Question

A veterinarian investigating possible causes of stomach distress in horses suspects that feeding alfalfa may be to blame. the veterinarian wishes to estimate the proportion of horses with stomach distress who are fed at least two flakes of alfalfa per day. in a sample of 62 horses with stomach distress, 42 are fed two or more daily flakes of alfalfa. for a 99% confidence interval, the margin of error is:

Answer 1

To estimate the proportion of horses with stomach distress who are fed at least two flakes of alfalfa per day and determine the margin of error, a confidence interval can be used. The margin of error in this case is 0.136.

Step-by-step explanation:

To estimate the proportion of horses with stomach distress who are fed at least two flakes of alfalfa per day, the veterinarian can use a confidence interval.

To find the margin of error for a 99% confidence interval, the formula is:

Margin of Error = Z * sqrt(p(1-p)/n)

where Z is the Z-score for a 99% confidence level, p is the proportion from the sample, and n is the sample size.

In this case, the sample proportion is 42/62, and you can use a Z-score of 2.576 for a 99% confidence level. Plugging in these values, the margin of error is:

Margin of Error = 2.576 * sqrt((42/62)(1-42/62)/62) = 0.136