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It randomly samples 120 of its alumni. The mean annual income was $58,700 with a standard deviation of $1,500. Find a 95% confidence interval for the true mean annual income of the nursing school alumni. Write a statement about the confidence level and the interval you find.
Tags: Alumni., annual, Estimate, Income, Mean, Nursing, School, True, Wants
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October 1st, 2009 at 11:01 pm
ANSWER: 95% Confidence Interval [$57,532 $58,968]
Why???
Large-Sample confidence interval for “mu” (true population mean) with a confidence level of (approximately) 95% (manufacturer’s claim .05) has:
lower confidence limit = x-bar – 1.96 * s/SQRT(n)
upper confidence limit = x-bar + 1.96 * s/SQRT(n)
x-bar = sample mean [58700]
s = sample standard deviation [1500]
n = number of samples [120]
October 2nd, 2009 at 4:08 am
This problem requires the z-distribution formula. Which is the sample mean plus and minus z times the standard deviation divided by the square root of the sample.
_
X+-((z*?)/?n)
NOTE: (the addition sign is suppose to be on top of the minus sign)
sample mean = X = $58,700
confidence= z = 95% = 1.96 (you get this from a z-chart)
standard deviation = ? = $1,500
sample = n = 120
58700+-(1.96*1500)/?120
NOTE: (the addition sign is suppose to be on top of the minus sign)
first don’t worry about 58700 and solve the other half, which you should get:
(1.96*1500)/?120 = 268.384 = 268.38
now you can add and subtract that from $58,700
58700+268.38 = 58968.4
58700-268.38 = 58431.6
Thus showing with 95% confidence the annual income is well within mean of $58,700 with the standard deviation of $1,500.