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 (p) of all Oregon whelks that will spontaneously drill into mussels are met.

To determine if the conditions for calculating a confidence interval for p are met, two criteria need to be considered: random sampling and large counts.

A. Random Sampling: The information provided does not specify if the whelk eggs were collected randomly. Therefore, this condition is not confirmed.

B. Large Counts: The condition of large counts is met if both np and n(1-p) are greater than or equal to 10, where n is the sample size and p is the proportion of interest. In this case, the sample size is 98, and the number of whelks that drilled into mussels is 9. Thus, np = 98 * (9/98) = 9, and n(1-p) = 98 * (89/98) = 89. Since both values are greater than or equal to 10, the condition of large counts is satisfied.

Based on the analysis, the condition of large counts is met, but the condition of random sampling is not confirmed. Therefore, the correct answer would be B. Fails Random and Large Counts.

Learn more about random sampling here: brainly.com/question/30759604

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