Question

Par Inc., is a major manufacturer of golfequipment. Management believes that Par’s market share could be increased with the introduction of a new ball designed to fly straighter and to be more durable. Therefore, the research group at Par has been investigating a new golf ball coating designed to achieve this goal. The tests with the new ball have been promising. One of the researchers voiced a concern about the potential for the new coating to have a negative effect on driving distances. Par would like the new ball to offer driving distances comparable to those of the current model golf ball. To compare the driving distances for the two balls, 70 balls of the new and 70 of the current models were subjected to distance tests. The testing was performed with a mechanical hitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the two models. The results of the tests, with distances measure to the nearest yard, have been provided to you. Managerial Report: 1) Provide a table of descriptive statistical summaries of the data for each model and discuss any initial observations. 2) Provide 95% confidence intervals for the two types of golf balls. 3) Formulate and present the rationale for a hypothesis test that Par could use to address the researchers concern. Use a 0.05 level of significance. 4) Analyze the data to provide a conclusion for your test. Provide the p-value for your test and discuss your findings. 5) Do you see a need for larger sample sizes and more testing with the golf balls? Discuss.

Current New
263 292
277 291
287 280
254 328
284 274
287 281
290 253
296 297
285 251
301 281
268 256
275 255
267 267
287 290
265 288
287 274
283 259
271 262
304 269
273 255
277 275
270 277
281 250
267 280
304 291
289 272
298 271
258 291
292 294
280 273
280 282
268 294
261 268
279 255
271 285
264 257
256 259
268 288
288 260
277 281
291 276
279 305
297 289
235 250
276 278
281 247
316 277
275 288
290 252
275 271
284 279
269 274
283 281
276 261
289 281
273 269
268 298
276 295
291 242
274 310
271 296
271 274
272 277
312 261
279 256
308 268
291 287
280 282
284 250
258 288

Answer 1

We use Minitab to solve this question-
MTB > Describe 'Current' 'New';
SUBC>   Mean;
SUBC>   SEMean;
SUBC>   Variance;
SUBC>   QOne;
SUBC>   Median;
SUBC>   QThree;
SUBC>   Minimum;
SUBC>   Maximum;
SUBC>   N;
SUBC>   NMissing.

Descriptive Statistics: Current, New

Variable   N N*    Mean SE Mean Variance    Minimum      Q1 Median      Q3 Maximum
Current   70   0 279.37     1.70    202.12 235.00    270.75 279.00    288.25   316.00
New    70   0 275.26     2.01    281.64 242.00    260.75 276.50    288.00   328.00

MTB > OneT 'Current' 'New';
SUBC>   Confidence 95.0;
SUBC>   Alternative 0.

One-Sample T: Current, New

Variable   N    Mean StDev SE Mean       95% CI
Current   70 279.37 14.22     1.70 (275.98, 282.76)
New       70 275.26 16.78     2.01 (271.26, 279.26)

Two-sample T for Current vs New

          N   Mean StDev SE Mean
Current 70 279.4   14.2      1.7
New      70 275.3   16.8      2.0

Difference = μ (Current) - μ (New)
Estimate for difference: 4.11
95% CI for difference: (-1.08, 9.31)
T-Test of difference = 0 (vs ≠): T-Value = 1.57 P-Value = 0.120 DF = 138
Both use Pooled StDev = 15.5526

______________________________________________________________________________

1)
Descriptive Statistics: Current, New

Variable   N N*    Mean SE Mean Variance    Minimum      Q1 Median      Q3 Maximum
Current   70   0 279.37     1.70    202.12 235.00    270.75 279.00    288.25   316.00
New    70   0 275.26     2.01    281.64 242.00    260.75 276.50    288.00   328.00

2)
95 % Confidence Interval for two golf balls for the variable Current and New is (275.98, 282.76) and (271.26, 279.26) respectively.
3)
Two-sample T for Current vs New

          N   Mean StDev SE Mean
Current 70 279.4   14.2      1.7
New      70 275.3   16.8      2.0

Difference = μ (Current) - μ (New)
Estimate for difference: 4.11
95% CI for difference: (-1.08, 9.31)
T-Test of difference = 0 (vs ≠): T-Value = 1.57 P-Value = 0.120 DF = 138
Both use Pooled StDev = 15.5526
4)
p-value = 0.120 > 0.05 , Do not reject Ho

5)
Need for larger sample sizes and more testing with the golf balls because give sample information doesnot provide enough evidence that par could use to address the researchers concern.