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: 277 278 287 271 283 271 279 275 263 267 Brand 2: 262 248 260 265 273 281 271 270 263 268 Is there evidence to show that there is a difference in the mean overall distance of brands?

Answer:1) Since the calculated value of t = -0.241640 does not fall in the critical region so we accept H0 and conclude that there is not enough evidence to show the difference in the mean overall distance of brands.2) The 95% CI  is [-18.587; 1.01242]Step-by-step explanation:The given data is Brand 1     257      276     260   262   287   271    260  265   283     271Brand 2     273     281      279    275   271    270   263   267   263    268Difference d -16   -5       -19       -13     -16    - 01       -3     -2      20     3 ∑ -98d²               256     25     361      169    256     1     9      4     400      9 ∑14901) Let the hypotheses be H0: ud= 0 against the claim Ha: ud ≠02) The degrees of freedom = n-1= 10-1= 93) The significance level is 0.05 4) The test statistic is t= d`/sd/√n5) The critical region is ║t║≤ t (0.025,9) = ±2.262Calculations6) d`= ∑di/n= -98/10= -9.8Sd²= ∑(di-d`)²/n-1 = 1/n-1 [∑di²- (∑di)²n]= 1/9[1490-(-9.8)²/10] =1/9 [1490-9.604]= 164.4884Sd= 12.825Therefore t= d`/ sd/√nt= -9.8/ 12.825/√10t= -0.764132/3.16227= -0.2416407) Conclusion: 1) Since the calculated value of t = -0.241640 does not fall in the critical region so we accept H0 and conclude that there is not enough evidence to show the difference in the mean overall distance of brands.2) The confidence interval for the difference of two samples can be calculated by d ` ± td sd/√nPutting the values-9.8 ±2.262* 12.825/√10[-18.587; 1.01242]