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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...

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


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 = 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 x ̅ 1 =270.275 and x ̅ 2 =267.50. Standard Deviation 1 = 8.64 and 2 = 9.77 and sample size for both is n=40. = The critical value rule is so 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
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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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