There is some evidence indicating that people with visible tattoos are viewed mo

Question

There is some evidence indicating that people with visible tattoos are viewed more negatively than people without visible tattoos (Resenhoeft, Villa, & Wiseman, 2008). In a similar study, a researcher first obtained overall ratings of attractiveness for a woman with no tattoos shown in a color photograph. On a 7-point scale, the woman received an average rating of = 4.9. The distribution of ratings was normal with a standard deviation of = 0.84. The researcher then modified the photo by adding a tattoo of a butterfly on the woman’s left arm. The modified photo was then shown to a sample of n = 16 students at a local community college. The students used the same 7-point scale to rate the attractiveness of the woman. The average score for the photo with the tattoo was M= 4.2.

Do the data indicate a significant difference in rated attractiveness of a woman who appeared to have a tattoo? Use a two-tailed test with = .05.

Compute Cohen’s d to measure the size of the effect.

Explanation / Answer

given we have a population with both mean and standard deviation, mu = 4.9, sigma = .84

sample, n = 16, with a mean of M = 4.2.

Step 1: let us state hypotheses
Hsub0: mu = 4.9
Hsub1: mu =/= 4.9 (not equal to)

Step 2: now we need to find the critical z value using the z-table

given alpha = .05  It's two-tailed test
the associated z-value: 1.96 as it is a two-tailed test, we have two critical z values
z(crit) = +/- 1.96

Step 3: we know that z formulea is given by
z = (M-mu)/Standard error
Standard error = sigma/sqrt(n) = .84/sqrt(16) = .84/4 = 0.21
z = (M-mu)/standard error = (4.2 - 4.9)/0.21 = (-0.7)/0.21 = -3.333...

Step 4:by Comparing z values to decide whether to reject or fail to reject the null.
z-observed = -3.33 < -1.96 = z-critical, so we reject the null.
Cohen’s d to measure the size of the effect.
Cohen's d = (mu(with treatment) - mu(without treatment))/sigma
so we get mu(with Tx) = 4.2 and mu(w/o Tx) = 4.9 and the given sigma was 0.84.
Cohen's d = (4.2-4.9)/0.84 = (-0.7)/0.84 = -0.8333... ~ -0.83

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.