In: Statistics and Probability
Consider the trash bag problem. Suppose that an independent laboratory has tested trash bags and has found that no 30-gallon bags that are currently on the market have a mean breaking strength of 50 pounds or more. On the basis of these results, the producer of the new, improved trash bag feels sure that its 30-gallon bag will be the strongest such bag on the market if the new trash bag’s mean breaking strength can be shown to be at least 50 pounds. The mean of the sample of 42 trash bag breaking strengths in Table 1.10 is x⎯⎯x¯ = 50.579. If we let µ denote the mean of the breaking strengths of all possible trash bags of the new type and assume that σ equals 1.60:
(a) Calculate 95 percent and 99 percent confidence intervals for µ. (Round your answers to 3 decimal places.)
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(a)
n = Sample Size = 42
= Sample Mean = 50.579
= 1.60
SE = /
= 1.60/
= 0.2469
= 0.05
From Table,critical values of Z = 1.96
95% confidence interval:
50.579 (1.96 X 0.2469)
= 50.579 0.4839
= [ 50.0951 ,51.0629]
(b)
= 0.01
From Table,critical values of Z = 2.576
95% confidence interval:
50.579 (2.576 X 0.2469)
= 50.579 0.6360
= [ 49.9430 ,51.2150]
So,
Answer is:
95 percent confidence interval for is | [ 50.095 ,51.063] |
99 percent confidence interval for is | [ 49.943 ,51.215] |