Question 822164
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Hi
The average age in the sample of 16 was 22 years with a standard deviation of six years.
 ME = {{{(highlight_green(1.96))(6/sqrt(16))}}}
95% confidence interval for the average age of the population. 
  22 - ME < {{{mu}}} < 22 + ME
Note:
80%	z = 1.28155
90%	z =1.645
92%	z = 1.751
95%	z = 1.96
98%	z = 2.326
99%	z = 2.576