What price do farmers get for their watermelon crops? In the third week of July,

Question

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 45 farming regions gave a sample mean of x = $6.88 per 100 pounds of watermelon. Assume that is known to be $1.82 per 100 pounds.

(a) Find a 90% confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (Round your answers to two decimal places.)

lower limit = $_____

upper limit = $_____

margin of error = $_____

(b) Find the sample size necessary for a 90% confidence level with maximal error of estimate E = 0.43 for the mean price per 100 pounds of watermelon. (Round up to the nearest whole number.)

_____ = farming regions

(c) A farm brings 15 tons of watermelon to market. Find a 90% confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds. (Round your answers to two decimal places.)

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 =    6.88          
z(alpha/2) = critical z for the confidence interval =    1.644853627          
s = sample standard deviation =    1.82          
n = sample size =    45          
              
Thus,              

Lower bound =    6.433735451   [ANSWER]              
Upper bound =    7.326264549   [ANSWER]              
Margin of Error E =    0.446264549   [ANSWER]      
  
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b)

Note that      
      
n = z(alpha/2)^2 s^2 / E^2      
      
where      
      
alpha/2 = (1 - confidence level)/2 =    0.05  
      
Using a table/technology,      
      
z(alpha/2) =    1.644853627  
      
Also,      
      
s = sample standard deviation =    1.82  
E = margin of error =    0.43  
      
Thus,      
      
n =    48.46858917  
      
Rounding up,      
      
n =    49   [ANSWER]

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