Question

5. A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130°F. A sample of 9 systems, when tested, yields a sample average activation temperature of 131.08°F. Assume that the distribution of activation times is normal with a population standard deviation of 1.5°F. a) Does the data contradict the claim at 0.01 level? Carry out a detailed test by writing the appro- priate hypotheses, rejection region and conclusion. [4] b) Find a 95% CI for then true average system-activation temperature. [3]

Answer 1

5) claim :- A manufacturer of sprinkler For fire protection in office systems used the dresage true that buildings claims is 130° F temperature system activation *) 으 here n = sample size = x 131.08 °F sample mean = - population standard deviation = 1.5°F

a ) Does the data contradict the claim 2 at oool level? here test can use we one sample 2 vs To Test . Hoi la Hoi H = 1300 F Hii U 130° F Test statistics rž -U)Jn z = ó 2= (131.08-130) * 59 2.16 = 1.5

*) critical value= 2212 zo 20일 +2.58 - (from table) Rejection Region Acceptanie Region Rejection Region 2.58 -2.58 *) Decision !- we reject Ho at Cl-d)* 100%. if or level of significance 27 Zala 1 < Zal2 we -2<-2212 here fail to reject Ho

evidence to *) conclusion There is l Insufficient conclude that the frue average system temperature is activation different than ( te 4 = 1300 F. 130°F () for true CI 95% system average activanon temperature,

=H E a (usle talk = dedy do 1-0.95 = 10.05 5002 11.967- ( foom tables sit # 9b-1) F 89.181 ) = N ; (or = (131.08 † (1-96* 0.5)) = (131.08 [ 0-98)

-C 130.1 132.06) +) | Interpretation]:- we can say 953. confident that true average system activation 130.1 to to 132:08 lies within is temperature