Question 884258: the IQ test is designed so that the mean is 100 and the standard deviation is 19 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 90% confidence that the sample mean is within 6 IQ points of the true mean. Assume σ=19 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! i think you meant the mean is 100 so i changed your 10 to 100.
if i got that wrong, then you can adjust the numbers because the process of arriving at the answer is the same.
mean is 100
standard deviation is 19.
the alpha and the critical z score for a confidence level of 90% is calculated as follows:
two sided alpha = (100% - 90%) / 200 = .05
critical z factor for two sided alpha of .05 is calculated as follows:
critical z factor = z factor for (1 - .05) = z factor for (.95) which becomes:
critical z factor = 2.58.
it is actuallyh smewhere between 2.57 and 2.58 but i chose 2.58 because that gets you an alpha just under .05.
the margin of error is equal to the critical z factor times the standard error.
the standard error is equal to the standard deviation divided by the square root of the sample size.
we are given that the standard deviation is equal to 19.
we are given that the margin of error is desired to be smaller than or equal to 6.
we calculated the critical z factor to be equal to 2.58.
we replace critical z factor with 2.58 and margin or error with 6 and we let se = standard error.
the formula for margin of error becomes:
6 = 2.58 * se
we solve for se to get:
se = 6/2.58
we replace se with the formula to derive it of se = psd / sqrt(ss), where se = standard error and psd = population standard deviation and ss = sample size to get:
psd / sqrt(ss) = 6/2.58
we replace psd with the population standard deviation of 19 to get:
19 / sqrt(ss) = 6/2.58
we multiply both sides of the equation by sqrt(ss) and multiply both sides of the equation by (2.58/6) to get:
19 * 2.58 / 6 = sqrt(ss)
we square both sides of thie equation to get:
(19 * 2.58 / 6)^2 = ss
we simplify to get:
ss = 66.7489 which we round to the next highest integer of 67.
if we are correct, the sample size needs to be 67 in order for us to get a margin of error that is within 6.
let's go through the drill to see if we're correct.
start with:
mean = 100
standard deviation = 19
sample size = 67
at 90% confidence level the critical z factor is 2.58
standard error = standard deviation divided by sqrt(sample size) = 19 / sqrt(67) = 2.321219443.
margin of error is equal to critical z factor * standard error which is equal to 2.58 * 2.321219443 which is equal to 5.988746162
that's pretty close to, and within, 6.
the confidence interval becomes 100 plus or minus the margin of error which becomes 100 plus or minus 5.988746162.
if we didn't round the critical z factor and we didn't round the sample size we would more than likely have been right on exactly at a margin of error of 6.
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