A nursing school wants to estimate the true mean annual income of its alumni.?

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.A nursing school wants to estimate the true mean annual income of its alumni.?
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]A nursing school wants to estimate the true mean annual income of its alumni.?
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*蟽)/鈭歯)


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 = 蟽 = $1,500


sample = n = 120





58700+-(1.96*1500)/鈭?20


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)/鈭?20 = 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.