A bark beetle infestation will kill off a pine forest in months. To deal with this problem rangers conduct controlled burns if the bark beetle infests more than 10% of all pine trees in the forest. 1500 randomly selected Tahoe pine trees in a very large national park were tested for traces of the Bark Beetle infestation. It was found that 153 of the trees showed such traces. They wish to test that more than 10% of the Tahoe trees have been infested. (Use a 5% level of significance). Are the conditions for the appropriate test met?单项选择题
A
No, some conditions are not met.
B
No, none of the conditions are met.
C
Yes all conditions are met.

Question

A bark beetle infestation will kill off a pine forest in months. To deal with this problem rangers conduct controlled burns if the bark beetle infests more than 10% of all pine trees in the forest. 1500 randomly selected Tahoe pine trees in a very large national park were tested for traces of the Bark Beetle infestation. It was found that 153 of the trees showed such traces. They wish to test that more than 10% of the Tahoe trees have been infested. (Use a 5% level of significance). Are the conditions for the appropriate test met?单项选择题

A

No, some conditions are not met.

B

No, none of the conditions are met.

C

Yes all conditions are met.

BlackTom AI · Accepted Answer

For a one-proportion z-test the key conditions are: a random sample, independence (often checked by the 10% condition when sampling without replacement), and the success–failure condition (both n·p0 and n·(1 − p0) are at least 10 so the sampling distribution is approximately normal).

1) Relevant knowledge points:
- Hypotheses concern a single population proportion; use the one-proportion z-test.
- Independence: sample should be random and size < 10% of the population if sampling without replacement.
- Success–failure: use the null proportion p0 to check n·p0 ≥ 10 and n·(1 − p0) ≥ 10.

2) Step-by-step reasoning:
- The problem states 1500 trees were randomly selected, so the random-sampling condition is explicitly satisfied.
- The park is described as “very large,” so it is reasonable to assume 1500 is less than 10% of the total number of Tahoe pines; thus the 10% condition for independence is met.
- Use p0 = 0.10 (the threshold in the ranger policy) to check the success–failure rule.

3) Calculations (intermediate steps shown):
- n·p0 = 1500 × 0.10 = 150
- n·(1 − p0) = 1500 × 0.90 = 1350
Both values are well above 10, so the sampling distribution of the sample proportion is approximately normal under the null.

4) Why the correct answer is correct:
Because the sample is random, the 10% independence condition is satisfied (given the population is very large), and both n·p0 and n·(1 − p0) are far greater than 10, all required conditions for the one-proportion z-test are met.

5) Why the other options are incorrect:
- “No, some conditions are not met” is wrong because we have randomness, independence, and the success–failure counts satisfied.
- “No, none of the conditions are met” is also wrong for the same reasons.

6) Conclusion:
All conditions for the appropriate one-proportion z-test are met. This confirms the correct choice: Yes all conditions are met.

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