Question

The efficiency for a steel specimen immersed in a phosphating tank is the weight of the phosphate coating divided by the metal loss (both in mg/ft2). An article gave the accompanying data on tank temperature (x) and efficiency ratio (y). Temp. 173 175 176 177 177 178 179 180 Ratio 0.74 1.21 1.36 1.09 1.09 1.18 1.06 1.74 Temp. 183 183 183 183 183 134 184 185 Ratio 1.53 1.70 1.63 2.23 2.23 0.86 1.39 0.82 Temp. 185 185 185 187 187 188 189 191 Ratio 1.87 2.02 2.76 1.43 2.60 2.98 1.85 3.00 12 USE SALT (a) Determine the equation of the estimated regression line. (Round all numerical values to four decimal places.) YE (b) Calculate a point estimate for true average efficiency ratio when tank temperature is 185. (Round your answer to four decimal places.) (c) Calculate the values of the residuals from the least squares line for the four observations for which temperature is 185. (Round your answers to two decimal places.) (185, 0.82) (185, 1.87) (185, 2.02) (185, 2.76)

Answer 1

n2x2-(2x) il y = 6 ) These pesiduals do not all have the same sign because in the cases of the first two pains of observations, the observed efficfeney the last two paine of observations, the observed efficieney Ans: a) By the given data it Temp. _X and Ratio=Y Hien we have x = 4380, 2 Y= 40.37 , IXY = 7415.98 , Ix?=799854,7424 Jateneept : a = IT IX²-[XEXY 40.378799954-1380 x 74 15.98 n2x2 - (2x)? 24 *799854 - 43802 -15. 86936 slope: b= nIXY-IXIY 24% 8415.98-4380 X 40.37 000 000 0.09614 24x799854 -(4380) Hince the eanation of the estimated regrassion line is y = atba -15.86367 0.096018 Tid y = -15.8636+0.0961X when 185, y = poste -15.8636 +0.0961 X185 1.9149 apoint estimate for tove average efficiency natio when tank temparature is 185 is 1.9149 from (6) } = 1.9149 le = y; - The values of the residuals from the 0.82 1.9149 -1.0949 least squares line for the four observations 1.87 1-9149 -0.0449 for which temparature is 185 ane. 12.02 11.9149 0.1051 (185, 0-82) = 6:10 | 2.76 1.9149 0.845) (185, 1-87) -0.04 (185,2.02)=0:11 gott 0.85 Haxy not all have the same sign smaller than the predicted value falte cases of larger than the predicted valve. yig (185, 2.76) webee natios natio were

d ) By hie above table we have. ps η ΣκΎ -ΣΧΣΥ vpms Ix2 - (2 x)2JC 272-(27)2] 24X7415.98 4380x90.37 (24 x 799851 - 43802) (24x78.2615-40.372) 0, 671 22 = On 450 Hence, the proportion of she obsepred vaniation in efficiency natio Can be attributed to the simple linear relation ship between the two raniableis is 0.450 lo