Ex. 13,14

in the fact that the ited to obtain an estimate of Iners whose blood pressure exceeds 150. The linear random walk model for the successive daily prices of a stock or commodity supposes that the successive differences of the end-of-day prices of a given stock constitute a random sample from a normal popu- lation. The following 20 data values represent the closing prices of crude oil on the New York Mercantile Exchange on 20 consecutive trading days in 1994. Assuming the linear random walk model, use these data to es timate the mean and standard deviation of the population distribution. (Note that the data give rise to 19 values from this distribution, the first being 17.60 - 17.50 = 0.10, the second being 17.81 - 17.60 = 0.21, and so on . ) 17.50, 17.60, 17.81, 17.67, 17.53, 17.39, 17.12, 16.71, 16.70, 16.83, 17.21, 17.24, 17.22, 17.67, 17.83, 17.67, 17.55, 17.68, 17.98, 18.39 (14. Due to a lack of precision in the scale used, the value obtained when a fish is weighed is normal with mean equal to the actual weight of the fish and with standard deviation equal to 0.1 grams. A sample of 12 different fish was chosen, and the fish were weighed, with the following results: 5.5, 6.2, 5.8, 5.7, 6.0, 6.2, 5.9, 5.8, 6.1, 6.0, 5.7, 5.6 Estimate the population standard deviation of the actual weight of a fish. Hint: First note that, due to the error involved in weighing a fish, each data value is not the true weight of a fish, but rather is the true weight plus an error term. This error term is an independent random variable that has mean O and standard deviation 0.1. Therefore, Data = true weight + error and so Var(data) = Var(true weight) + Var(error) To determine the variance of the true weight, first estimate the variance of the data