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 39 trash bag breaking strengths in Table 1.9 is x⎯⎯x¯ = 50.573. 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 1.61:

  

(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	[, ].

(b) Using the 95 percent confidence interval, can we be 95 percent confident that µ is at least 50 pounds? Explain.

(Click to select)NoYes , 95 percent interval is (Click to select)belowabove 50.

(c) Using the 99 percent confidence interval, can we be 99 percent confident that µ is at least 50 pounds? Explain.

(Click to select)NoYes , 99 percent interval extends (Click to select)abovebelow 50.

(d) Based on your answers to parts b and c, how convinced are you that the new 30-gallon trash bag is the strongest such bag on the market?

(Click to select)FairlyNot confident, since the 95 percent CI is (Click to select)belowabove 50 while the 99 percent CI contains 50.

rev : 08_22_2016_QC_CS-57697

Answer 1

a)

sample mean, xbar = 50.573
sample standard deviation, σ = 1.61
sample size, n = 39

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, Zc = Z(α/2) = 1.96

ME = zc * σ/sqrt(n)
ME = 1.96 * 1.61/sqrt(39)
ME = 0.51

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (50.573 - 1.96 * 1.61/sqrt(39) , 50.573 + 1.96 * 1.61/sqrt(39))
CI = (50.068 , 51.078)

95 percent confidence intervals for µ is (50.068 , 51.078)

sample mean, xbar = 50.573
sample standard deviation, σ = 1.61
sample size, n = 39

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, Zc = Z(α/2) = 2.58

ME = zc * σ/sqrt(n)
ME = 2.58 * 1.61/sqrt(39)
ME = 0.67

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (50.573 - 2.58 * 1.61/sqrt(39) , 50.573 + 2.58 * 1.61/sqrt(39))
CI = (49.908 , 51.238)

99 percent confidence intervals for µ is (49.908 , 51.238)

b)

No , 95 percent interval extends above 50.

Yes , 99 percent interval extends below 50.

c)

Not confident, since the 95 percent CI is above 50 while the 99 percent CI contains 50.