A major manufacturer of golf equipment seeks to improve its market share by introducing a golf ball that is less likely to slice when hit. The company's research and development group has been experimenting with dimple patterns that promote straight flight, and have achieved some degree of success. The vice president of marketing, however, is worried about the effect that the new pattern might have on driving distance.

To investigate this issue, the research and development group did a test to compare the driving distances for the two balls: 40 balls of the new type, and 40 of the old. They used a mechanical hitting machine to control all factors other than the balls themselves. The resulting data of distance measured to the nearest yard for each of the 80 balls that were hit is provided in the Golf Ball Distance Test Excel file (copied and pasted below).

•What is the null hypothesis that the research and development group is testing?

**** I came up with my null hypothesis as H_0:μ_1- μ_2= 0 and H_a: μ_1-μ_2 ≠0

•Analyze the data in the Excel file.

■What is the appropriate statistical test for determining the null hypothesis?

The appropriate test for testing this hypothesis is comparing two population means by using independent samples: variances unknown. I don’t know if this is the correct test to use?

■What is the p-value?

Not sure how to determine the p-value.

■What should you tell the vice president of marketing?

** So fare what I have is x̅1 =270.275 and x̅2 =267.50. Standard Deviation σ1= 8.64 and σ2= 9.77 and sample size for both is n=40.

z=((x̅1 – x̅2) – 0)/(√(〖(σ)〗^2/40)+ 〖(σ)〗^(2 )/40) = ((270.275-267.50)-0)/(√((8.64)^2 )/40+(9.77)^(2 )/40)=2.775/2.9107=.953

The critical value rule is z>z_(α/2) or z<〖-z〗_(α/2) so α(.05)/2= .025×.953= .023

I think what this means is that we can reject the null hypothesis because .953 is greater than .023. I would tell the vice president of marketing that there is no difference in the new design and the effects on the performance of the new golf ball. If you could give me advice as to how I’m doing solving the problem so fare and help me to determine the p-value that would be great.

Distance Measured in Yards

Current New

264 277

261 269

267 263

272 266

258 262

283 251

258 262

266 289

259 286

270 264

263 274

264 266

284 262

263 271

260 260

283 281

255 250

272 263

266 278

268 264

270 272

287 259

289 264

280 280

272 274

275 281

265 276

260 269

278 268

275 262

281 283

274 250

273 253

263 260

275 270

267 263

279 261

274 255

276 263

262 279

To investigate this issue, the research and development group did a test to compare the driving distances for the two balls: 40 balls of the new type, and 40 of the old. They used a mechanical hitting machine to control all factors other than the balls themselves. The resulting data of distance measured to the nearest yard for each of the 80 balls that were hit is provided in the Golf Ball Distance Test Excel file (copied and pasted below).

•What is the null hypothesis that the research and development group is testing?

**** I came up with my null hypothesis as H_0:μ_1- μ_2= 0 and H_a: μ_1-μ_2 ≠0

•Analyze the data in the Excel file.

■What is the appropriate statistical test for determining the null hypothesis?

The appropriate test for testing this hypothesis is comparing two population means by using independent samples: variances unknown. I don’t know if this is the correct test to use?

■What is the p-value?

Not sure how to determine the p-value.

■What should you tell the vice president of marketing?

** So fare what I have is x̅1 =270.275 and x̅2 =267.50. Standard Deviation σ1= 8.64 and σ2= 9.77 and sample size for both is n=40.

z=((x̅1 – x̅2) – 0)/(√(〖(σ)〗^2/40)+ 〖(σ)〗^(2 )/40) = ((270.275-267.50)-0)/(√((8.64)^2 )/40+(9.77)^(2 )/40)=2.775/2.9107=.953

The critical value rule is z>z_(α/2) or z<〖-z〗_(α/2) so α(.05)/2= .025×.953= .023

I think what this means is that we can reject the null hypothesis because .953 is greater than .023. I would tell the vice president of marketing that there is no difference in the new design and the effects on the performance of the new golf ball. If you could give me advice as to how I’m doing solving the problem so fare and help me to determine the p-value that would be great.

Distance Measured in Yards

Current New

264 277

261 269

267 263

272 266

258 262

283 251

258 262

266 289

259 286

270 264

263 274

264 266

284 262

263 271

260 260

283 281

255 250

272 263

266 278

268 264

270 272

287 259

289 264

280 280

272 274

275 281

265 276

260 269

278 268

275 262

281 283

274 250

273 253

263 260

275 270

267 263

279 261

274 255

276 263

262 279

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