A sample of 250 observations is selected from a normal popul

A sample of 250 observations is selected from a normal population for which the population standard deviation is known to be 23. The sample mean is 18.

Determine the standard error of the mean. (Round your answer to 3 decimal places.)

Determine the 99% confidence interval for the population mean. (Use z Distribution Table.) (Round your answers to 3 decimal places.)

Solution

a)

SE = s/sqrt(n) = 23/sqrt(250) = 1.454647724 [ANSWER]
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B)

Note that              
      
Lower Bound = X - z(alpha/2) * s / sqrt(n)              
Upper Bound = X + z(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    18          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    23          
n = sample size =    250          
              
Thus,              
Lower bound =    14.25307577          
Upper bound =    21.74692423          
              
Thus, the confidence interval is              
              
(   14.25307577   ,   21.74692423   ) [ANSWER]