1) To what extent is blushing attractive? To help find out Pazda et al. (2016) asked male participants to look at photos of women. Each participant saw a photo of the same women, but one photo was manipulated to have redder checks. Each participant specified which photo they found most attractive, the one with extra red (1) or the unaltered original photo (0). Each participant rated 6 different pairs of photos. The researchers found that the redder photo was picked 72% of the time. They then used the NHST approach to test this result against a null hypothesis of 50%, which represents even preference. They found: t(132) = 7.59, p < 0.001.
a. What would it mean to make a Type I error in this study? Describe the researcher's belief and reality.
b. What would it mean to make a Type II error in this study? Describe the researcher's belief and reality.
c. What does the 132 represent?
d. Why did the researchers calculate a t value rather than a z value?
e. Using α = 0.05, is this a statistically significant finding?
f. Using α = 0.01, is this a statistically significant finding?
g. Using α = 0.05, will the researchers reject the null hypothesis of fail to reject the null hypothesis?
h. The redder photo was preferred 72% of the time. Will the 95% CI for preference include 50%?
i. Will the 95% CI for preference include 40%?
2) To what extent is seeking help a matter of mindset? One recent study gave students a reminder that everyone struggles; this mindset intervention caused the students to seek more help and do better in school (Yeager et al., 2016). A chemistry teacher tries this intervention in her organic chemistry class. All students were given the mindset intervention at the beginning of the semester (the reminder that everyone struggles). At the end of the semester, all students took the American Chemistry Society Organic Chemistry Exam, a standardized test administered throughout the USA: µ = 36.99, σ = 10.62. For the students with the mindset intervention, the professor finds: M = 41.2, N = 25.
a. Use the NSHT approach with σ = 0.05 to test the results the professor obtained against the null hypothesis that the intervention has no benefit (H0: µintervention = 36.99). First calculate a z value by hand, then use the Normal tab in ESCI to look up the two-tailed p value, then make a decision to reject or fail to reject the null.
b. Now use the estimation approach to analyse the extent to which the intervention might be useful. Calculate SE for the sample, then the 95% MoE, and then the 95% CI. Interpret your result.
c. Compare and contrast the two approaches. Which do you feel would be more informative for others who are considering implementing this intervention? Why?
For thought: How well does this study fit the assumptions for calculating a CI? Is it fair for the professor to consider any difference from the national norm to be caused by the intervention? Would it matter at what university the professor worked at?