The overall distance traveled by a golf ball is tested by hitting the ball with

Question

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: 285 280 287 27 283 271 279 275 263 267 Brand 2: 269 236 260 265 273 281 271 270 263 268 a) Is there evidence to show that there is a difference in the mean overall distance of brands? Use alpha = 0.05. a 95% two-sided confidence interval on the difference in mean overall distance. (e.g. 98.7) lessthanorequalto mu_1 - mu_2 lessthanorequalto

Explanation / Answer

(a)

Data:        

n1 = 10       

n2 = 10       

x1-bar = 276.1       

x2-bar = 265.6       

s1 = 8       

s2 = 11.89       

Hypotheses:        

Ho: 1 = 2        

Ha: 1 2        

Decision Rule:        

= 0.05       

Degrees of freedom = 10 + 10 - 2 = 18      

Lower Critical t- score = -2.100922037       

Upper Critical t- score = 2.100922037       

Reject Ho if |t| > 2.100922037       

Test Statistic:        

Pooled SD, s = [{(n1 - 1) s1^2 + (n2 - 1) s2^2} / (n1 + n2 - 2)] =   (((10 - 1) * 8^2 + (10 - 1) * 11.89^2)/(10 + 10 -2)) =     10.13

SE = s * {(1 /n1) + (1 /n2)} = 10.1334125545149 * ((1/10) + (1/10)) = 4.531799863      

t = (x1-bar -x2-bar)/SE = 2.316960218       

p- value = 0.032492035       

Decision (in terms of the hypotheses):        

Since 2.316960218 > 2.100922037 we reject Ho and accept Ha    

Conclusion (in terms of the problem):    

There is evidence of a significant difference between the means of the two brands

(b)

n1 = 10

n2 = 10

x1-bar = 276.1

x2-bar = 265.6

s1 = 8

s2 = 11.89

% = 95

Degrees of freedom = n1 + n2 - 2 = 10 + 10 -2 = 18

Pooled s = (((n1 - 1) * s1^2 + (n2 - 1) * s2^2)/DOF) = (((10 - 1) * 8^2 + ( 10 - 1) * 11.89^2)/(10 + 10 -2)) = 10.13341255

SE = Pooled s * ((1/n1) + (1/n2)) = 10.1334125545149 * ((1/10) + (1/10)) = 4.531799863

t- score = 2.100922037

Width of the confidence interval = t * SE = 2.10092203686118 * 4.53179986318902 = 9.520958199

Lower Limit of the confidence interval = (x1-bar - x2-bar) - width = 10.5 - 9.52095819921829 = 1.0

Upper Limit of the confidence interval = (x1-bar - x2-bar) + width = 10.5 + 9.52095819921829 = 20.0

The 95% confidence interval is [1.0, 20.0]   

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