In a sample of 101 Atlanta residents average income is $56,560. If the standard

Question

In a sample of 101 Atlanta residents average income is $56,560. If the standard deviation of income in Atlanta area is $12,000 then please do the following:

- 90% confidence interval for population mean?

-95% confidence interval for population mean?

- 99% confidence interval for population mean?

-If a statistician report Atlanta median income is $55,733 since you can assume population mean is Normally distributed can you accept this number based on this number? Is there a confidence interval where $55,733 is smaller than lower bound or higher than higher bound?

Explanation / Answer

a)

90% confidence:

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
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 =    0.05          
X = sample mean =    56560          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    12000          
n = sample size =    101          
              
Thus,              
Margin of Error E =    1964.028638          
Lower bound =    54595.97136          
Upper bound =    58524.02864          
              
Thus, the confidence interval is              
              
(   54595.97136   ,   58524.02864   ) [ANSWER]

******************

b)

95% confidence:

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
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 =    56560          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    12000          
n = sample size =    101          
              
Thus,              
Margin of Error E =    2340.284467          
Lower bound =    54219.71553          
Upper bound =    58900.28447          
              
Thus, the confidence interval is              
              
(   54219.71553   ,   58900.28447   ) [ANSWER]

********************

c)

Note that              
Margin of Error E = z(alpha/2) * s / sqrt(n)              
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 =    56560          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    12000          
n = sample size =    101          
              
Thus,              
Margin of Error E =    3075.655143          
Lower bound =    53484.34486          
Upper bound =    59635.65514          
              
Thus, the confidence interval is              
              
(   53484.34486   ,   59635.65514   ) [ANSWER]

************************

YES, THIS CAN BE ACCEPTED, AS 55733 IS INSIDE ALL THE CONFIDENCE INTERVALS HERE. [ANSWER]

There is no confidence interval (among the three) that has a smaller lower bound or higher higher bound than 55733. [ANSWER]

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