Question

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 37 trash bag breaking strengths in Table 1.9 is x⎯⎯ x ¯ = 50.560. 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 1.67: (a) Calculate 95 percent and 99 percent confidence intervals for µ. (Round your answers to 3 decimal places.) 95 percent confidence intervals for µ is [, ]. 99 percent confidence intervals for µ is [, ].

Answer 1

mean = 50.560 , sigma = 1.67 , n = 37

The z value at 95% confidence interval is,

alpha = 1 - 0.95 = 0.05
alpha/2 = 0.05/2 = 0.025
Zalpha/2 = Z0.025 = 1.96

Margin of error = z *(s/sqrt(n))
= 1.96 *(1.67/sqrt(37))
= 0.538

CI = xbar +/ E
= 50.560 +/- 0.538

= (50.022 , 51.098)

The 95% CI for mu is (50.022 , 51.098)

The z value at 99% confidence interval is,

alpha = 1 - 0.99 = 0.01
alpha/2 = 0.01/2 = 0.005
Zalpha/2 = Z0.005 = 2.58

Margin of error = z *(s/sqrt(n))
= 2.58 *(1.67/sqrt(37))
= 0.708

CI = xbar +/ E
= 50.560 +/- 0.708
= (49.852 , 51.268)

The 99% CI for mu is (49.852 , 51.268)