# In Exercise 14 (Section 7.2),

In Exercise 14 (Section 7.2), the sample mean and standard deviation of the dye-layer density of aerial photographs of 69 forest trees were found to be 1.028 and .163, respectively. Because the raw data is not available, a researcher suggests using a computer to generate a random sample of 69 observations from a normal distribution whose mean and standard deviation are 1.028 and .163, respectively. If necessary, after obtaining the sample, the data are adjusted so that their sample mean and standard deviation coincide exactly with 1.028 and .163. A 95% bootstrap interval is then generated using this simulated data.

a. Under what conditions will this procedure provide a reliable interval estimate?

b. Use the procedure outlined in this exercise to generate a 95% bootstrap interval for the average dye-layer density.

c. Compare your result in part (b) to the 95% confidence interval found in Exercise 14(a).

Exercise 14

The article “Extravisual Damage Detection? Defining the Standard Normal Tree” (Photogrammetric Engr. and Remote Sensing, 1981: 515–522) discusses the use of color infrared photography in identification of normal trees in Douglas fir stands. Among data reported were summary statistics for green-filter analytic optical densitometric measurements on samples of both healthy and diseased trees. For a sample of 69 healthy trees, the sample mean dye-layer density was 1.028, and the sample standard deviation was .163.

a. Calculate a 95% two-sided CI for the true average dye-layer density for all such trees.

b. Suppose the investigators had made a rough guess of .16 for the value of s before collecting data. What sample size would be necessary to obtain an interval width of .05 for a confidence level of 95%?

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