*.8/2516) = .0079745222.
Z = (X-M)/S on the high side becomes:
1.88079361 = (X - .2) / .0079745222.
solve for X to get X = 1.88079361 * .0079745222 + .2 = .2149984304.
subtract .2 from that to get E = .0149984304 which is less than the required margin of error of .015.
if you had left the sample size at 3931 when P = .2, you would have gotten:
S = sqrt(.2 * .8 / 3931) = .0063798206.
X = Z * S + .2 would become X = .2119991259.
that minus .2 would results in a margin of error of .0119991259 which is much less than the required margin of error of less than or equal to .015.
bottom line is:
if you know what P is assumed to be, use that.
otherwise, use P = .5
when P = .5 you will get the largest minimum sample size required.
when P is something other than .5, your sample size will be larger than required unless you redo the formula using P equal to something other than .5
there are various references on the web, some easier to understand than others.
ones that i used are:
http://sphweb.bumc.bu.edu/otlt/mph-modules/bs/bs704_power/BS704_Power_print.html
https://onlinecourses.science.psu.edu/stat500/node/31/
https://www.dummies.com/education/math/statistics/how-to-calculate-the-margin-of-error-for-a-sample-proportion/
https://www.unc.edu/~rls/s151-2010/class23.pdf
good luck with them.
your solution, as best i can determine, is a minimum sample size of 3931 would be required, assuming the population proportion is .5.
if the population proportion is assumed to be something other than that, you could require a smaller sample size, but you should not require a larger sample size.
|
|
|
