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Chap 8 Chapter 8 problems 8 #3 Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40-amp fuses wants to make sure that the true average amperage at which its fuses burn out is indeed 40. If the average amperage is lower than 40, purchasers will complain because the fuses will have to be replaced too frequently, whereas if the average exceeds 40, the manufacturer might be liable for damage to an electrical system due to fuse malfunction. After obtaining data from a sample of fuses, what null and alternative hypotheses would be of interest to the manufacturer? Answer: Let denote the average amperage in the population of all such fuses, then we are interested in the two hypotheses: H 0 : 40 H a : 40 (null hypothesis) (alternative hypothesis) 8 #5 A new design for the braking system on a certain type of car has been proposed. For the current system, the true average braking distance at 40 mph under specified conditions is known to be 120 ft. It is proposed that the new design be implemented only if sample data strongly indicates a reduction in true average braking distance for the new design. State the relevant hypotheses, and describe the type I and type II errors in the context of this situation. Answer: Type I error is the error of reject the H o when H o is actually true. Type II error consists of not rejecting H o when H o is false. The hypotheses are: H 0 : 120 vs H a : 120 A type I error is to conclude that the new system really does reduce the average braking distance (rejecting H o ) when in fact it doesn’t ( H o is true). A type II error is to conclude that the new system does not achieve a reduction in average braking distance (not rejecting H o ) when in fact it does ( H o is false). Chap8 #9 For which of the given P-values will the null hypothesis be rejected when using a test with a significance level of .05? a. .001 b. .021 c. .078 d. .047 e. .156 Answer: Since options c and e are larger than the .05 significance level, so H o should be rejected for these two P-values. 17. Use table Ⅵ (page544) (a) Df:8, P(t>2.0)=> 0.040 (b) Df:11, P(t<-2.4)= P(t>2.4)=> 0.018 (c) Df:15, 2*P(t>1.6) = 0.13 (d) Df:19, P(t>-0.4) =P(t<0.4)=1-P(t>0.4) = 1-0.347 = 0.653 (e) Df:5, P(t>5.0)= approximately 0 (f) Df:40, 2*P(t>4.8) = approximately 0 19. (a) Let μ denote the true average writing lifetime. The wording in this exercise indicates that the investigators believe, a priori, that μ can’t be less than 10 (i.e. μ ≥ 10), so the relevant hypotheses are H0: μ = 10 versus H1 : μ < 10. (b) The degrees of freedom are d.f.=n-1=18-1=17, so the P-value =P(t<-2.3) = P(t>2.3) =0.017. Since P-value = 0.017<α =0.05, we should reject H0 and conclude that the design specification has not been satisfied. (c) For t=-1.8, P-Value = P(t<-1.8) = P(t>1.8) = 0.045, which exceeds α =0.01. There fore, in this case H0 would not be rejected. That is, there is not sufficient evidence to conclude that the design specification is not satisfied. (d) For t=-3.6, P-Value = P(t<-3.6) = P(t>3.6) = 0.001. This P-value is smaller than either α =0.01 or α =0.05, so in either case H0 would be rejected. In fact, H0 would be rejected for any value of α that exceeds 0.001. For such values of α, we would conclude that the design specification has not been satisfied. Chap 11. 9. 3 Let β denote the true average change in runoff(for each 1 m increase in rainfall). To test the hypotheses H0: β = 0 versus H1 : β ≠0, the calculated t statistics is t=b/sb = 0.82697/0.03652 = 22.64 which(from the printout) has an associated P-value of P=0.000. Therefore, since the P-value is so small, H0 is rejected and we conclude that there is a useful linear relationship between runoff and rainfall. A confidence interval for β is based on n-2 =15-2=13 degrees of freedom. The t critical value for, say, 95% confidence is 2.160 (from Table Ⅳ), so the interval estimates : b ±(t critical)* sb= 0.82697 ±(2.160)* (0.03652) = 0.82697±0.07888 = [0.748,0.906]. Therefore, we can be confident that the true average change in runoff, 3 3 3 for each 1 m increase in rainfall, is somewhere between 0.748 m and 0.906 m . 11. let denote true average change in flux (for each 1 cm -1 increase in inverse foil thickness). To test the gypotheses H0 : = 0 versus Ha : 0 , the calculateed t statistic is t = b/sb = .26042/.01502 = 17.34 which ( from the printout) has an associated P- value of P = 0.000. Therefore, we reject H0 and conclude that there is a useful relationship between the x and y variables. 17. How does Lateral acceleration-side forces experienced in turns that are largely under driver control-affect nausea as perceived by bus passengers? The article “Motion Sickness in Public Road Transport: The Effect of Driver, Route, and Vehicle” reported data on x= motion sickness dose (calculated in accordance with a British standard for evaluating similar motion at sea) and y = reported nausea (%).Relevant summary quantities are n = 17 x i =221.1 y i =193 x i 2 =3056.69 x y =2759.6 i i y i 2 =2975 Values of dose in the sample ranged from 6.0 to 17.6 a. Assuming that the simple linear regression model is valid for relating these t wo variables (this supported by the raw data), calculate and interpret an esti mate of the slope parameter that conveys information about the precision a nd reliability of estimation. b = Sxy/Sxx = 1.378 .There is , on average, a 1.378% incre ase in reported nausea for ech unit increse in motion sickness dose b. Does it appear that there is a useful linear relationship between these two v ariables? 2 Sxx=3056.69-221.1 /17 =181.089 2 Se =(Syy –b Sxy )/(n-2)=(783.882 – 1.378 * 249.465)/15= 440.12/1 5 = 29.34 Se 2 Sb= 0.4025 Sxx t =b/Sb= 3.423 Yes . There is a useful relationship between two variables c. Would it be sensible to use the simple linear regression model as a basis fo r prediction % nausea when dose = 5.0? Explain your reasoning. It would be possible , but not advisable, since x = 5 is outside t he range of the x data. d. When MINITAB was used to fit the simple linear regression model to the ra w data, the observation(6.0,2.50)was flagged as possibly having a substantial impact on the fit .Eliminate this observation from the sample and recalculat e the estimate of part (a). Based on this, does the observation appear to be exerting an undue influence? n =16 x i =215.1 y i =190.5 x i 2 =3020.69 x y =2744.6 i i y i 2 =2968.75 Sxx = 128.939 Sxy = 183.566 b= Sxy/Sxx = 1.424 No undue influence. Ch11#27, #31(a,b) #37 Chapter 11 27. Let y = sales at a fast food outlet (1000’s of $) x1 = number of competing outlets within a 1-mile radius, x2 = population within a 1-mile radius (1000’s of people) x3 = an indicator variable that equals 1 if the outlet has a drive-up window and 0 otherwise. Suppose that the true regression model is y = 10.00 - 1.2x1 + 6.8x2 + 15.3x3 + e a. What is the mean value of sales when the number of competing outlets is 2, there are 8000 people within a 1-mile radius, and the outlet has a drive-up window? b. What is the mean value of sales for an outlet without a drive-up window that has 3 competing outlets and 5000 people within a 1-mile radius? Solution: a. For x1 = 2, x2 = 8 and x3 = 1, the average sales are y = 10.00 – 1.2(2) + 6.8(8) + 15.3(1) = 77.3 (ie. $ 77,300). b. For x1 = 3, x2 = 5 and x3 = 0, the average sales are y = 10.00 – 1.2(3) + 6.8(5) + 15.3(0) = 40.4 (ie, $40,400). 31. See question details in text a. Assuming that the sample size was n = 25, state and test the appropriate hypothesis to decide whether the fitted model specifies a useful linear relationship between the dependent variable and at least one of the four model predictors. b. Again using n = 25, calculate the value of the adjusted R2. c. Calculate a 99% confidence interval for true mean yarn tenacity when yarn count is 16.5, yarn contains 50% polyester, first nozzle pressure is 3, and second nozzle pressure is 5 if the estimated standard deviation of predicted tenacity under these circumstances is .350. Solution: a. The appropriate hypotheses are H 0 : 1 2 3 4 0 versus Ha: at least one of the i's is not zero. R2 / k .946 / 4 The test statistic is F 87.6 (1 R ) /( n (k 1)) (1 .946) /(25 (4 1)) 2 The test is based on df1 = 4, df2 = 20. From Table IX, the P-value associated with F = 6.59 is .001, so the P-value associated with 87.6 is .000. Therefore H0 can be rejected at any reasonable level of significance. We conclude that at least one of the four predictor variables appears to provide useful information about tenacity. b. The adjusted R2 value is n 1 SSResid n 1 24 1 1 [1 R 2 ] 1 [1 .946] .935 n (k 1) SSTo n (k 1) 20 c. The estimated average tenacity when x1 = 16.5, x2 = 50, x3 = 3, and x4 = 5 is: y 6.121 .082 x1 .113x2 .256 x3 .219 x4 10.091 ˆ For a 99% confidence interval based on 20 d.f., the t-critical value is 2.845. The desired interval is: 10.091 (2.845)(.350) = 10.091 .996, or, about [9.095, 11.087]. Therefore, when the four predictors are as specified in this problem, the true average tenacity is estimated to be between 9.095 and 11.087. 37. See question details in text. a. Construct a normal quantile plot of the standardized residuals to see whether it is plausible that the random deviations in the fitted model come from a normal distribution. b. Plot the standardized residuals against depth and against water content, and comment on the plots. Solutions: a. Plot Y = stresid, X= nquant and you should see a reasonably straight line, with the possible exception of the point in the upper right corner, (1.70991, 2.16543). It seems reasonable that the standardized residuals could have come from a normal distribution b. Neither plots of the standardized residuals versus x1 (depth) and x2 (water content) shows any obvious pattern (ie. non-random behavior), so we have no evidence in these plots that the model should be modified.

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Chap nausea

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