Use a confidence interval formula for proportions. Here's one:
CI99 = p + or - (2.58)(√pq/n)
...where p = .206, q = 1 - p, and n = 165.
Plug the values into the formula and calculate.
I hope this will help get you started on the first part of your problem.
wildlife biologists inspect 165 deer taken by hunters and find 34 or 20.6% of them carry ticks that test positive for Lyme disease. Calculate the 99% margin of error and confidence interval for the proportion of deer that may carry ticks. How many deer are needed to be inspected if the margin of error were to be cut in half? Comment on the statistical concerns about this study if it were reported in the news.
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