Question

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The overall distance traveled by a golf ball is tested by hitting the ball with Iron Byron, a mechanical golfer with a swing that is said to emulate the legendary champion, Byron Nelson. Ten randomly selected balls of two brands are tested and the overall distance measured. Assume that the variances are equal. The data follow: Brand 1: 288 285 287 271 283 271 279 275 263 267 Brand 2: 261 246 260 265 273 281 271 270 263 268 (a) Is there evidence to show that there is a difference in the mean overall distance of brands? Use a = 0.05. (b) Calculate a 95% two-sided confidence interval on the difference in mean overall distance. Round your answer to one decimal place (e.g. 98.7). 3.Mi - M2 SL

Answer 1

for brand I n-10 Ex = 2769 medio, sé, = 276.9 Standard deviation, si (7:23)2 W 8.7990 For brand 2 me =10 Ex: 2658 mean, ry= 265.8 Standard deviation, S₂ = 9.3666 Hypothesis test Ho: There is no difference in the mean overall distance between two brand Ha : There is difference in the mean overall distance between two beremos. NOO Sp2 (mini) si? + (2-1) mitn2-2 = (10-1)x 8.7998?+ (10-1)x9.36662 10 +10-2 82.5831

Test statistic t=ło-zła Sý limitlina) - 276.9. 265.8 92.5831 (0+110) - 2.7313 = 2.731 Degree of freedom dfan, t02-2 = 10 +10-2 : 18 م له عام p-value with test statistic to 2:73) and dfa18 is p-value = 0.0137 SO Since P-value is less than significance level, a=0.os we reject null hypothesis. Therefore we have sufficient evidence to conclude that there is difference in the overall distance of brande. mean

b) Coritical value eft with significance level, 2=0.05 and df=18 is tdie,df=t0.0512, 18 = 2.1009 95% confidence interval CI = ( 51-az) I tdie,df * Sp² / Ynitline) = (276.9- 265.8) + 2.1009* 82.5831 ( 10+110) = 12:5618, 19.6382) -|(2.6, 19:6)