Exercises for Section 6.1

Transcription

1 Exercie for Section The ph of an acid olution ued to etch aluminum varie omewhat from batch to batch. In a ample of 50 batche, the mean ph wa 2.6, with a tandard deviation of 0.3. Let μ repreent the mean ph for batche of thi olution. a. Find the P-value for teting H 0 : μ 2.5 veru H 1 : μ > 2.5. b. Either the mean ph i greater than 2.5 mm, or the ample i in the mot extreme % of it ditribution. n = 50; X = 2.6; = 0.3 a) Z = X μ 0 = = P(Z>2.3570) = P-value = b) Either the mean ph i greater than 2.5 mm, or the ample i in the mot extreme % of it ditribution. 4. In a proce that manufacture tungten-coated ilicon wafer, the target reitance for a wafer i 85 m. In a imple random ample of 50 wafer, the ample mean reitance wa 84.8 m, and the tandard deviation wa m. Let μ repreent the mean reitance of the wafer manufactured by thi proce. A quality engineer tet H 0 : μ = 85 veru H 1 : μ 85. a. Find the P-value. b. Do you believe it i plauible that the mean i on target, or are you convinced that the mean i not on target? Explain your reaoning. n = 50; X = 84.8; = a) Z = X μ 0 = = P(Z< ) = P-value = b) The mean i NOT on target ince the p-value i extremely mall. 8. Laer can provide highly accurate meaurement of mall movement. To determine the accuracy of uch a laer, it wa ued to take 100 meaurement of a known quantity. The ample mean error wa 25 μm with a tandard deviation of 60 μm. The laer i properly calibrated if the mean error i μ = 0. A tet i made of H 0 : μ = 0 veru H 1 : μ 0. a. Find the P-value. b. Do you believe it i plauible that the laer i properly calibrated, or are you convinced that it i out of calibration? Explain your reaoning. n = 100; X = 25; = 60 a) Z = X μ 0 = = P(Z>4.1667) = P-value = b) The laer i out of calibration ince the p-value i extremely mall. Exercie for Section George performed a hypothei tet. Lui checked George work by redoing the calculation. Both George and Lui agree that the reult wa tatitically ignificant at the 5% level, but they got different P-value. George got a P-value of, and Lui got a P-value of a. I it poible that George work i correct? Explain. b. I it poible that Lui work i correct? Explain. a) No, if p-value i, the reult i NOT tatitically ignificant at the 5% level. b) Ye, if p-value i 0.02, the reult i tatitically ignificant at the 5% level. 12. A machine that fill cereal boxe i uppoed to be calibrated o that the mean fill weight i 12 oz. Let μ denote the true mean fill weight. Aume that in a tet of the hypothee H 0 : μ = 12 veru H 1 : μ 12, the P-va1ue i a. Should H 0 be rejected on the bai of thi tet? Explain. b. Can you conclude that the machine i calibrated to provide a mean fill weight of 12 oz.? Explain. a) No, H 0 can NOT be rejected ince p-value i not mall. b) No, we can NOT conclude that the machine i calibrated. 18. A hipment of fiber i not acceptable if the mean breaking trength of the fiber i le than 50 N. A large ample of fiber from thi hipment wa teted, and a 98% lower confidence bound for the mean breaking trength wa computed to be 50.1 N. Someone ugget uing thee data to tet the hypothee H 0 : μ 50 veru H 1 : μ > 50. a. I it poible to determine from the confidence bound whether P < 0.01? Explain.

2 b. I it poible to determine from the confidence bound whether P < 0.05? Explain. a) No, we only know p-value < 0.02, but Not ure if p-value < b) Ye, we know p-value < 0.02, therefore p-value < Exercie for Section A random ample of 80 bolt i ampled from a day production, and 4 of them are found to have diameter below pecification. It i claimed that the proportion of defective bolt among thoe manufactured that day i le than I it appropriate to ue the method of thi ection to determine whether we can reject thi claim? If o, tate the appropriate null and alternate hypothee and compute the P-value. If not, explain why not. No. n = 80; p 0 = 0.10; np 0 = 8.0 which i le than A grinding machine will be qualified for a particular tak if it can be hown to produce le than 8% defective part. In a random ample of 300 part, 12 were defective. On the bai of thee data, can the machine be qualified? H 0 : p 0.08 v. H 0 : p < 0.08 n = 300; X = 12; p = = 0.04 p = p 0(1 p 0 ) = n Z = p p = p = P(Z< ) = P-value = Reject H 0. The machine be qualified. Exercie for Section A certain manufactured product i uppoed to contain 23% potaium by weight. A ample of 10 pecimen of thi product had an average percentage of 23.2 with a tandard deviation of 0.2. If the mean percentage i found to differ from 23, the manufacturing proce will be recalibrated. a. State the appropriate null and alternate hypothee. b. Compute the P-value. c. Should the proce be recalibrated? Explain. n = 10; X = 23.2; = 0.2 a) H 0 : μ = 23 v. H 0 : μ <> 23 b) t = X μ 0 = = P(t>3.1623) = P-value = c) Ye, the proce hould be recalibrated ince the p-value i very mall. 6. The thicknee of ix pad deigned for ue in aircraft engine mount were meaured. The reult, in mm, were 40.93, 41.11, 41.47, 40.96, 40.80, and a. Can you conclude that the mean thickne i greater than 41 mm? b. Can you conclude that the mean thickne i le than 41.4 mm? c. The target thickne i 41.2 mm. Can you conclude that the mean thickne differ from the target value? n = 6; X = ; = a) H 0 : μ 41 v. H 0 : μ > 41 t = X μ = = P(t>0.9464) = P-value = No, we can NOT conclude that the mean thickne i greater than 41 mm. b) H 0 : μ 41 v. H 0 : μ < 41.4 t = X μ = = P(t< ) = P-value = Ye, we can conclude that the mean thickne i le than 41.4 mm. c) H 0 : μ = 41.2 v. H 0 : μ <> 41.2 t = X μ = = P(t< ) = P-value = No, we can NOT conclude that the mean thickne differ from 41.2 mm. 8. The article Solid-Phae Chemical Fractionation of Selected Trace Metal in Some Northern Kentucky Soil (A. Karathanai and J. Pil, Soil and Sediment Contamination, 2005: ) report that in a ample of 26 oil pecimen taken in a region of northern Kentucky, the average concentration of

3 chromium (Cr) in mg/kg wa with a tandard deviation of a. Can you conclude that the mean concentration of Cr i greater than 20 mg/kg? b. Can you conclude that the mean concentration of Cr i le than 25 mg/kg? n = 26; X = 20.75; = 3.93 a) H 0 : μ 20 v. H 0 : μ > 20 t = X μ = = P(t>0.9731) = P-value = No, we can NOT conclude that the mean concentration of Cr i greater than 20 mg/kg. b) H 0 : μ 25 v. H 0 : μ < 25 t = X μ = = P(t< ) = P-value = Ye, we can conclude that that the mean concentration of Cr i le than 25 mg/kg. Exercie for Section The article referred to in Exercie 3 categorized firm by ize and percentage of full-operatingcapacity labor force currently employed. The number of firm in each of the categorie are preented in the following table. Percent of Full Operating- Capacity Labor Force Small Large Currently Employed >100% % % % % % % <70% Can you conclude that the ditribution of labor force currently employed differ between mall and large firm? Compute the relevant tet tatitic and P-value. Ob. Small Large Subtotal >100% % % % % % % <70% Subtotal Exp. Small Large Subtotal >100% % % % % % % <70% Subtotal k χ 2 = (O i E i ) 2 = i=1 E i ν = (8 1)(2 1) = 7 p value = P(χ 2 > ) =.2611 Fail to reject the H 0. We can not conclude that the ditribution of labor force currently employed differ between mall and large firm. 8. For the given table of oberved value: a. Contruct the correponding table of expected value. b. If appropriate, perform the chi-quare tet for the null hypothei that the row and column outcome are independent. If not appropriate, explain why. Oberved Value A B C Ob Subtotal A B C Subtotal Exp Subtotal A B C Subtotal

4 χ 2 -tet hould not be performed ince ome expected value are le than The article Determination of Carboxyhemoglobin Level and Health Effect on Officer Working at the Itanbul Bophoru Bridge (G. Kocaoy and H. Yalin, Journal of Environmental Science and Health, 2004: ) preent aement of health outcome of people working in an environment with high level of carbon monoxide (CO). Following are the number of worker reporting variou ymptom, categorized by work hift. The number were read from a graph. Shift Morning Evening Night Influenza Headache Weakne Shortne of Breath Can you conclude that the proportion of worker with the variou ymptom differ among the hift? Ob. Morning Evening Night Subtotal Influenza Headache Weakne Sh of Breath Subtotal Ob. Morning Evening Night Subtotal Influenza Headache Weakne Sh of Breath Subtotal k χ 2 = (O i E i ) 2 = i=1 E i ν = (4 1)(3 1) = 6 p value = P(χ 2 > ) = Reject the H 0. We can conclude that the proportion of worker with the variou ymptom differ among the hift. Exercie for Section A hypothei tet i to be performed, and the null hypothei will be rejected if P If H 0 i in fact true, what i the maximum probability that it will be rejected? The max. probability i the level of ignificance, which i A watewater treatment program i deigned to produce treated water with a ph of 7. Let μ repreent the mean ph of water treated by thi proce. The ph of 60 water pecimen will be meaured, and a tet of the hypothee H 0 : μ = 7 veru H 1 : μ 7 will be made. Aume it i known from previou experiment that the tandard deviation of the ph of water pecimen i approximately. a. If the tet i made at the 5% level, what i the rejection region? b. If the ample mean ph i 6.87, will H 0 be rejected at the 10% level? c. If the ample mean ph i 6.87, will H 0 be rejected at the 1% level? d. If the value 7.2 i a critical point, what i the level of the tet? n = 60; = a) α = 0.05 C. V. = μ 0 ± Zα 2 = 7 ± = (6.8734, ) The rejection region i >7.1265, and < b) α = 0.10 C. V. = μ 0 ± Zα 2 = 7 ± = (6.8938, ) H 0 will be rejected. c) α = 0.01 C. V. = μ 0 ± Zα = 7 ± = (6.8337, ) H 0 will not be rejected. d) 7.2 = μ 0 ± Zα = 7 ± 2 Zα 2 Zα = α = 2( ) = Exercie for Section A proce that manufacture gla heet i uppoed to be calibrated o that the mean thickne of the heet i more than 4 mm. The tandard deviation of the heet thicknee i known to be well approximated by = mm. Thicknee of each heet in a ample of heet will be meaured, and a tet of the hypothei H 0 : μ 4 veru H 1 : μ > 4 will be performed. Aume that, in fact, the true mean thickne i 4.04 mm.

5 a. If 100 heet are ampled, what i the power of a tet made at the 5% level? b. How many heet mut be ampled o that a 5% level tet ha power 0.95? c. If 100 heet are ampled, at what level mut the tet be made o that the power i 0.90? d. If 100 heet are ampled, and the rejection region i X 4.02, what i the power ofthe tet? a) α = 0.05; n = 100; = C. V. = μ 0 + Z α = = If μ = 4.04, β = Prob(X < ) = Prob(Z < ) = Power = 1 β = b) α = 0.05; n =? ; = C. V. = μ 0 + Z α = If μ = 4.04, Power = 1 β = 0.95; β = 0.05 Z β = C. V. = μ + Z β = n = 271 c) α =? ; n = 100; = C. V. = μ 0 + Z α = 4 + Z α 100 If μ = 4.04, Power = 1 β = 0.90; β = 0.10 Z β = C. V. = μ + Z β = = Z α = α = Prob(Z > ) = d) If μ = 4.04, β = Prob(X < 4.02) = Prob(Z < 1.00) = Power = 1 β = Water quality in a large etuary i being monitored in order to meaure the PCB concentration (in part per billion). a. If the population mean i 1.6 ppb and the population tandard deviation i 0.33 ppb, what i the probability that the null hypothei H 0 : μ 1.50 i rejected at the 5% level, if the ample ize i 80? b. If the population mean i 1.6 ppb and the population tandard deviation i 0.33 ppb, what ample ize i needed o that the probability i 0.99 that H 0 : μ 1.50 i rejected at the 5% level? a) α = 0.05; n = 80; = C. V. = μ 0 + Z α = = If μ = 1.6, β = Prob(X < ) = Prob(Z < ) = Power = 1 β = b) α = 0.05; n =? ; = C. V. = μ 0 + Z α = If μ = 1.6, Power = 1 β = 0.99; β = 0.01 Z β = C. V. = μ + Z β = n = 172 Exercie for Section Five different variation of a bolt-making proce are run to ee if any of them can increae the mean breaking trength of the bolt over that of the current proce. The P-value are 0.13, 0.34, 0.03, 0.28, and Of the following choice, which i the bet thing to do next? i. Implement the proce whoe P-value wa 0.03, ince it performed the bet. ii. Since none of the procee had Bonferroniadjuted P-value le than 0.05, we hould tick with the current proce. iii. Rerun the proce whoe P-value wa 0.03 to ee if it remain mall in the abence of multiple teting. iv. Rerun all the five variation again, to ee if any of them produce a mall P-value the econd time around. 4. Five new paint additive have been teted to ee if any of them can reduce the mean drying time from the current value of 12 minute. Ten pecimen have been painted with each of the new type of

6 paint, and the drying time (in minute) have been meaured. The reult are a follow: Additive A B C D E For each additive, perform a hypothei tet of the null hypothei H 0 : μ 12 againt the alternate H 1 : μ < 12. You may aume that each population i approximately normal. a. What are the P-value for the five tet? b. On the bai of the reult, which of the three following concluion eem mot appropriate? Explain your anwer. i. At leat one of the new additive reult in an improvement. ii. None of the new additive reult in an improvement. iii. Some of the new additive may reult in improvement, but the evidence i inconcluive. A B C D E X n μ Tet < < < < < t-tat p B-adj P