A radio talk show invites listeners to enter a dispute about

A radio talk show invites listeners to enter a dispute about proposed pay increase for city council members. \"What yearly pay do you think council members should get? Call us with your number.\" In all, 958 people call in. The mean pay they suggest is x = 8740 dollars and the standard deviation of the population is known to be 1125. We wish to construct a 95% confidence interval for mean pay mu, which table should we check for critical value ? Select one from the two below: Calculate the 95% confidence interval for the mean pay mu that all citizens would propose for the council members to be.

Solution

A)

We use a Z TABLE, as our sample size is very large here, 958 (n > 30).

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 =    0.025          
X = sample mean =    8740          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    1125          
n = sample size =    958          
              
Thus,              
Lower bound =    8668.760993          
Upper bound =    8811.239007          
              
Thus, the confidence interval is              
              
(   8668.760993   ,   8811.239007   ) [ANSWER]