- Submitted By: msval78
- Date Submitted: 02/27/2014 1:35 PM
- Category: Miscellaneous
- Words: 411
- Page: 2

Statistics for Managers

Par, Inc.

At this time the population standard deviations is unknown, but the two samples are independent. This will mean that I would use a two-sample t-test with independent samples. The sample standard deviations for the new and current balls are 8.75 and 9.90. These values are close, again meaning that the two-sample t-test with independent samples will be used to determine the outcome.

(40-1) (8.75²) + (40-1) (9.90²) = 2985.84 + 3822.39 = 6808.23 / 78 = 87.285

√87.285 = 9.343

t = 270.28 – 267.50 = 2.78 / 9.343 ( √(2/40) ) = 0.224 = 9.343 * 0.224 = 2.09

/ 2.09 = 1.33 The p value is 0.094

The null hypothesis will not be rejected. There is also not enough evidence to indicate that the new ball travels further than the old ball.

Current ball sample averages = 270.28

Current ball variance = 76.6147436

Current ball standard error = 1.38396842

Current ball standard deviation = 8.75

Old ball sample averages = 267.50

Old ball variance = 97.9487179

Old ball standard error = 1.56483799

Old ball standard deviation = 9.90

Current = 95% Confidence level is 54.08, 324.50

New = 95% Confidence level is 53.93, 323.59

Confidence level of the two populations = -0.69, 6.25

I do see a need for a larger sample size and more testing with the golf balls. The data that we have from the samples provided are too similar and you cannot come up with a conclusive answer as to whether the newer golf balls are better than the current ones.

Air Force Training Program

Current Proposed

Mean 75.06557 75.42623

Median 76 76

Mode 76 76

Standard Error 0.0505094 0.32091

Standard Deviation 3.944907...