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# Suppose the number of successes in n = 200 binomial trials is 20. (a) Find a 90% confidence interval (CI) for p, the probability of success at each...

Suppose the number of successes in n = 200 binomial trials is 20. (a) Find a 90% confidence interval (CI) for p, the probability of success at each trial. (b) Find a 98% CI for p. Why is this interval wider than the previous one?

Confidence interval for proportion is given by CI = p + z*sqrt(pq/n) p=successes/ n... View the full answer

1 comment
• Note that the interval in b is wider because as the precision of the confidence interval increases (ie CI width decreasing), the reliability of an interval containing the actual population proportion decreases (less of a range to possibly cover the P).
• Stat.Solve
• Apr 04, 2018 at 12:17pm

a) ﻿ 0 . 0 6 5 ≤ p ≤ 0 . 1 3 5 ﻿ b) ﻿ 0 .... View the full answer

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