Question 1120030: An IQ test is designed so that the mean is 100 and the standard deviation is 12 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 95% confidence that the sample mean is within 7 IQ points of the true mean. Determine the required sample size using the TI-83/84.
The required sample size is________
(Round UP to the nearest integer.)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean is 100.
standard deviation is 12.
you want the sample mean to be within plus or minus 7 points at 95% confidence level.
first you want to know what the critical z-score is.
at 95% confidence level, 2.5% of the area under the normal distribution curve is outside the confidence interval on the low side and 2.5% of the normal distribution curve is outside the confidence interval on the high side.
look for a z-score that has .025 proportion of the area under the normal distribution curve to the left of it.
that z-score will be minus 1.959963986.
since the normal distribution curve is symmetric about the mean, then the z-score that has .025 proportion of the area under the normal distribution curve to the right of it would be plus 1.959963986.
your confidence interval is between the critical z-scores of plus or minus 1.959963986.
that z-score value was found using the TI-83/84.
you would select 2d vars (this gets you distr) and then select 3.
that give you the inverse norm function.
you then enter .025) and the calculator tells you that the z-score is minus 1.959963986.
you can use the concept of symmetry to determine that the area of .025 to the right of the z-score leads to plus 1.959963986, or you can do the following:
1 - .025 = .975
find the inverse norm for .975.
you select 2d vars and then select 3 and then enter .975).
the calculator tells you that the z-score is plus 1.959963986.
the inverse norm function in the TI-83/84 is designed to give you the z-score that has the area to the left of it.
if you want the area to the right of the z-score, you take 1 minus the area to the left.
for .975 area to the left, 1 - .975 = .025 area to the right.
now that you know the critical z-score, you can find the critical raw score and/or the sample size required so that the interval is plus or minus 7 points at 95% confidence level.
the z-score formula is:
z = (x-m)/s
z is the z-score
x is the raw score
m is the raw mean
s is the standard deviation or standard error.
in this case, when you are dealing with the mean of a sample of a specified size, standard error is used.
the formula for standard error is:
s = standard deviation / square root of sample size.
in your problem, the population mean is 100 and standard deviation is 12.
the formula for standard error becomes:
s = 12 / sqrt(n), where n is the sample size.
you want your x to be within plus or minus 7 of the mean.
your mean is 100.
that means you want your x to be between 93 and 107.
when x is 93, the z-score formula becomes:
-1.959963986 = (93-100) / (12/sqrt(n)).
multiply both sides of this equation by 12/sqrt(n)) to get:
-1.959963986 * 12/sqrt(n) = 93 - 100.
this results in -1.959963986 * 12/sqrt(n) = -7
divide both sides of this equation by -7 and multiply both sides of this equation by sqrt(n) to get:
-1.959963986 / -7 * 12 = sqrt(n)
square both sides of this equation to get:
(-1.959963986 / -7 * 12) ^ 2 = n
solve for n to get:
n = 11.28918512
when n = 11.28918512, your standard error becomes 12 / sqrt(11.28918512) = 3.571494196.
your z-score formula used a standard error of 3.571494196.
to confirm this will give you an interval of 93 to 107, you do the following.
use z = (x - m) / s
when z = -1.959963986 and m = 100 and s = 3.571494196, this formula becomes:
-1.959963986 = (x - 100) / 3.571494196.
solve for x to get:
x = -1.959963986 * 3.571494196 + 100 = 93.
on the high side, .....
x = 1.959963986 * 3.571494196 + 100 = 107.
the sample size of 11.28918512 leads to a 95% confidence interval of 93 to 107 which is plus or minus 7 from the mean of 100.
since simple size has to be integer, you would normally round up to the next highest integer which would be 12.
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