Question 1076061: Question Help:
A television sports commentator wants to estimate the proportion of citizens who "follow professional football." What sample size should be obtained if he wants to be within 2 percentage points with 96% confidence if he uses an estimate of 54% obtained from a poll?
I think I use the equation E = z*sqrt[p*q/n] but I keep coming up with the wrong answer so I think I'm plugging in the wrong numbers
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! E is equal to sqrt(p*q/n), E being the standard error.
the formula for the z-score is z = (x-m)/s
z is the z-score
x is the sample mean
m is the population mean
s is the standard error
if you want the sample mean to be within 2 percentage points of .45, then you are looking for x to be equal to .45 plus or minus .009.
.45 * .02 = .009
if your confidence limit is 96% and you are dealing with a two sided confidence interval, then your critical alpha has to be 2% on each side.
you will be looking for a high z-score that has 98% of the area under the distribution curve to the left of it and a low z-score that has 2% of the area under the distribution curve to the left of it.
using the z-score calculator at http://davidmlane.com/hyperstat/z_table.html, i found the high critical z-score to be equal to 2.054 and the low critical z-score to be equa to -2.054
that's rounded to 3 decimal digits.
if you want to get more detailed, my ti-84 plus calculator said that the critical z-score was plus or minus 2.053748911.
using the high z-score, the formula for z-score is equal to (x-m)/s
the high raw score is .45 + .009 = .459
the low raw score is .45 - .009 = .441
the formula is z = (x-m) / s
using the high raw score, this becomes z = .009 / s
since the high critical z-score is 2.054, this becomes 2.054 = .009 / s
solve for s to get s = .009/2.054
since s = sqrt (p*q/n), you get sqrt(p*q/n) = .009/2.054
square both sides of this to get p*q/n = (.009/2.054)^2
p*q is equal to .54 * .46 = .2484
p*q /n becomes .2484/n, so the formula becomes .2484/n = (.009/2.054)^2
multiply both sides of the equation by n and divide both sides of the equation by (.009/2.054)^2 to get .2484 / (.009/2.054)^2 = n
solve for n to get n = 12938.00907
we should be able to round that to 12939.
going back to the original problem, we can now confirm whether the solution is correct or not.
you have p = .54 and q = .46 and n = 12939
m = p = .54
s = sqrt(p*q/n) = sqrt(.54*.46/12939) = .0043815265
your high critical z-score is 2.054.
your low critical z-score is -2.054.
high z = 2.054 = (x - .45)/.0043815265
low z = -2.054 = (x - .45)/.0043815265
in general, z = (x-m)/s is solved for x to get x = z*s + m
your high x-score becomes x = 2.054 * .0043815265 + .45 = .4589999... which can be rounded to .459
your low x-score becomes x = -2.054 * .0043815265 + .45 = .4410000.... which can be rounded to .441.
.459 - .45 = .009
.441 - .45 = -.009
.009 is equal to 2% of .45
both your high and low score are within 2% of .45.
i believe your solution needs to be that you sample size needs to be greater than or equal to 12939.
hopefully this gets you closer to the correct answer, if not right on.
the online calculator i used makes life easy for you.
a picture of the results i got from this calculator is shown below:
you just set the calculator to find the z-score from the area and it does all the dirty work for you.
using the ti-84 plus calculator, i looked for the high z-score with the knowledge that the low z-score would be the negative of that.
i put an area of .98 in the calculator and it told me the high z-score of 2.05374891 which i rounded to 2.054.
since the normal distribution is symmetrical about the mean, the low z-score had to be -2.054.
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