Question

BrAC BAC 0.7 0.5 1.13 0.22 0.82 Question 2. Prevention and detection of drinking and driving are major concerns of traffic Driver police. In the U.S. in 2012, nearly one-third of all traffic deaths involved 0.34 drivers with excessive blood alcohol concentrations (BAC). Investigative 0.26 authorities employ breath-alcohol analyzers, i.e. breathalyzers, to measure breath alcohol concentrations (BrAC) because the tests can be easily applied and do not require an attending physician to extract blood. Thresholds for 0.29 alcohol-impaired driving are based on blood alcohol concentration, so it is important to understand the relationship between breath and blood alcohol concentrations. Researchers in Germany measured both the blood and breath alcohol concentrations for 17 drivers that were stopped for traffic violations. The data are shown in the table and are also available in a separate file on Canvas. (Note: German authorities use different units of measurements than U.S. 0.64 authorities, so the numbers presented in the table are not equivalent to the 0.19 0.38 U.S. threshold of 0.08 BAC for drunk driving.) Use the data to develop a regression equation that will predict the value of BAC (blood alcohol) by the using value of BrAC (breath alcohol). Note: We predict Y by using X. Make sure you have X and Y correct in your calculations. You can lose a lot of points if X and Y are backward. 1 2 3 4 5 6 7 8 9 10 11 12 0.57 0.14 0.46 0.16 0.51 0.39 0.53 0.34 0.32 0.2 0.58 0.17 1.1 0.82 1.08 0.68 0.66 0.44 1.22 0.32 1.35 0.82 13 14 15 16 17 0.41 (a) What percent of the variability in blood alcohol is explained by the regression? (b) Fit a linear regression equation that can be used to predict BAC from BrAC. (You will have to provide the estimated intercept and the estimated slope.) (c) Use the estimated regression equation to predict BAC when BrAC = 0.28. (You will have to calculate this by hand. Please round to 3 decimals.)

Answer 1

sr.no.	BrAC (X)	BAC (Y)	X^2	Y^2	XY
1	0.34	0.7	0.1156	0.49	0.238
2	0.26	0.5	0.0676	0.25	0.13
3	0.57	1.13	0.3249	1.2769	0.6441
4	0.14	0.22	0.0196	0.0484	0.0308
5	0.46	0.82	0.2116	0.6724	0.3772
6	0.16	0.29	0.0256	0.0841	0.0464
7	0.51	1.1	0.2601	1.21	0.561
8	0.39	0.82	0.1521	0.6724	0.3198
9	0.53	1.08	0.2809	1.1664	0.5724
10	0.34	0.68	0.1156	0.4624	0.2312
11	0.32	0.66	0.1024	0.4356	0.2112
12	0.2	0.44	0.04	0.1936	0.088
13	0.58	1.22	0.3364	1.4884	0.7076
14	0.17	0.32	0.0289	0.1024	0.0544
15	0.64	1.35	0.4096	1.8225	0.864
16	0.41	0.82	0.1681	0.6724	0.3362
17	0.19	0.38	0.0361	0.1444	0.0722
total	6.21	12.53	2.6951	11.1923	5.4845

Solution ; a We have, EX= 6-21, Ey = 12.53, {X? 2.6951 Ey? - 11.1923; 5.4845. EXY = We have to find what percent of variability vis explained in blood alcohol by regression: R² SSR SST habagal where, SSR = b. Say no b - slope of equation. seg.is Y = atbx. Say Sax (EX) (27) Say EXT- 5.4845 (6:21) (12.53) 17 0.9074. son n (6.21) 2 2.6951

Date Sax 0.4266 say ba San 0.9074 0.4266 b 2:1269 SSR = b. Say 2.1269 x 0.9074 .: SSR 1.9298. SST 1923 - (12.53² 17 SST 1.9570 RE SSR SST 1.9298 1.9570 R? 0.9861 98.61% of variability in blood alcohol is explained.

Date b Regression eg. is. y = atbx. b = SX 5x3 i b = 2.1269. Now, a- j - by EJ 12.53 17 0.737 6.21 X = 0.3653 fb7 0.3 2.1269 x 0.3653 -0.0399 eg. is regression ☺ - 2.12692 0.0399

where, y-intercept a a -0.0399 slope = b = 2.1269 To predict BAR when BRAC -0.28 À 2.1269 x = 0.0399 - 2.1269 X 0.28 – 0.0399 I 0.556.