A sample of 20 small bags of the same brand of candies was s

A sample of 20 small bags of the same brand of candies was selected. Assume that the population distribution of bag weights is normal. The weight of each bag was then recorded. The mean weight was 5 ounces with a sample standard deviation s=0.5 ounces. We would like to calculate an 80% confidence interval for the average weight of a sample of size 20 .

sx=   

The critical   t   value for an 80% confidence interval is tcrit=   

  EBM=sxtcrit=   

An   80%   confidence interval for the population average weight of the candies is from

   to   

Solution

a)

sx = s/sqrt(n) = 0.5/sqrt(20) = 0.111803399 [ANSWER]

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b)

Here,

alpha/2 = (1 - confidence level)/2 =    0.1          
Thus, as df = n - 1 = 19,

t(alpha/2) = critical t for the confidence interval =    1.327728209   [ANSWER]

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c)

Note that              

Margin of Error EBM = t(alpha/2) * s / sqrt(n)  

where              

alpha/2 = (1 - confidence level)/2 =    0.1          
t(alpha/2) = critical t for the confidence interval =    1.327728209          
s = sample standard deviation =    0.5          
n = sample size =    20          

Thus,              

Margin of Error EBM =    0.148444527 [ANSWER]

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d)
          
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              

alpha/2 = (1 - confidence level)/2 =    0.1          
X = sample mean =    5          
t(alpha/2) = critical t for the confidence interval =    1.327728209          
s = sample standard deviation =    0.5          
n = sample size =    20          
df = n - 1 =    19          
Thus,              

Lower bound =    4.851555473          
Upper bound =    5.148444527          
              
Thus, the confidence interval is              
              
(   4.851555473   ,   5.148444527   ) [ANSWER]