Standard deviation of mean lucanarity measurement

Refer to the Geographical Analysis (Oct. 2006) study of a new method for analyzing remote sensing data from satellite pixels, Exercise 6.25 (p. 335). Recall that the method uses a numerical measure of the distribution of gaps or hole sizes in the pixel, called lucanarity. Summary statistics for the lacunarity measirements in a sample of 100 grassland pixels are x* (this should be an x with a line over it (the mean))= 225 and s = 20. As stated in Exercise 6.25 it is known that the mean lucanarity measurement for all grassland pixels is 220. The method will be effective in identifying land cover if the standard deviation of the measurements is 10% (or less) of the true mean, i.e., if the standard deviation is less than 22.

a) Give the null and alternative hypothesis for a test to determine if, in fact, the standard deviation of all grassland pixels is less than 22.

b) A Minitab analysis of the data is shown at the bottom of the page. Locate and interpret the p-value of the test. Use ? = .10.

NEW: Problem 6.25 from page 335: (this does not need to be completed--only the problem above needs to be completed. I just added the problem below for reference if needed)

For planning purposes, urban land-cover must be identified as either grassland, commercial, or residential. This is typically done using remote sensing data from satellite pictures. In "Geographical Analysis" (Oct. 2006), researchers from Arizona State, Florida State, and Louisiana State Universities collaborated on a new method for analyzing remote sensing data. A satellite photograph of an urban area was divided into 4x4 meter areas (called pixels). Of interest is a numerical measure of the distribution of gaps or hole sizes in the pixel, called lucanarity. The mean and standard deviation of the lucanarity measurements for a sample of 100 pixels randomly selected from a specific urban area are 225 and 20, respectively. It is knowh that the mean lucanarity measurement for all grassland pixels is 220. Do the data suggest that the area sampled is grassland? Test at ? = .01