SOLUTION: A random sample of 100 healthy residents has a mean chloride level of 102 mEq/L. If it is known that the chloride levels in healthy individuals have a standard deviation of 40 mEq/
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Question 1119755: A random sample of 100 healthy residents has a mean chloride level of 102 mEq/L. If it is known that the chloride levels in healthy individuals have a standard deviation of 40 mEq/L, find a 95% confidence interval for the true mean chloride level of all healthy residents. Carry your intermediate computations to at least three decimal places. Round your answers to one decimal place.
What is the lower limit of the 95% confidence interval?
What is the upper limit of the 95% confidence interval? Answer by Theo(13342) (Show Source):
standard deviation of chloride level with mean of 102 is 40.
standard error of the distribution of sample means is equal to standard deviation of the population divided by the sample size.
that makes the standard error of the distribution means equal to 40 / sqrt(100) = 4.
95% confidence interval for the true mean requires an alpha of .05/2 = .025 on each end of the confidence interval.
that leads to a critical z-score of plus or minus 1.959963986.
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 error of the distribution of sample means.
for the low end of the confidence interval, you get:
-1.959963986 = (x - 102) / 4.
solve for x to get x = 4 * -1.959963986 + 102 = 94.16014406.
for the high end of the confidence interval, you get:
1.959963986 = (x - 102) / 4.
solve for x to get x = 4 * 1.959963986 + 102 = 109.8398559.
round to 1 decimal place and your 95% confidence interval for the true mean will be between 94.2 and 109.8.
use of the online normal distribution calculator at http://davidmlane.com/hyperstat/z_table.html confirms that the calculations are correct within minor differences due to rounding.
here's what the results look like from that calculator after determining that the standard error is 40 / sqrt(100) = 4.
