Statistical Sciences 2035 Study Guide - Midterm Guide: Binomial Distribution, Confidence Interval, Random Variable
12 Oct 2013
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Answer: d. exactly 5 out of the 6 employees cannot have the correct checks. If 5 have the correct ones then the 6th person must automatically have the correct one also. Let u be the event that the person is under 25 and let u be the event that the person is not under 25. Also, let a be the event that the driver was in at least one accident and let u be the event that the driver was not in at least one accident. P(u) = 0. 35 => p(u ) = 0. 65 and p(a|u) = 280/700 and p(a|u ) = 130/1300. This gives us p(a) = 0. 35*280/700 + 0. 65*130/1300 = 0. 205. make use of our answer from question #3 here. I will use a p-value approach to explain this answer. If we cannot reject ho in the one-sided test for a single mean then this would mean that our one-sided p-value is larger than .
