Homework Answers
(a) Here large value of X2 indicate the percentage of holes scored better or worse than par different among rounds because half could be eliminated after two rounds, placement of holes etc. Option A is correct here.
(b) here option B is correct here.
(c) Here option C is correct here.
(d) Yes, all conditions are satisfied here.
(e) Here option B is correct here.
Observed table
| Round | |||||
| 1 | 2 | 3 | 4 | Total | |
| Birdie | 346 | 364 | 198 | 196 | 1104 |
| Bogey | 731 | 654 | 391 | 396 | 2172 |
| Double Bogeey | 113 | 94 | 62 | 53 | 322 |
| Par | 1604 | 1653 | 824 | 832 | 4913 |
| Total | 2794 | 2765 | 1475 | 1477 | 8511 |
Expected Table
| Round | |||||
| 1 | 2 | 3 | 4 | Total | |
| Birdie | 362.4223 | 358.6606 | 191.3289 | 191.5883 | 1104 |
| Bogey | 713.0264 | 705.6257 | 376.4188 | 376.9292 | 2172 |
| Double Bogeey | 105.7065 | 104.6093 | 55.80425 | 55.87992 | 322 |
| Par | 1612.845 | 1596.104 | 851.4481 | 852.6026 | 4913 |
| Total | 2794 | 2765 | 1475 | 1477 | 8511 |
Chi-square value
| Round | |||||
| 1 | 2 | 3 | 4 | Total | |
| Birdie | 0.744135 | 0.079489 | 0.232605 | 0.101588 | 1.157817 |
| Bogey | 0.453067 | 3.777086 | 0.56483 | 0.964896 | 5.75988 |
| Double Bogeey | 0.503235 | 1.075983 | 0.687892 | 0.148424 | 2.415534 |
| Par | 0.048505 | 2.028127 | 0.884845 | 0.49785 | 3.459327 |
| Total | 1.748942 | 6.960686 | 2.370172 | 1.712758 | 12.79256 |
X2 = 12.79256
Degree of freedom = dF = (4-1) * (4-1) = 9
p - value = CHIINV(12.79256, 9) = 0.1722
The p value is valid and it losely confirms the similarlty of the percentages.
Add Answer to:
Birdie Bogey Double Bogey Par Strokes Round 2 3 364 198 654 391 94 62 1,653...
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