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 the data are normally distributed. The data follow:

Brand 1 257 276 260 262 287 271 260 265 283 271
Brand 2 273 281 279 275 271 270 263 267 263 268
1) Is there evidence to show that there is a difference in the mean overall distance of brands? Use alpha = 0.05.
2) Calculate a 95% two-sided confidence interval on the difference in mean overall distance.

Answer 1

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]

Explanation:

The given data is

Brand 1 257 276 260 262 287 271 260 265 283 271

Brand 2 273 281 279 275 271 270 263 267 263 268

Difference d -16 -5 -19 -13 -16 - 01 -3 -2 20 3 ∑ -98

d² 256 25 361 169 256 1 9 4 400 9 ∑1490

1) Let the hypotheses be

H0: ud= 0 against the claim Ha: ud ≠0

2) The degrees of freedom = n-1= 10-1= 9

3) The significance level is 0.05

4) The test statistic is

t= d`/sd/√n

5) The critical region is ║t║≤ t (0.025,9) = ±2.262

Calculations

6) d`= ∑di/n= -98/10= -9.8

Sd²= ∑(di-d`)²/n-1 = 1/n-1 [∑di²- (∑di)²n]

= 1/9[1490-(-9.8)²/10] =1/9 [1490-9.604]= 164.4884

Sd= 12.825

Therefore

t= d`/ sd/√n

t= -9.8/ 12.825/√10

t= -0.764132/3.16227= -0.241640

7) 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/√n

Putting the values

-9.8 ±2.262* 12.825/√10

[-18.587; 1.01242]