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Thirty small communities in Connecticut (population near 10,000 each) gave an av

ID: 3159269 • Letter: T

Question

Thirty small communities in Connecticut (population near 10,000 each) gave an average of x = 138.5 reported cases of larceny per year. Assume that is known to be 44.3 cases per year.

(a) Find a 90% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit upper limit margin of error

(b) Find a 95% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit upper limit margin of error

(c) Find a 99% confidence interval for the population mean annual number of reported larceny cases in such communities. What is the margin of error? (Round your answers to one decimal place.) lower limit upper limit margin of error

Explanation / Answer

a)

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 =    138.5          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    44.3          
n = sample size =    30          
              
Thus,              
Margin of Error E =    13.30363606   [ANSWER]
      
Lower bound =    125.1963639   [ANSWER]      
Upper bound =    151.8036361   [ANSWER]

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

b)

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 =    138.5          
z(alpha/2) = critical z for the confidence interval =    1.959963985          
s = sample standard deviation =    44.3          
n = sample size =    30          
              
Thus,              
Margin of Error E =    15.85226011       [ANSWER]      
Lower bound =    122.6477399   [ANSWER]          
Upper bound =    154.3522601   [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 =    138.5          
z(alpha/2) = critical z for the confidence interval =    2.575829304          
s = sample standard deviation =    44.3          
n = sample size =    30          
              
Thus,              
Margin of Error E =    20.83340125       [ANSWER]      
Lower bound =    117.6665987   [ANSWER]          
Upper bound =    159.3334013   [ANSWER]      
  

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