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Feminized faces in TV commercials. Television commercials most often employ females or 鈥渇eminized鈥 males to pitch a company鈥檚 product. Research published in Nature (August 27 1998) revealed that people are, in fact, more attracted to 鈥渇eminized鈥 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 鈥渕ost attractive鈥 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

Expert verified

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

Step by step solution

01

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.

02

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.

03

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.

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Most popular questions from this chapter

Consider the test \({H_0}:\mu = 70\) versus \({H_a}:\mu \ne 70\) using a large sample of size n = 400. Assume\(\sigma = 20\).

a. Describe the sampling distribution of\(\bar x\).

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c. Refer to part b. Find the p-value of the test.

d. Find the rejection region of the test for\(\alpha = 0.01\).

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a. Construct a hypothesis test with the alternative hypothesis that\(\mu < 198\)at a 1% significance level. Interpret your results.

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at a 1% significance level. Interpret your results.

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