Check to see if the conditions for calculating a confidence interval for $ p $ are met, where $ p $ is the proportion of all Oregon whelks that will spontaneously drill into mussels.

A. Random? Collected 98 whelk eggs (not sure if random)
Large Counts? $ 98 \times \frac{9}{98} = 9 \geq 10 $ and $ 98 \times \frac{89}{98} = 89 \geq 10 $

B. Fails Random and Large Counts

C. Fails Large Counts

D. Not Random

The conditions for calculating a confidence interval for the proportion of all Oregon whelks that will spontaneously drill into mussels (p) are met. The answer is A. Random? √ Large Counts? √

To calculate a confidence interval for a proportion, certain conditions need to be met. A. Random: The question mentions that the whelk eggs were collected from the coast of Oregon. While it is not explicitly mentioned whether the collection was random, we can assume that the researchers took random samples or used some randomization method to select the whelk eggs. Therefore, the condition for randomness is likely met. B. Large Counts: The condition for large counts is also met. According to the given information, there were 98 whelk eggs collected, and out of those, 9 whelks drilled into a mussel. The counts of both successes (9) and failures (89) are greater than or equal to 10. This satisfies the condition for large counts. Since both conditions for random sampling and large counts are met, the conditions for calculating a confidence interval for p are satisfied. Therefore, the answer is A. Random? √ Large Counts? √

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