Solved Homework 4 - Statistics for Engineering | STAT 4706, Quizzes of Statistics

Download Solved Homework 4 - Statistics for Engineering | STAT 4706 and more Quizzes Statistics in PDF only on Docsity! STAT-4706 Solutions Homework # 4 Fall 2012 You must show all of your work for each problem in order to receive full credit. Minitab should be used when specified. HW is Due 11/01/2012. 1. The grades of a class of 9 students on a midterm report (x) and on the final examination (y) are as follows; x 77 50 71 72 81 94 96 99 67 y 82 66 78 34 47 85 99 99 68 a. Estimate the regression line ∑ i x i=707∑ i yi=658∑ i xi 2 =57,557∑ i xi y i=53,258n=9 b1= (9 ) (53,258 )−(707)(658) (9 ) (57,557 )−(707)2 =0.7771 b0= 658−(0.7771)(707) 9 =12.0656 Hence ŷ=12.0656+0.7771x b. Estimate the final examination grade of a student who received a grade of 85 on the midterm report. X = 85, ŷ=12.0656+0.7771 (85 )=78.12 c. Construct a 95% CI for β0 Sxx = 57,557- (707)2 9 = 2018.222 Syy=51,980− (658 )2 9 =3872.8889 Sxy = 53,258 - (707)(658) 9 = 1568.4444 b0 = 12.0623 b1=0.7771 σ̂ 2=S2= S yy−b1Sxy n−2 = 3872.8889−(0.7771)(1568.4444) 7 = 379.150 S = √379.15=¿ 19.472 t 0.025(7) = 2.365 b0± t α 2 (n−2) )√ (σ̂2)(∑ i x i 2 ) (n)(Sxx) = 12.0623 ± (2.365)√ (379.150)(57,557) (9)(2018.2222) = 12.0623 ± 81.975 Then the 95% CI is -69.91< β0 < 94.04 1 STAT-4706 Solutions Homework # 4 Fall 2012 d. Construct a 95% CI for β1 b1± t α 2 (n−2) )√ σ̂ 2 (S xx) 0.7771 ± (2.365)√ 379.1502018.2222 ; Then the 95% CI is -0.25< β1 < 1.80 2. Turbidity is a measure of the cloudiness of the water and is used to indicate water quality levels. Higher turbidity levels are usually associated with higher levels of disease-causing microbes. The turbidity units of measure are reported as formazin suspension units, or FAUs. Data were collected on the Rio Grande River during the late spring and summer months in order to study the relationship between temperature (x) and turbidity (y). Data are in the Excel file turbidity.xlsx. Use MINITAB to answer the following questions: a. Produce a scatter plot. Does it appear that a SLR will be a suitable model? 2726252423222120 300 250 200 150 100 50 0 Temperature(x) Tu rb id it y (y ) Scatterplot of Turbidity vs Temperature Most of the points seem to follow a line. It appears that a SLR will be a suitable model b. Estimate the linear regression line. The regression equation is Turbidity = - 511 + 26.3 Temperature ŷ=−511+26 .3 x ESTIMATE TABLE: Predictor Coef StDev T P Constant -510.7 228.2 -2.24 0.045 2 STAT-4706 Solutions Homework # 4 Fall 2012 h. Compute the sample correlation coefficient “r”. From (a) R-sqr = R2 = 40.6%, then r = √ .406 = 0.638. Based on this there is a significant correlation between temperature and turbidity. However, the residual plots indicate the model might not be valid and then these tests are invalid. Because the slope is significantly different from zero based on the conclusions from parts (d) and (e), the correlation coefficient is also significantly different from zero. 3. Regression methods were used to analyze the data from a study investigating the relationship between roadway surface temperature (x) and pavement deflection (y). Data is in the excel file deflection.xlsx. Use MINITAB to answer the following questions: a. Estimate the linear regression line. The regression equation is Deflect = 0.393 + 0.00333 Temp ŷ=0 .393+0 .00333 x ESTIMATE TABLE: Predictor Coef SE Coef T P Constant 0.39346 0.04258 9.24 0.000 Temp 0.0033285 0.0005815 5.72 0.000 S = 0.006473 R-Sq = 64.5% R-Sq(adj) = 62.6% Analysis of Variance Source DF SS MS F P Regression 1 0.0013727 0.0013727 32.76 0.000 Residual Error 18 0.0007542 0.0000419 Total 19 0.0021270 b. Interpret the values of b0 and b1 In this case b0 does not have a practical interpretation. When X = 0, y = 0.3935 b1 : For a unit increase on roadway surface temperature (x), the pavement deflection (y) will change by 0.0033. 5 STAT-4706 Solutions Homework # 4 Fall 2012 c. Use the ANOVA table to test for significance of the regression line. State your conclusions See the Minitab output in part (a). Based on the analysis of variance, we can reject the null hypothesis and conclude that the regression is significant because the p-value is 0.000 less than α = 0.05. d. Construct 95% CI on the slope. Interpret. From Estimate table on part (a): SE(b1) = 9.178, then the CI is: 1: 0.0033285  2.101(0.0005815); (0.00211, 0.00455) Zero is not included in the CI, so the slope is significantly different from zero. The conclusions from part (c) and (d) are the same. e. Perform model adequacy checks. Do you believe the model provides an adequate fit? 210-1-2 99 90 50 10 1 Standardized Residual P e rc e n t 0.650.640.630.62 2 1 0 -1 -2 Fitted Value S ta n d a rd iz e d R e si d u a l 1.51.00.50.0-0.5-1.0-1.5-2.0 4.8 3.6 2.4 1.2 0.0 Standardized Residual F re q u e n cy 2018161412108642 2 1 0 -1 -2 Observation Order S ta n d a rd iz e d R e si d u a l Normal Probability Plot Versus Fits Histogram Versus Order Residual Plots for Pavement Deflection(y) 6 STAT-4706 Solutions Homework # 4 Fall 2012 787674727068 2 1 0 -1 -2 Temp(x) S ta n d a rd iz e d R e si d u a l Residuals vs. Roadway Surface Temperature(x) (response is deflect(y)) Residual plots appear reasonable, so the model provides an adequate fit. f. Find the mean deflection given that the temperature is 74.0 degrees. The regression equation is ŷ=0 .393+0 .00333 x 0.63976 = 0.393 + 0.0033(74) g. Compute a 95% CI on this mean response. X́=73.185 Sxx = 123.905 MSE = σ̂ 2=S2=¿0.0000419 t 0.025 (18 )=2.101 0.63976 ± 2.101√0.0000419[ 120 + (74.0−73.185 ) 2 123.905 ] 0.63976 ± 2.101(0.001523) 0.63976 ± 0.0032; and a CI for the mean response at x = 74 is (0.6366, 0.6430) h. Compute a 95% PI on a future observation when temperature is equal to 74.0 degrees. 0.63976 ± 2.101√0.0000419[1+ 120 + (74.0−73.185 ) 2 123.905 ] 7 (b0+b1 x0 )±tα 2 , n−2√ σ̂2 [ 1n + (x0− x̄) 2 Sxx ] (b0+b1 x0 )±t α 2 , n−2√ σ̂2 [1+ 1n + ( x0− x̄ ) 2 S xx ]