Question 1122419: n order to estimate the mean amount of time computer users spend on the internet each month, how many computer users must be surveyed in order to be 95% confident that your sample mean is within 13 minutes of the population mean? Assume that the standard deviation of the population of monthly time spent on the internet is 221 min. What is a major obstacle to getting a good estimate of the population mean? Use technology to find the estimated minimum required sample size.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the critical z-score for 95% confidence interval is z = plus or minus 1.959963986 according to my TI-84 Plus calculator.
you could use this z-score or you could round it off to plus or minus 1.96 which is probably what you'd do if you used the z-score table.
i'll use plus or minus 1.96.
the differences should be very small.
at 95% confidence interval, the z-score needs to be plus or minus 1.96.
the formula for z-score is z = (x - m) / s
x is the raw score.
m is the mean
s is the standard error of the distribution of sample means.
s = standard deviation / sqrt (sample size)
you want your estimate of population mean to be plus or minus 13 at 95% confidence interval.
this means the critical z-score should result in a critical raw score that is plus or minus 13 from the sample mean.
the formula for z-score is z = (x - m) / s
you want (x - m) to be plus or minus 13.
working on the high side, you get 1.96 = 13 / s
s = standard deviation / sqrt(sample size) = 221 / sqrt(sample size)
formula becomes 1.96 = 13 / (221 / sqrt(sample size)
multiply both sides of this equation by 221 / sqrt(sample size) gets you:
1.96 * 221 / sqrt(sample size) = 13
divide both sides of this equation by 13 and multiply both sides of this equation by sqrt(sample size) gets you:
1.96 * 221 / 13 = sqrt(sample size)
solve for sqrt(sample size) gets you sqrt(sample size) = 33.32
solve for sample size gets you sample size = 33.32^2 = 1110.2224
round to the next highest integer gets you sample size = 1111.
that's the sample size required to get your margin of error to be within plus or minus 13.
to confirm, use the z-score formula again.
on the high side.....
1.96 = (x-m) / (221 / sqrt(1111)
multiply both sides of this equation by 221 / sqrt(1111) to get:
1.96 * 221 / sqrt(1111) = (x-m)
solve for (x-m) to get (x-m) = 12.995..... which is less than or equal to 13.
on the low side, -1.96 = (x-m) / 221 / sqrt(1111)
multiply both sides of this equation by 221 / sqrt(1111) to get:
-1.96 * 221 / sqrt(1111) = (x-m)
solve for (x-m) to get (x-m) = -12.995..... which is less than 13 from the mean.
looks like you got the sample size right.
the margin of error is less than or equal to plus or minus 13 at 95% confidence interval.
i believe this is accurate to the best of my knowledge.
the general formula for margin of error is z * standard deviation / sqrt (sample size).
in your problem, this becomes 13 = 1.96 * 221 / sqrt(sample size)
solve for sqrt(sample size) to get sqrt(sample size) = 1.96 * 221 / 13 = 33.32
sample size is 33.32^2 = 1110.something = 1111 rounded to next highest integer.
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