Hw 2 Prob 1 Two new methods for producing a tire have been proposed. To ascertain which is superior, a tire manufacturer produces a sample of 10 tires using the first method and a sample of 8 using the second. The first set are to be road tested at location A and the second at location B. It is known from past experience that the lifetime of a tire that is road tested at one of these locations is normally distributed with a mean life due to the tire but with a variance due (for the most part) to the location. Specifically, it is known that the lifetimes of tires tested at location A are normal with standard deviation equal t kilometers, whereas those tested at location B are normal with σ = 6000 kilometers. If the manufacturer is interested in testing the hypothesis that there are no appreciable difference in the mean life of tires produced by either method, what conclusion should be drawn at the 5% level of significance if the resulting data are as given in the table below? Tire Lives in Units of 100 Kilometers Tires tested at ATires tested at B

Hw 2 Prob 1 Two new methods for producing a tire have been proposed. To ascertain which is superior, a tire manufacturer produces a sample of 10 tires using the first method and a sample of 8 using the second. The first set are to be road tested at location A and the second at location B. It is known from past experience that the lifetime of a tire that is road tested at one of these locations is normally distributed with a mean life due to the tire but with a variance due (for the most part) to the location. Specifically, it is known that the lifetimes of tires tested at location A are normal with standard deviation equal t kilometers, whereas those tested at location B are normal with σ = 6000 kilometers. If the manufacturer is interested in testing the hypothesis that there are no appreciable difference in the mean life of tires produced by either method, what conclusion should be drawn at the 5% level of significance if the resulting data are as given in the table below?  Since Z=0.066 is neither in the right tail nor the left tail, we have no statistical evidence they are different and we accept Ho.

Hw 2 Prob 2 A public health official claims that the mean home water use is 350 gallons a day. To verify this claim, a study of 20 randomly selected homes was instigated with the result that the average daily water uses of these 20 homes were as follows: Do the data contradict the official’s claim?  Z=0.78 lies between +/ We do not have enough evidence to conclude our hypothesis is wrong. Accept Ho t 19

Hw 2 Prob 3 An industrial safety program was recently instituted in the computer chip industry. The average weekly loss (averaged over one month) in man-hours due to accidents in 10 similar plants both before and after the program are as follows: PlantBeforeAfter Determine at the 5 percent level of significance, whether the safety program has been proven to be effective.  T = is left of Therefore, we have enough evidence to conclude there is an improvement. Reject H o, accept H A t9t9

Hw 2 Prob 4 A machine that automatically controls the amount of ribbon on a tape has recently been installed. This machine will be judged to be effective if the standard deviation σ of the amount of ribbon on a tape is less than 0.15 cm. If a sample of 20 tapes yields a sample variance of S 2 =.025 cm 2, are we justified in concluding that the machine is ineffective? < <30.14; therefore we do not have enough evidence to reject the null hypothesize. Accept Ho, machine is ineffective. X

Hw 2 Prob 5 There are two different choices of a catalyst to stimulate a certain chemical process. To test whether the variance of the yield is the same no matter which catalyst is used, a sample of 10 batches id produced using the first catalyst, and 12 using the second. If the resulting data is S 2 1 =.14 and S 2 2 =.28, can we reject, at the 10 percent level, the hypothesis of equal variance?.05 F 9,11.05 F.05,9,11 =2.90 Note that I changed  so we could look up two tail. F.95,9,11 =1/F.95,11,9 =1/3.2 = < F=.5 < 2.90; therefore, we do not have enough evidence to reject Ho. Variances are equal.