Question

Feminized faces in TV commercials. Television commercials most often employ females or “feminized” males to pitch a company’s product. Research published in Nature (August 27 1998) revealed that people are, in fact, more attracted to “feminized” faces, regardless of gender. In one experiment, 50 human subjects viewed both a Japanese female face and a Caucasian male face on a computer. Using special computer graphics, each subject could morph the faces (by making them more feminine or more masculine) until they attained the “most attractive” face. The level of feminization x (measured as a percentage) was measured.

a. For the Japanese female face, x = 10.2% and s = 31.3%. The researchers used this sample information to test the null hypothesis of a mean level of feminization equal to 0%. Verify that the test statistic is equal to 2.3.

b. Refer to part a. The researchers reported the p-value of the test as p = .021. Verify and interpret this result.

c. For the Caucasian male face, x = 15.0% and s = 25.1%. The researchers reported the test statistic (for the test of the null hypothesis stated in part a) as 4.23 with an associated p-value of approximately 0. Verify and interpret these results.

Short answer

It is verified that the value of the test statistic is 2.30.

Step-by-step solution

Given information

It is given that the sample mean,\(\overline x = 10.2\% \)and the standard deviation\(s = 31.3\% \).

Also, the sample size is 50.

Concept of testing of hypothesis

Hypothesis testing is a technique for drawing statistical conclusions from population data. It is a tool for analyzing assumptions and determining how likely something is within a particular level of accuracy. Hypothesis testing allows us to see if the outcomes of an experiment are correct.

Testing the hypothesis

a.

The null hypothesis:

\({H_0}:{\mu _0} = 0\)

The null hypothesis:

\({H_a}:{\mu _0} \ne 0\)

Now, consider the test statistic,

\(z = \frac{{\overline x - \mu }}{{\frac{s}{{\sqrt n }}}}\)

Therefore,

\(\begin{aligned}z &= \frac{{0.102 - 0}}{{\frac{{0.313}}{{\sqrt {50} }}}}\\ &= \frac{{0.102}}{{0.442649}}\\ &= 2.30\end{aligned}\)

It is verified that the value of test statistic is 2.30.