Masculinizing human faces. Nature (Aug. 27, 1998) published a study of facial characteristics that are deemed attractive. In one experiment, 67 human subjects viewed side by side an image of a Caucasian male face and the same image 50% masculinized using special computer graphics. Each subject was asked to select the facial image they deemed more attractive. Fifty-eight of the 67 subjects felt that masculinization of face shape decreased attractiveness of the male face. The researchers used this sample information to test whether the subjects showed a preference for either the unaltered or the morphed male face.

a. Set up the null and alternative hypotheses for this test.

b. Compute the test statistic.

c. The researchers reported a p@value ≈ 0 for the test. Do you agree?

d. Make the appropriate conclusion in the words of the problem. Use a = .01.

*8.146 Identifying type of urban land cover. Geographical Analysis (Oct. 2006) published a study of a new method for analyzing remote-sensing data from satellite pixels in order to identify urban land cover. The method uses a numerical measure of the distribution of gaps, or the sizes of holes, in the pixel, called lacunarity. Summary statistics for the lacunarity measurements in a sample of 100 grassland pixels are x = 225 and s = 20. It is known that the mean lacunarity 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 hypotheses for a test to determine whether, in fact, the standard deviation of all grassland pixels is less than 22.

b. A MINITAB analysis of the data is provided below.
Locate and interpret the p-value of the test. Use a = .10.

Test for One Standard Deviation Method Null hypothesis Alternative hypothesis Sigma 22 Sigma - < 22 The standard method is only for the normal distribution. Statistics N StDev Variance 100 20.0 400 Tests Method Chi-Square Standard 81.82 99 DF P-Value 0.105