Supplementary Exercise 7.43 THE TRASH BAG CASE Suppose that (unknown to the manu

Question

Supplementary Exercise 7.43 THE TRASH BAG CASE

      Suppose that (unknown to the manufacturer) the breaking strengths of the new 30-gallon bag are normally distributed with a mean of µ = 50.3 pounds and a standard deviation of ? = 1.61 pounds.

Find an interval containing 95.44 percent of all possible sample means if the sample size employed is
n = 4.

Find an interval containing 95.44 percent of all possible sample means if the sample size employed is
n = 46.

If the trash bag manufacturer hopes to obtain a sample mean that is at least 50 pounds (so that it can provide evidence that the population mean breaking strength of the new bags is at least 50), which sample size (n = 4 or n = 46) would be best?

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 = 50.3 pounds and a standard deviation of ? = 1.61 pounds.

Explanation / Answer

a)

p = 95.44/2 = 47.72

Calculate the first p:
2 = (X - ?)/(?/?n)
2 = (X - 50.3)/(1.61/?4)
1.61 = X-50.3
X = 51.91

Calculate the first p:
-2 = (X - ?)/(?/?n)
-2 = (X - 50.3)/(1.61/?4)
-1.61 = X-50.3
X = 48.69

Interval is [48.69,51.91]

b)

p = 95.44/2 = 47.72 , z value is 2

Calculate the first p:
2 = (X - ?)/(?/?n)
2 = (X - 50.3)/(1.61/?46)
(2*1.61)/6.78 = X-50.3
X = 50.3+0.47 = 50.77

Calculate the first p:
-2 = (X - ?)/(?/?n)
-2 = (X - 50.3)/(1.61/?46)
-(2*1.61)/6.78 = X-50.3
X = 50.3-0.47 = 49.83

Interval is [49.83,50.77]

c)

Sample size of 46 is better

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