As part of quality-control program for a catalysismanufacturing line, the raw ma

Question

As part of quality-control program for a catalysismanufacturing line, the raw materials (alumina and a binder) aretested for purity. The process requireds that the purity ofthe purity of the Alumina be greater than 85 percent. Arandomsample from a recent shipment of alumina yielded the followingresults (in percent). 93.2; 87.0; 92.1; 90.1; 87.3; 93.6 A hypothesis test will be done to determine wheter or not toaccept the shipment. You may assume individual percentages ofpurity follow a normal distribution. a) State the appropriate null and alternativehypotheses. b) Conduct the hypothesis test at the =0.05 level, stateyour conclusion and wheter or not the shipment should be acceptedbased on this data. c) Compute a 90% confidence on the standard deviation of thetrue purity percent of alumina. As part of quality-control program for a catalysismanufacturing line, the raw materials (alumina and a binder) aretested for purity. The process requireds that the purity ofthe purity of the Alumina be greater than 85 percent. Arandomsample from a recent shipment of alumina yielded the followingresults (in percent). 93.2; 87.0; 92.1; 90.1; 87.3; 93.6 A hypothesis test will be done to determine wheter or not toaccept the shipment. You may assume individual percentages ofpurity follow a normal distribution. a) State the appropriate null and alternativehypotheses. b) Conduct the hypothesis test at the =0.05 level, stateyour conclusion and wheter or not the shipment should be acceptedbased on this data. c) Compute a 90% confidence on the standard deviation of thetrue purity percent of alumina.

Explanation / Answer

H0: p = 0.85 H1: p 0.85 ps = 0.9055 t = 0.9055 - 0.85 /[(0.85)(1-0.85)/6] = 0.381 = 0.05 df = 5 t = 2.015 t < 2.015(t = 0.318 do ot fall inside the rejection region) Decision: Do not reject H0 Conclusion: There are not enough statistical evidence to prove thatpurity of the Alumina be greater than 85 percent. c) Range = 0.936 - 0.87 = 0.066 Range = 2t0.1 /n 0.066 = 2 *1.475884 /6 standard deviation = 0.055

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