Suppose that (unknown to the manufacturer) the breaking strengths of the new 30-

Question

      Suppose that (unknown to the manufacturer) the breaking strengths of the new 30-gallon bag are normally distributed with a mean of µ = 50.2 pounds and a standard deviation of = 1.66 pounds.

Find an interval containing 95.44 percent of all possible sample means if the sample size employed is
n = 6.

(b)

Find an interval containing 95.44 percent of all possible sample means if the sample size employed is
n = 48.

Recall that the trash bag manufacturer has concluded that its new 30-gallon bag will be the strongest such bag on the market if its mean breaking strength is at least 50 pounds. In order to provide statistical evidence that the mean breaking strength of the new bag is at least 50 pounds, the manufacturer randomly selects a sample of n bags and calculates the mean of the breaking strengths of these bags. If the sample mean so obtained is at least 50 pounds, this provides some evidence that the mean breaking strength of all new bags is at least 50 pounds.

      Suppose that (unknown to the manufacturer) the breaking strengths of the new 30-gallon bag are normally distributed with a mean of µ = 50.2 pounds and a standard deviation of = 1.66 pounds.

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 =    0.0228          
X = sample mean =    50.2          
z(alpha/2) = critical z for the confidence interval =    1.999077215          
s = sample standard deviation =    1.66          
n = sample size =    6          
              
Thus,              
Margin of Error E =    1.35475896          
Lower bound =    48.84524104          
Upper bound =    51.55475896          
              
Thus, the confidence interval is              
              
(   48.84524104   ,   51.55475896   ) [ANSWER]

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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 =    0.0228          
X = sample mean =    50.2          
z(alpha/2) = critical z for the confidence interval =    1.999077215          
s = sample standard deviation =    1.66          
n = sample size =    48          
              
Thus,              
Margin of Error E =    0.478979624          
Lower bound =    49.72102038          
Upper bound =    50.67897962          
              
Thus, the confidence interval is              
              
(   49.72102038   ,   50.67897962   ) [ANSWER]

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