Question
A graduate student believes that people consider faces with more contrast between eye color and skin tone as more feminine. He identifies the null and
A graduate student believes that people consider faces with more contrast between eye color and skin tone as more feminine. He identifies the null and alternative hypotheses as:
H: The level of contrast between eye color and skin tone does notaffect how feminine a face is considered.
H: The level of contrast between eye color and skin tone affects how feminine a face is considered.
He chooses a significance level of 0.05. After he collects the data and computes the sample statistics, it is time for him to make a decision about H.
Check the two possible decisions that the graduate student can make given his choices of H and H.Check all that apply.
There is not enough evidence to reject the hypothesis that the contrast between eye color and skin tone affects how feminine a face is considered.
There is enough evidence to reject the hypothesis that the contrast between eye color and skin tone does notaffect how feminine a face is considered.
There is enough evidence to reject the hypothesis that the contrast between eye color and skin tone affects how feminine a face is considered.
There is not enough evidence to reject the hypothesis that the contrast between eye color and skin tone does notaffect how feminine a face is considered.
What decision should the graduate student make if the test statistic is in the critical region?
The graduate student should reject the alternative hypothesis.
The graduate student should reject the null hypothesis.
The graduate student cannot reject the null hypothesis.
Suppose that the test statistic is 2.63 and the boundary to the critical region is 1.96. The test statistic is the critical region. Therefore, the graduate student reject the null hypothesis, and he conclude that the level of contrast between eye color and skin tone affects how feminine a face is considered.
You may use the Distributions tool if you find it helpful.
Standard Normal Distribution
Mean = 0.0
Standard Deviation = 1.0
-4-3-2-101234z.2500.5000.2500-0.6740.674
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