A business magazine targeting recent college graduates condu

A business magazine targeting recent college graduates conducted a study to estimate its subscribers annual household income and the proportion who plan to make a real estate purchase within the next year. The results from a random sample of n=36 subscribers are as follows:

annual income:x= $70,400; s=$16,500

sixteen subscribers plan to make a real estate purchase within the next year.

a. construct a 98% confidence interval estimate for the subscrivers\' population mean annual household income.

b. explain/interpret in words the meaning associated with the confidence interval from part a.

d. using 90% confidence and p* equal to your point estimate from part c., how large a sample is needed if the magazine desires a margin of error equal to 0.07 when estimating the population proportion of customers who plan to make a real estate purchase within the next year?

Solution

a)

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 =    0.01          
X = sample mean =    70400          
z(alpha/2) = critical z for the confidence interval =    2.326347874          
s = sample standard deviation =    16500          
n = sample size =    36          
              
Thus,              
              
Lower bound =    64002.54335          
Upper bound =    76797.45665          
              
Thus, the confidence interval is              
              
(   64002.54335   ,   76797.45665   ) [ANSWER]

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b)

We are 98% confidence that the true population mean is between $64002.54 and $76797.46.

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c)

Note that              
              
p^ = point estimate of the population proportion = x / n =    0.444444444          
              
Also, we get the standard error of p, sp:              
              
sp = sqrt[p^ (1 - p^) / n] =    0.082817332          
              
Now, for the critical z,              
alpha/2 =   0.05          
Thus, z(alpha/2) =    1.644853627          
Thus,              
              
lower bound = p^ - z(alpha/2) * sp =   0.308222055          
upper bound = p^ + z(alpha/2) * sp =    0.580666834          
              
Thus, the confidence interval is              
              
(   0.308222055   ,   0.580666834   ) [ANSWER]

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D)

Note that      
      
n = z(alpha/2)^2 p (1 - p) / E^2      
      
where      
      
alpha/2 =    0.05  
       
      
Using a table/technology,      
      
z(alpha/2) =    1.644853627  
      
Also,      
      
E =    0.07  
p =    0.444444444  
      
Thus,      
      
n =    136.3337593  
      
Rounding up,      
      
n =    137   [ANSWER]