Question

In a test of braking​ performance, a tire manufacturer measured the stopping distance for one of its tire models. On a test​ track, a car made repeated stops from 60 miles per hour. The test was run on both dry and wet​ pavement, with results as shown below. Stopping Distance​ (ft) Wet 209 187 214 203 198 197 196 182 192 224 Dry 153 151 142 143 134 154 127 141 140 128 ​a) Construct a​ 95% confidence interval for the mean dry pavement stopping distance.

The confidence interval is?

Answer 1

given that the data ģ aloly 198/197/196 182 het 209 (187/214 1203 Poy 153 151 142 143 1134 115411271141 For wet COUNTN=10 Som Ex= 2002 mean, y = 2002 10 200.2 ple = { inte) 2 S, = Standard deviation Het EC200 200.2)+ ----+(224 -2002) Si= 160.8444 s = 12.68244 Now fol Dry COUNT N=10 Sum Ex=1413 mean X₂ = 1413 1413 X = 1413

Standard deviation Š Ce;-) E = (153–141.3) +----+C128-141.372 3) +-- S2 to-1 S₂ = V 92.45555 S, = 9.615381 Confidence Intaval find 95.1. Now, we need to for the mean ve CI - Coto-17)+ tua, se 10 10 = (-200.2–141.3 ) + too,10-1X/12,68244 9.61532 58.9 2.262*( 5603286) 70.28432 ,47,51567 & Opper limit = 70.28432 lower limit = 47.515 67 . (47.575, 70,084 ) pavemer (or) for for the mean dry the 951. Confidence autewal Stopping distance & # 27 x 9.61538) 141.34 2005 * Vio = 141.3 1.96* 9.61538) (135.340,147,299